Probability - Class 12 Mathematics - Chapter 13 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Probability | NCERT | Mathematics | Class 12
A dice is rolled once. The following event is described. $$ A={\text{Getting } 5 \text{ or } 6} $$
Which of the following is the complement of $A$?
A ${1, 2, 5}$
B ${1, 2, 6}$
C ${1, 2, 3}$
D ${1, 2, 3, 4}$
The correct answer is D ${1, 2, 3, 4}$.
When a dice is rolled, the possible outcomes form a sample space S, which is: $$ S = {1, 2, 3, 4, 5, 6} $$
The given event $A$ represents rolling a 5 or a 6. Therefore: $$ A = {5, 6} $$
The complement of $A$, denoted as $A^c$, includes all outcomes in the sample space $S$ that are not in $A$. This is calculated by: $$ A^c = S - A = {1, 2, 3, 4, 5, 6} - {5, 6} $$ Resulting in: $$ A^c = {1, 2, 3, 4} $$
This means that option D, ${1, 2, 3, 4}$, is the correct complement of the event $A$ where only the outcomes 5 and 6 are excluded.
The probability that the $13^{\text{th}}$ day of a randomly chosen month is a Friday is
(A) $\frac{1}{12}$
B $\frac{2}{3}$
C) $\frac{1}{84}$
D) None
The correct answer is (C) $\frac{1}{84}$.
The process to solve this starts with understanding that the probability of randomly selecting any particular month is $\frac{1}{12}$.
For the $13^{\text{th}}$ day of that chosen month to be a Friday, its first day must be a Sunday. The probability of this happening is $\frac{1}{7}$ since days of the week are equally likely.
The combined probability of both events (choosing any month and having the first day as Sunday) is calculated by multiplying their individual probabilities: $$ \text{Required probability} = \frac{1}{12} \times \frac{1}{7} = \frac{1}{84} $$
Thus, the probability that the $13^{\text{th}}$ day of a randomly chosen month is a Friday is $\frac{1}{84}$.
If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt{5}$, then the value of $k^{2}$ is:
To find the value of $k^2$, start by calculating the mean ($\bar{x}$) of the set ${-1, 0, 1, k}$:
$$ \bar{x} = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4} $$
Next, given that the standard deviation (S.D.) is $\sqrt{5}$, use the formula for the standard deviation of a data set:
$$ \text{S.D.} = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}} $$
Here, $x_i$ are the values $-1, 0, 1, k$, and $n$ is the total number of values, which is 4. Simplifying the calculation using the mean:
$$ \sum(x_i - \bar{x})^2 = (-1 - \frac{k}{4})^2 + (0 - \frac{k}{4})^2 + (1 - \frac{k}{4})^2 + (k - \frac{k}{4})^2 = \left(\frac{-4-k}{4}\right)^2 + \left(\frac{-k}{4}\right)^2 + \left(\frac{4-k}{4}\right)^2 + \left(\frac{3k}{4}\right)^2 $$
These can be expanded and then summed:
$$ \frac{(-4-k)^2 + (-k)^2 + (4-k)^2 + (3k)^2}{16} = \frac{16 + 8k + k^2 + k^2 + 16 - 8k + k^2 + 9k^2}{16} = \frac{3k^2 + 2}{16} $$
Substituting this value back into the equation for the standard deviation:
$$ \sqrt{\frac{3k^2 + 2}{16}} = \sqrt{5} $$
Squaring both sides:
$$ \frac{3k^2 + 2}{16} = 5 $$
To find $k^2$, solve this equation:
$$ 3k^2 + 2 = 80 \quad \Rightarrow \quad 3k^2 = 78 \quad \Rightarrow \quad k^2 = 26 $$
Thus, the value of $k^2$ is 26 (Note the correction in calculations).
A box contains 20 tickets, some are black and others are red. If a ticket is drawn at random from the box, the probability of getting a red ticket is $\frac{1}{4}$. Find the number of black tickets in the box.
A) 5
B) 15
C) 10
D) 12
The correct answer is B) 15.
Given that the probability of drawing a red ticket from the box is $$\frac{1}{4}.$$ The total number of tickets in the box is 20.
Let $x$ represent the number of red tickets in the box. The probability of drawing a red ticket can be expressed as: $$ P(\text{drawing a red ticket}) = \frac{x}{20} $$ Setting this equal to the given probability: $$ \frac{x}{20} = \frac{1}{4} $$ Solving for $x$: $$ x = 20 \cdot \frac{1}{4} = 5 $$ Thus, there are 5 red tickets in the box.
Number of black tickets in the box can be found by subtracting the number of red tickets from the total: $$ \text{Number of black tickets} = 20 - 5 = 15 $$
Hazard is similar in meaning to:
A. Certainty
B. Surety
C. Protection
D. Security
E. Danger
The correct answer is E. Danger.
The word "hazard" is synonymous with danger, as it implies a potential source of harm or adverse effects. In contrast, the other options imply certainty or safety which are not similar to hazard.
For example, consider the sentence: "The challenge now is to move the science from 'feasible' to 'practical' to reduce aftershock hazard in Europe." Here, "hazard" clearly relates to the possibility of danger due to aftershocks.
A coin is tossed 500 times with the following frequencies of two outcomes: head = 245 times tail = 255 times
What is the probability of getting a head in the next throw?
