Algebra

Probability

Medium

applied

If the difference between mean and variance of a Binomial distribution is 1 and the difference of their squares is 5, then the probability of success is

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{2}{5}$

$\frac{3}{5}$

✅ Correct Option: 2

For a Binomial distribution:

Mean $= np$

Variance $= np(1-p) = npq$

where $n$ is the number of trials, $p$ is the probability of success, and $q = 1-p$ .

Given conditions:

Mean - Variance $= 1$ ... (i)

Mean² - Variance² $= 5$ ... (ii)

Using the algebraic identity $a^2 - b^2 = (a+b)(a-b)$ on condition (ii):

Mean² - Variance² $= ($ Mean $+$ Variance $)($ Mean $-$ Variance $)$

$5 = ($ Mean $+$ Variance $) \times 1$

Mean $+$ Variance $= 5$ ... (iii)

From equations (i) and (iii):

Mean - Variance $= 1$

Mean + Variance $= 5$

Adding these equations:

$2 \times$ Mean $= 6$

Mean $= 3$

Subtracting equation (i) from (iii):

$2 \times$ Variance $= 4$

Variance $= 2$

Using the relationship between mean and variance:

$\frac{\text{Variance}}{\text{Mean}} = \frac{np(1-p)}{np} = 1-p$

$\frac{2}{3} = 1-p$

$p = 1 - \frac{2}{3}$

$p = \frac{1}{3}$

Therefore, the probability of success is $\frac{1}{3}$ .

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The probabilities of occurrance of two events A and B are 0.45 and 0.20 respectively. The probability of their simultaneous occurrence is 0.06. The probability that neither A nor B occurs is

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If A and B are two independent events such that P(A) = 0.5, P(B) = 0.3 then

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. $P(A/B)$	I. 0.35
B. $P(B/A)$	II. 0.5
C. $P(A\cap B)$	III. 0.3
D. $P(\bar{A}\cap\bar{B})$	IV. 0.15

Choose the correct answer from the options given below:

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A coin is tossed K times. If the probability of getting $3$ heads is equal to the probability of getting $7$ heads, then the probability of getting $8$ tails is :

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If A and B are independent events, then which of the following is/are true?

(A) $\bar{A}$ and B are independent events

(B) $P(A \cap B) = 0$

(D) $P(A \cap B) = P(A) + P(B)$

Choose the correct answer from the options given below:

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Two cards are drawn simultaneously at random from a well shuffled pack of 52 Cards. Let X be the random variable which denotes number of kings in the draw. Then the probability distribution of X is

2026: 22 May Shift 1

Two dice are thrown simultaneously. If X denotes the number of sixes, then the mean of X is :

If the difference between mean and variance of a Binomial distribution is 1 and the difference