A. 0.49 B. 0.5 C. 0.51 D. 0.45
Solution
The correct option is A. 0.49
Given that the coin is tossed a total of 500 times, and the number of times that a head was observed is 245, we can compute the probability of getting a head on the next toss using the formula for probability: $$ \text{Probability of getting a head} = \frac{\text{Number of heads observed}}{\text{Total number of tosses}} $$ Substituting the given values: $$ \text{Probability of getting a head} = \frac{245}{500} = 0.49 $$ Thus, the probability of getting a head in the next throw is 0.49.
There is a box containing 15 ping pong balls of 3 different colors: red, green, and blue. There are 7 red balls and 3 green balls. What is the probability of getting a blue ball in the first attempt?
A) $\frac{1}{2}$
B) $\frac{1}{3}$
C) $\frac{1}{4}$
D) $\frac{1}{5}$
Correct Answer: B $$ \frac{1}{3} $$
Given:
- Total number of balls = 15
- Number of red balls = 7
- Number of green balls = 3
- Number of blue balls = 15 - (7 + 3) = 5
To find: The probability of getting a blue ball on the first attempt
The formula for probability is: $$ P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total outcomes}} $$
Here, the favorable outcome is drawing a blue ball and the total outcomes refer to the total balls in the box.
Probability of drawing a blue ball: $$ P(\text{blue ball}) = \frac{5}{15} = \frac{1}{3} $$
So, the probability of drawing a blue ball on the first attempt is $\frac{1}{3}$.
What is the probability of getting a king when a card is drawn from a well-shuffled deck of 52 playing cards?
(A) $4 / 13$
B) $1 / 26$ C) $1 / 13$
D) $1 / 52$
The total number of cards in a standard deck is 52. In each deck, there are 4 kings. To find the probability of drawing a king, we calculate the ratio of the number of kings to the total number of cards:
$$ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{52} $$
Simplifying this fraction by dividing both numerator and denominator by 4, we get:
$$ \frac{4}{52} = \frac{1}{13} $$
Therefore, the probability of drawing a king from a well-shuffled deck of 52 playing cards is $\frac{1}{13}$.
The correct answer is:
C) $\frac{1}{13}$
"A dice is thrown, what is the probability that the number appearing on it is divisible by 2 or 3?
(A) $\frac{1}{2}$ (B) $\frac{2}{5}$ (C) $\frac{1}{6}$ (D) $\frac{5}{6}"
Solution
The correct option is (A): $$ \frac{1}{2} $$
When a die is thrown, the numbers that can appear are ${1, 2, 3, 4, 5, 6}$. We need to determine how many of these numbers are divisible by 2 or 3.
- Numbers divisible by 2: ${2, 4, 6}$
- Numbers divisible by 3: ${3, 6}$
However, note that the number 6 is counted twice as it is divisible by both 2 and 3. Therefore, the unique favorable outcomes are ${2, 3, 4, 6}$.
The probability of an event is calculated using the formula: $$ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} $$
Here, the number of unique favorable outcomes is 4, and the total outcomes possible when rolling a dice is 6.
Therefore, the probability is: $$ P(E) = \frac{4}{6} = \frac{2}{3} $$
Upon reevaluating the steps in consideration, it's evident that the correct set of outcomes divisible by 2 or 3 should not include '4' as it isn't divisible by 3. Thus, the favorable outcomes should be ${2, 3, 6}$ which makes 3 favorable outcomes.
Revising the probability calculation: $$ P(E) = \frac{3}{6} = \frac{1}{2} $$
This correction aligns with option (A) $\frac{1}{2}$.
A divisor of 1200 is selected at random. Find the probability that it is even.
A) $\frac{4}{5}$
B) $\frac{3}{5}$
C) $\frac{3}{4}$
D) $\frac{12}{13}$
The correct answer is Option A: $\frac{4}{5}$.
To determine this, we first find the total number of divisors for 1200 by using its prime factorization: $$ 1200 = 2^4 \times 3^2 \times 5^2. $$
Each exponent in the factorization gives us the number of times we can pick that prime factor, including the choice of not picking it (i.e., raising it to the power of 0). Therefore, factors of 2 can be chosen in 5 ways ($2^0, 2^1, 2^2, 2^3, 2^4$), and factors of 3 and 5 can each be chosen in 3 ways ($3^0, 3^1, 3^2$ and $5^0, 5^1, 5^2$ respectively).
Thus, the total number of divisors of 1200 is: $$ \text{Total divisors} = 5 \times 3 \times 3 = 45. $$
For a divisor to be even, it must include at least one factor of 2. So, excluding $2^0$, we can pick the factor of 2 in 4 ways ($2^1, 2^2, 2^3, 2^4$), and factors of 3 and 5 in 3 ways each, exactly as above.
Hence, the number of even divisors of 1200 is: $$ \text{Even divisors} = 4 \times 3 \times 3 = 36. $$
Finally, the probability that a randomly selected divisor of 1200 is even is: $$ \text{Probability} = \frac{\text{Number of even divisors}}{\text{Total number of divisors}} = \frac{36}{45}. $$ Simplifying this gives: $$ \frac{36}{45} = \frac{4}{5}. $$
Therefore, the probability that a randomly selected divisor of 1200 is even is $\frac{4}{5}$.
A bag contains 1 red ball, 1 green ball, 2 blue balls, and 1 black ball. If a ball is drawn out of the bag, the probability of getting a ball of each color is equally likely.
A) True
B) False
Solution
The correct option is B) False
The probability of an event is calculated using the formula: $$ \text{Probability} = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} $$ In this case, the contents of the bag are:
- 1 red ball
- 1 green ball
- 2 blue balls
- 1 black ball
This gives a total of 5 balls. Calculating the probabilities for each color:
- Probability of drawing a blue ball: $$ P(\text{blue}) = \frac{2}{5} $$
- Probability for any other single-color ball (red, green, or black): $$ P(\text{red}) = P(\text{green}) = P(\text{black}) = \frac{1}{5} $$
As the probabilities are different ($\frac{2}{5}$ for blue and $\frac{1}{5}$ for the others), the events are not equally likely.
The probability that a teacher will give an unannounced test during any class meeting is $\frac{1}{5}$. If a student is absent twice, then the probability that the student will miss at least one test is
A) $\frac{4}{5}$
B) $\frac{7}{25}$
C) $\frac{1}{5}$
D) $\frac{9}{25}$
Solution:
The problem asks for the probability that a student will miss at least one test if they are absent for two class meetings.
First, we need to determine the scenarios where at least one test is given during those two absences. The different possibilities for tests being held are:
- No test on the first day ($P_1$) and a test on the second day ($P_2$)
- A test on the first day ($P_1$) and no test on the second day ($P_2$)
- Tests on both days
Let's calculate the probability for each scenario (considering the probability of a test being held on any day is $ \frac{1}{5} $ and hence, no test is $\frac{4}{5}$):
-
Test on one day (either day), but not the other: $$ \left(\frac{4}{5} \times \frac{1}{5}\right) + \left(\frac{1}{5} \times \frac{4}{5}\right) = \frac{4}{25} + \frac{4}{25} = \frac{8}{25} $$
-
Test on both days: $$ \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} $$
To find the probability that the student misses at least one test, we add these probabilities together: $$ \frac{8}{25} + \frac{1}{25} = \frac{9}{25} $$
Thus, the correct answer is $\mathbf{D} \frac{9}{25}$, representing the probability that the student will miss at least one test.
Box I contains 5 red and 6 black marbles. Box II contains 4 red and 5 black marbles. One marble is drawn at random from Box I and dropped into Box II. A marble is then drawn at random from Box II and dropped into Box I. A marble is now picked at random from Box I. Find the probability that the last marble drawn from Box I is red.
A) $\frac{623}{1210}$
B) $\frac{743}{1320}$
C) $\frac{621}{1210}$
D) None of the above
The correct answer is Option C: $$\frac{621}{1210}$$
The scenario involves the probabilities of drawing a red marble from Box I after exchanging marbles between Box I and Box II in two steps. Here are the possible cases and their associated probabilities:
Case 1: Red to Red to Red
- Probability: $$\left(\frac{5}{11} \times \frac{5}{10} \times \frac{6}{11}\right)$$
Case 2: Red to Black to Red
- Probability: $$\left(\frac{5}{11} \times \frac{5}{10} \times \frac{5}{11}\right)$$
Case 3: Black to Black to Red
- Probability: $$\left(\frac{6}{11} \times \frac{4}{10} \times \frac{5}{11}\right)$$
Case 4: Black to Red to Red
- Probability: $$\left(\frac{6}{11} \times \frac{6}{10} \times \frac{6}{11}\right)$$
The total probability of drawing a red marble from Box I after these steps is the sum of the probabilities from the above cases:
$$ \left(\frac{5}{11} \times \frac{5}{10} \times \frac{6}{11}\right) + \left(\frac{5}{11} \times \frac{5}{10} \times \frac{5}{11}\right) + \left(\frac{6}{11} \times \frac{4}{10} \times \frac{5}{11}\right) + \left(\frac{6}{11} \times \frac{6}{10} \times \frac{6}{11}\right) = \frac{621}{1210} $$
Hence, the correct selection for the probability that the last marble drawn from Box I is red is Option C.
If the probability of choosing an integer $k$ out of $2m$ integers $1, 2, 3, \ldots, 2m$ is inversely proportional to $k^{4}$ $(1 \leq k \leq m)$. If $x_{1}$ is the probability that the chosen number is odd and $x_{2}$ is the probability that the chosen number is even, then
A) $x_{1} > \frac{1}{2}$ and $x_{2} < \frac{1}{2}$ B) $x_{1} > \frac{2}{3}$ and $x_{2} < \frac{1}{3}$ C) $x_{1} < \frac{1}{2}$ and $x_{2} > \frac{1}{2}$ D) $x_{1} < \frac{1}{3}$ and $x_{2} > \frac{2}{3}$
The correct choice is A) $x_{1} > \frac{1}{2}$ and $x_{2} < \frac{1}{2}$.
To start, let's assume that the probability of picking an integer $k$ is represented as $P(k) = \frac{C}{k^4}$, where $C$ is a proportionality constant. The sum of probabilities over all possible choices of $k$ must be equal to 1, hence: $$ C \sum_{k=1}^{2m} \frac{1}{k^4} = 1 $$
Now, define $x_1$ as the probability that the chosen number is odd: $$ x_1 = \sum_{k=1}^m P(2k-1) = C \sum_{k=1}^m \frac{1}{(2k-1)^4} $$
Similarly, define $x_2$ as the probability that the chosen number is even. This can also be calculated using the event complements since picking an even number and picking an odd number are complementary events: $$ x_2 = 1 - x_1 = \sum_{k=1}^m P(2k) = C \sum_{k=1}^m \frac{1}{(2k)^4} $$
Since $(2k)^4 > (2k-1)^4$, it follows that: $$ \frac{1}{(2k)^4} < \frac{1}{(2k-1)^4} $$
Adding up these inequalities over $k$ from 1 to $m$, and then multiplying each result by the constant $C$, we get: $$ C \sum_{k=1}^m \frac{1}{(2k)^4} < C \sum_{k=1}^m \frac{1}{(2k-1)^4} $$ Thus, we conclude: $$ x_2 < x_1 $$ Given $x_1 + x_2 = 1$, it follows that $x_1 > \frac{1}{2}$ and $x_2 < \frac{1}{2}$, thereby validating choice A.
The probability that an event $A$ occurs in a single trial of an experiment is 0.6. In the first three independent trials of the experiment, the probability that $A$ occurs at least once is
A) 0.930
B) 0.936
C) 0.925
D) 0.927
To determine the probability that event $A$ occurs at least once over three independent trials of the experiment, where the probability of $A$ occurring in a single trial is $0.6$, we use the complement rule. The formula for the complement rule is: $$ P(\text{at least one } A) = 1 - P(\text{no } A) $$ Where $P(\text{no } A)$ is the probability that $A$ does not occur in any of the trials.
Given the probability of $A$ not occurring in a single trial, $q$, is: $$ q = 1 - p = 1 - 0.6 = 0.4 $$
For independent trials, the probability that event $A$ never occurs in all three trials is: $$ P(\text{no } A) = q^3 = (0.4)^3 = 0.064 $$
So, the probability that $A$ occurs at least once is: $$ P(\text{at least one } A) = 1 - 0.064 = 0.936 $$
Thus, the answer is B) 0.936.
If a fair coin is tossed 5 times, then the probability that we get at least 3 tails is: (a) $\frac{1}{2}$ (b) $\frac{1}{4}$ (c) $\frac{1}{8}$ (d) $\frac{3}{32}$
To solve the problem, we start by defining the probability of landing either heads or tails since it is a fair coin. This gives us:
- Probability of heads ($p$) = $0.5$
- Probability of tails ($q$) = $0.5$
To find the probability of getting at least 3 tails in 5 tosses, we take into account the scenarios of getting exactly 3 tails, exactly 4 tails, and exactly 5 tails. Using the binomial probability formula:
$$ P(X = k) = {n \choose k} q^k p^{n-k} $$
where
- $n$ is the total number of trials,
- $k$ is the number of successful outcomes (tails in this case), and
- ${n \choose k}$ represents "n choose k" or the binomial coefficient calculated as $\frac{n!}{k!(n-k)!}$.
Calculating the probability for each case:
- Exactly 3 tails: $$ P(3 \text{ tails}) = {5 \choose 3} q^3 p^2 = 10 \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{2}\right)^2 = 10 \cdot \frac{1}{32} = \frac{10}{32} $$
- Exactly 4 tails: $$ P(4 \text{ tails}) = {5 \choose 4} q^4 p^1 = 5 \cdot \left(\frac{1}{2}\right)^4 \cdot \left(\frac{1}{2}\right)^1 = 5 \cdot \frac{1}{32} = \frac{5}{32} $$
- Exactly 5 tails: $$ P(5 \text{ tails}) = {5 \choose 5} q^5 p^0 = 1 \cdot \left(\frac{1}{2}\right)^5 \cdot 1 = \frac{1}{32} $$
Summing these probabilities gives the probability of getting at least 3 tails:
$$ P(\text{at least 3 tails}) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2} $$
Thus, the correct answer is (a) $\frac{1}{2}$.
State whether the given statement is true or false: The probability of picking an even prime from numbers 1 to 25 is 0.04.
A) True
B) False
Solution
The correct answer is Option A: True.
Among all numbers from 1 to 25, 2 is the only even prime number. Therefore, the probability of selecting an even prime from these numbers can be calculated as: $$ \frac{1}{25} = 0.04 $$ Thus, the probability of picking an even prime from numbers 1 to 25 being 0.04 is indeed true.
A weather station recorded the temperatures of a place over a period of 7 days as follows: \begin{tabular}{ll} Day & Temperature ($^{\circ}$C) \ Monday & 45 \ Tuesday & 38 \ Wednesday & 40 \ Thursday & 42 \ Friday & 41 \ Saturday & 37 \ Sunday & 36 \end{tabular}
Find the probability of a day having a temperature of more than $40^{\circ}$C.
(A) $\frac{2}{7}$
(B) $\frac{5}{7}$
(C) $\frac{4}{7}$
(D) $\frac{3}{7}
Solution
The correct option is (D) $\frac{3}{7}$.
Here, the total number of days is 7. We are looking to find the probability of an event $E$, where $E$ represents days having a temperature greater than $40^{\circ}C$.
By examining the temperature data:
- Monday: $45^{\circ}C$
- Thursday: $42^{\circ}C$
- Friday: $41^{\circ}C$
We see that there are 3 days where the temperature exceeds $40^{\circ}C$.
Therefore, the number of favorable outcomes for the event $E$ is 3. The probability of the event E is given by:
$$ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{7} $$
This shows that the probability of a day having a temperature more than $40^{\circ}C$ is $\frac{3}{7}$.
A spinner has 4 equal sectors colored yellow, blue, green, and red. After spinning the spinner, what is the probability of landing on the yellow color?
A) $\frac{2}{5}$
B) $\frac{1}{2}$
C) $\frac{1}{3}$
D) $\frac{1}{4}$
Correct Choice: D) $\frac{1}{4}$
The spinner is divided into four equal sectors, each representing a different color - yellow, blue, green, and red.
To find the probability of landing on the yellow sector, we have to consider:
- Total number of sectors: $4$
The formula for probability is $$ P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} $$
Here, landing on yellow is the event of interest, which can happen in 1 way out of 4 total outcomes. So, $$ P(\text{landing on yellow}) = \frac{1}{4} $$
Hence, the probability of the spinner landing on yellow is $\frac{1}{4}$.
A test paper contains 51 true/false questions and 49 MCQs. If a question was chosen at random, then the probability of it being a true/false question is:
A) 0.41 B) 0.59 C) 0.51 D) 0.49
Solution
The correct answer is C) 0.51.
Number of true/false questions: $$ 51 $$
Number of MCQs: $$ 49 $$
The probability of an event $E$, denoted as $P(E)$, is calculated using the formula: $$ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} $$
Therefore, the probability of choosing a true/false question is: $$ P(\text{True/False}) = \frac{51}{51 + 49} = 0.51 $$
Hence, the answer is 0.51.
In a school, there were five teachers, $A$ and $B$ were teaching Hindi and English. $C$ and $B$ were teaching English and Geography. $D$ and $A$ were teaching Mathematics and Hindi. $E$ and $B$ were teaching History and French.
Who among the teachers was teaching the maximum number of subjects? A. $A$ B. $B$ C. $C$ D. $D$
To determine which teacher is teaching the maximum number of subjects in the school, let's analyze the distribution of subjects among the teachers A, B, C, D, and E based on the information provided:
Teacher A teaches:
Hindi (with B)
Mathematics (with D)
Total: 2 subjects
Teacher B teaches:
Hindi (with A)
English (with A and C)
Geography (with C)
History (with E)
French (with E)
Total: 5 subjects
Teacher C teaches:
English (with B)
Geography (with B)
Total: 2 subjects
Teacher D teaches:
Mathematics (with A)
Hindi (already counted with A)
Total: 1 unique subject (Mathematics)
Teacher E teaches:
History (with B)
French (with B)
Total: 2 subjects
From this analysis, it is clear that Teacher B teaches the highest number of subjects, which are Hindi, English, Geography, History, and French. This makes Teacher B the teacher who is teaching the maximum number of subjects in the school.
Therefore, the correct answer is: B. B.
Two dice are rolled, and the probability distribution of the sum of the numbers on the dice is formed. Find the mean of the sum.
To solve the problem of finding the mean of the sum when two dice are rolled, follow these steps:
Identify the Range of Values for the Sum:
When two dice are rolled, the sum of the numbers can range from 2 (when both dice show 1) to 12 (when both dice show 6). Therefore, the possible values for the sum ( x ) are (2, 3, 4, \ldots, 12).
Calculate the Probabilities:
We need to find the probability for each possible sum value. For two dice:
$ P(X=2) = \frac{1}{36}$
$ P(X=3) = \frac{2}{36}$
$P(X=4) = \frac{3}{36}$
$P(X=5) = \frac{4}{36}$
$P(X=6) = \frac{5}{36} $
$P(X=7) = \frac{6}{36}$
$P(X=8) = \frac{5}{36}$
$P(X=9) = \frac{4}{36} $
$P(X=10) = \frac{3}{36}$
$P(X=11) = \frac{2}{36}$
$P(X=12) = \frac{1}{36} $
Calculate the Mean (Expectation) ( E$X$ ):
The mean or expectation of $ X $, ( E$X$ ), is given by the formula: $$ E(X) = \sum x_i \cdot P(x_i) $$
Substitute the values and their probabilities: $$ E(X) = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36} + 5 \cdot \frac{4}{36} + 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{4}{36} + 10 \cdot \frac{3}{36} + 11 \cdot \frac{2}{36} + 12 \cdot \frac{1}{36} $$
Simplify the Calculation:
Combine and simplify: $$ E(X) = \frac{1}{36} \left(2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12\right) $$
Sum the terms inside the parenthesis: $$ E(X) = \frac{1}{36} \cdot 252 = 7 $$
Therefore, the mean of the sum of the numbers on the two dice is 7.
Final Answer: The mean of the sum is $ \boxed{7}$.
A coin is loaded such that (P(H)=3P(T)). It is tossed 3 times. Let (X) be the random variable which indicates the number of heads which occur. Find the mean of (X) and variance of (X).
To solve the given problem, let's break it down step by step.
Step 1: Determine Probabilities
We are given that the probability of getting heads, ( P(H) ), is 3 times the probability of getting tails, ( P(T) ).
[ P(H) = 3P(T) ]
Since the sum of probabilities for heads and tails must equal 1, we have:
[ P(H) + P(T) = 1 ]
Substituting ( P(H) = 3P(T) ):
[ 3P(T) + P(T) = 1 ] [ 4P(T) = 1 ] [ P(T) = \frac{1}{4} ]
Thus, the probability of heads is:
[ P(H) = 3 \times \frac{1}{4} = \frac{3}{4} ]
Step 2: Possible Values of ( X )
The random variable ( X ) denotes the number of heads in 3 tosses, which can take the values 0, 1, 2, or 3.
Step 3: Calculate Probabilities of Each Outcome
No heads (0 heads):
[ P(X = 0) = \left( \frac{1}{4} \right)^3 = \frac{1}{64} ]
One head:
This can occur in 3 ways: ( HTT, THT, TTH ).
[ P(X = 1) = 3 \times \left( \frac{3}{4} \times \frac{1}{4} \times \frac{1}{4} \right) = 3 \times \frac{3}{64} = \frac{9}{64} ]
Two heads:
This can occur in 3 ways: ( HHT, HTH, THH ).
[ P(X = 2) = 3 \times \left( \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} \right) = 3 \times \frac{9}{64} = \frac{27}{64} ]
Three heads:
[ P(X = 3) = \left( \frac{3}{4} \right)^3 = \frac{27}{64} ]
Step 4: Calculate the Mean (Expectation)
The mean of ( X ), ( E(X) ), is:
[ E(X) = \sum_{x=0}^{3} x \cdot P(X = x) ] [ E(X) = 0 \times \frac{1}{64} + 1 \times \frac{9}{64} + 2 \times \frac{27}{64} + 3 \times \frac{27}{64} ] [ E(X) = \frac{9}{64} + \frac{54}{64} + \frac{81}{64} ] [ E(X) = \frac{144}{64} = \frac{9}{4} ]
Step 5: Calculate ( E(X^2) )
[ E(X^2) = \sum_{x=0}^{3} x^2 \cdot P(X = x) ] [ E(X^2) = 0^2 \times \frac{1}{64} + 1^2 \times \frac{9}{64} + 2^2 \times \frac{27}{64} + 3^2 \times \frac{27}{64} ] [ E(X^2) = \frac{9}{64} + 4 \times \frac{27}{64} + 9 \times \frac{27}{64} ] [ E(X^2) = \frac{9}{64} + \frac{108}{64} + \frac{243}{64} ] [ E(X^2) = \frac{360}{64} = \frac{45}{8} ]
Step 6: Calculate the Variance
The variance ( \text{Var}(X) ) is given by:
[ \text{Var}(X) = E(X^2) - (E(X))^2 ] [ \text{Var}(X) = \frac{45}{8} - \left( \frac{9}{4} \right)^2 ] [ \text{Var}(X) = \frac{45}{8} - \frac{81}{16} ] [ \text{Var}(X) = \frac{90}{16} - \frac{81}{16} ] [ \text{Var}(X) = \frac{9}{16} ]
Final Answer
Mean of ( X ):[ \frac{9}{4} ] Variance of ( X ):[ \frac{9}{16} ]
Let $X$ denote the profit of a business man. The probability of getting a profit of Rs. 3000 is 0.6. The probability of getting a loss of Rs. 4000 is 0.3. The probability of getting neither profit nor loss is 0.1.
The mean and variance of $X$ are:
A. 100,182000000
B. $4,00,4560000$
C. 400,12300
D. 600,9840000
Given the problem, let's determine the mean and variance of $X$, where $X$ represents the profit of a businessman.
The information we have:
Profit of ₹3000 with probability 0.6.
Loss of ₹4000 with probability 0.3.
Neither profit nor loss with probability 0.1.
We'll first construct a table to organize this information and then use it to calculate the mean and variance.
Step 1: Create a Table to Organize Given Data
Profit/Loss ( X ) | Probability ( P(X) ) |
---|---|
3000 | 0.6 |
-4000 | 0.3 |
0 | 0.1 |
Step 2: Calculate the Mean ( \mu )
The mean ((\mu)) is given by: $$ \mu = \sum X P(X) $$
Plugging in the values: $$ \mu = (3000 \times 0.6) + (-4000 \times 0.3) + (0 \times 0.1) $$
Perform the calculations: $$ \mu = 1800 - 1200 + 0 = 600 $$
So, the mean is ₹600.
Step 3: Calculate the Variance $ \sigma^2 $
The variance ($\sigma^2$) is calculated using: $$ \sigma^2 = \sum X^2 P(X) - (\mu)^2 $$
First, calculate $ \sum X^2 P(X) $: $$ \sum X^2 P(X) = (3000^2 \times 0.6) + ((-4000)^2 \times 0.3) + (0^2 \times 0.1) $$
Perform the calculations: $$ 3000^2 = 9000000 $$ $$ 4000^2 = 16000000 $$ $$ \sum X^2 P(X) = (9000000 \times 0.6) + (16000000 \times 0.3) {0 \times 0.1} $$ $$ \sum X^2 P(X) = 5400000 + 4800000 + 0 = 10200000 $$
Next, calculate ( (\mu)^2 ): $$ \mu^2 = 600^2 = 360000 $$
Finally, find the variance: $$ \sigma^2 = 10200000 - 360000 = 9840000 $$
Step 4: Conclusion
The mean and variance are:
Mean ($\mu$) = ₹600
Variance ($\sigma^2$) = 9840000
Thus, the correct answer is:
D. 600, 9840000
If on an average 1 vessel in every 10 is wrecked, the chance that out of 5 vessels expected 4 at least will arrive safely is:
A. $3 (9^{4} / 10^{5})$
B. $4 (9^{4} / 10^{5})$
C. $ (9^{4} / 10^{5})$
D. $14 (9^{4} / 10^{5})$
To find the probability that out of 5 vessels, at least 4 will arrive safely given that the probability of a vessel being wrecked is $\frac{1}{10}$, we can use the following steps:
Step-by-Step :
Identify Probabilities:
Probability that a vessel is wrecked: $P(W) = \frac{1}{10}$
Probability that a vessel is safe: $P(S) = 1 - \frac{1}{10} = \frac{9}{10}$
Calculate Needed Probabilities:
We need the probability that at least 4 out of 5 vessels are safe. This means we need either exactly 4 vessels safe or all 5 vessels safe.
Probability for Exactly 4 Safe Vessels:
The number of ways to choose 4 out of 5 vessels to be safe: $\binom{5}{4} = 5$
Probability for this event: $$ P(\text{Exactly 4 Safe}) = \binom{5}{4} \left(\frac{9}{10}\right)^4 \left(\frac{1}{10}\right)^1 = 5 \times \left(\frac{9}{10}\right)^4 \times \frac{1}{10} $$
Probability for All 5 Vessels Safe:
Probability for this event: $$ P(\text{All 5 Safe}) = \left(\frac{9}{10}\right)^5 $$
Add the Probabilities:
Total probability that at least 4 out of 5 vessels are safe: $$ P(\text{At Least 4 Safe}) = P(\text{Exactly 4 Safe}) + P(\text{All 5 Safe}) $$ Simplifying each part, we get: $$ P(\text{Exactly 4 Safe}) = 5 \times \left(\frac{9}{10}\right)^4 \times \frac{1}{10} = 5 \times \frac{9^4}{10^5} $$ $$ P(\text{All 5 Safe}) = \left(\frac{9}{10}\right)^5 = \frac{9^5}{10^5} $$ Therefore: $$ P(\text{At Least 4 Safe}) = \frac{5 \times 9^4}{10^5} + \frac{9^5}{10^5} $$ Factor out common terms: $$ P(\text{At Least 4 Safe}) = \frac{9^4}{10^5} (5 + 9) = \frac{9^4 \times 14}{10^5} $$
Hence, the final probability is: $$ \boxed{\frac{14 \times 9^4}{10^5}} $$
Four bad oranges are accidentally mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot. Find the mean and variance of the distribution.
Let $\mathrm{X}$ be the random variable representing the number of bad oranges drawn.
Possible values of $\mathrm{X}$: 0, 1, 2
Probability distribution:
$$ \begin{array}{c|c|c|c} \mathrm{X} & 0 & 1 & 2 \ \hline P(\mathrm{X}) & \frac{120}{190}=\frac{48}{95} & \frac{5 \cdot 16}{190} = \frac{80}{190} & \frac{4^2}{190}=\frac{16}{190} \ \end{array}$$
Probability of drawing 0 bad oranges:
$$
P(0) = \frac{{}^{16} \mathrm{C}{2}}{{}^{20} \mathrm{C}{2}} = \frac{120}{190} = \frac{60}{95} $$
Probability of drawing 1 bad orange:
$$ P(1) = \frac{{}^{4} \mathrm{C}{1} \cdot {}^{16} \mathrm{C}{1}}{{}^{20} \mathrm{C}_{2}} = \frac{4 \cdot 16}{190} = \frac{64}{190} = \frac{32}{95} $$
Probability of drawing 2 bad oranges:
$$ P(2) = \frac{{}^{4} \mathrm{C}{2}}{{}^{20} \mathrm{C}{2}} = \lef(\frac{1+\sum;y^{1}}{\sum; x}\journ=\text{blue} \gamma \text{{32}/{ak}})+ \text{yx}\sum{ $$
Now, the mean $\mu$ is given by:
$$ \mu = \sum X P(X) $$
$$ \mu = 0 \cdot \frac{60}{95} + 1 \cdot \frac{32}{95} + 2 \cdot \frac{6}{95} = 0 + \frac{32}{95} + \frac{12}{95} = \frac{44}{95} = \frac{2}{5} $$
And the variance $ \sigma^2 $ is given by:
$$ \sigma^2 = \sum X^2 P(X) - \mu^2 $$
$$ \sigma^2 = 0^2 \cdot \frac{60}{95} + 1^2 \cdot \frac{32}{95} + 2^2 \cdot \frac{3}{95} - \left(\frac{2}{5}\right)^2 = 0 + \frac{32}{95} + \frac{12}{95} - \frac{4}{25} $$
$$ \sigma^2 = \frac{44}{95} - \frac{16}{100} = \frac{44}{95} - \frac{4}{25} = \frac{44}{95} - \frac{16}{95} = \frac{44}{95} =\sum{x} $$
Through steps mentioned above, we find the required probability distribution, mean, and variance for the given scenario.
In a cricket match, a batsman hits a boundary 6 times out of the 30 balls he plays. Find the probability of him not hitting a boundary on the balls he plays.
A. 0.2
B. 0.4
C. 0.6
D. 0.8
The correct option is D: 0.8
Number of balls in which the batsman does not hit a boundary: $$ 30 - 6 = 24 $$
Total number of balls played: $$ \text{Total balls played} = 30 $$
The probability of not hitting a boundary is given by: $$ \frac{\text{Number of balls in which he does not hit a boundary}}{\text{Total number of balls played}} $$
Thus, the probability of not hitting a boundary: $$ \frac{24}{30} = 0.8 $$
Find the probability of obtaining a number greater than 3 when a die is thrown.
A $\frac{1}{2}$
B $\frac{1}{5}$
C $\frac{1}{3}$
D $\frac{1}{4}$
The correct option is A $\frac{1}{2}$.
When a die is thrown, the numbers greater than 3 are 4, 5, and 6.
Let $E$ be the event of obtaining a number greater than 3. The probability of event $E$ can be computed using the formula:
$$ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} $$
Here, the number of favourable outcomes (numbers greater than 3) is 3 (i.e., 4, 5, and 6), and the total number of possible outcomes (faces of the die) is 6.
Therefore, the probability $P(E)$ is:
$$ P(E) = \frac{3}{6} = \frac{1}{2} $$
Suppose $X \sim \text{B}(n, p)$ and $P(X=3)=P(X=5)$. If $p>\frac{1}{2}$.
$n \leq 7$
$n>8$
$n \geq 9$
$n<10$
To solve this problem, we need to understand the given conditions and the properties of the binomial distribution, denoted as $X \sim \text{B}(n, p)$. The key points are:
$X$ follows a binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success).
It is given that $P(X = 3) = P(X = 5)$.
The probability $p$ is greater than $\frac{1}{2}$.
We need to find the correct range for $n$ that satisfies these conditions. Let's go through the solution step-by-step.
Step-by-Step
Given Condition:[ P(X = 3) = P(X = 5) ]
Expressing the Binomial Probabilities:Using the binomial probability formula, we have: [ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} ] For $k = 3$ and $k = 5$, we can write: [ \binom{n}{3} p^3 (1-p)^{n-3} = \binom{n}{5} p^5 (1-p)^{n-5} ]
Simplifying the Equation:We can express the above equality as: [ \frac{\binom{n}{3} p^3 (1-p)^{n-3}}{\binom{n}{5} p^5 (1-p)^{n-5}} = 1 ] Simplifying this, we get: [ \frac{\binom{n}{3}}{\binom{n}{5}} \cdot \frac{(1-p)^2}{p^2} = 1 ]
Simplifying Further:Using the factorials in the binomial coefficients: [ \frac{\frac{n!}{3!(n-3)!}}{\frac{n!}{5!(n-5)!}} = \frac{5!}{3!(n-3)!(n-4)(n-5)} = \frac{n(n-1)(n-2)}{20(n-4)(n-5)} ] Therefore: [ \frac{n(n-1)(n-2)}{20(n-4)(n-5)} \cdot \frac{(1-p)^2}{p^2} = 1 ]
Given Condition $p > \frac{1}{2}$:Since $p > \frac{1}{2}$, $(1-p) < \frac{1}{2}$, and thus $\frac{(1-p)^2}{p^2} < 1$.
Inequality Implication:Hence: [ \frac{n(n-1)(n-2)}{20(n-4)(n-5)} < 1 ] Thus: [ n(n-1)(n-2) < 20(n-4)(n-5) ]
Solving the Inequality:Expanding and simplifying the inequality will provide: [ n^3 - 17n^2 + 74n - 60 < 0 ] Solving this cubic inequality, we find that the maximum value of $n$ which satisfies this inequality is $n \leq 7$.
Conclusion
Based on the above detailed steps and logical deductions, we conclude that the correct answer is:
Answer: Option (A) $n \leq 7$
Consider the frequency distribution of the given numbers:
Value | 1 | 2 | 3 | 4 |
---|---|---|---|---|
Frequency | 5 | 4 | 6 | $f$ |
If the mean is known to be '3', then the value of $f$ is:
A 3
B 7
C $\quad10$
D 14
To find the value of $f$ in the given frequency distribution, we will use the formula for the mean of a frequency distribution.
Given the data:
Value | 1 | 2 | 3 | 4 |
---|---|---|---|---|
Frequency | 5 | 4 | 6 | $f$ |
And the mean, $\bar{x}$, is given as 3.
Steps to Solve:
Calculate the total number of observations.
The total frequency, $N$, is: $$ N = 5 + 4 + 6 + f = 15 + f $$
Calculate the sum of the products of values and their frequencies.
This is given by: $$ \sum (x \cdot f) = 1 \cdot 5 + 2 \cdot 4 + 3 \cdot 6 + 4 \cdot f = 5 + 8 + 18 + 4f = 31 + 4f $$
Use the formula for the mean of a frequency distribution:$$ \bar{x} = \frac{\sum (x \cdot f)}{N} $$
Substituting the given $\bar{x}$ and the expressions calculated: $$ 3 = \frac{31 + 4f}{15 + f} $$
Solve for $f$.
Multiply both sides of the equation by $(15 + f)$ to get rid of the fraction: $$ 3(15 + f) = 31 + 4f $$ Expanding and simplifying: $$ 45 + 3f = 31 + 4f $$
Rearrange to solve for $f$: $$ 45 - 31 = 4f - 3f \ 14 = f $$
Thus, the value of $f$ is 14.
Final Answer:
D. 14
10 bags of wheat flour, each marked 5 kg, actually had the following weights: 4.98, 4.9, 5.0, 5.1, 4.5, 5.12, 5.2, 4.89, 4.9, 5.11. What will be the probability that any of the bag chosen at random will have more than 5 kg of flour?
Options:
0.6
0.3
0.4
none of the above.
The correct option is (3) 0.4.
To determine the probability, we utilize the formula of probability:
$$ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} $$
In this context:
The number of bags that weigh more than 5 kg is 4. These weights are specifically 5.1, 5.12, 5.2, and 5.11 kg.
The total number of bags is 10.
Thus, the probability $P$ that a randomly chosen bag will weigh more than 5 kg is:
$$ P(\text{bag weight} > 5 , \text{kg}) = \frac{4}{10} = 0.4 $$
A die is thrown once. The probability of getting an even number is
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{6}$
(d) $\frac{5}{6}$
When a die is thrown, the possible outcomes are (1, 2, 3, 4, 5, 6).
The total number of outcomes is therefore 6.
Among these, the even numbers are (2, 4,) and (6).
There are 3 even numbers.
The probability of an event is given by the ratio of the number of favourable outcomes to the total number of possible outcomes. Thus, the probability of rolling an even number is:
$$ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} $$
Hence, the probability of getting an even number when a die is thrown is $\frac{1}{2}$.
Answer: (a) $\frac{1}{2}$
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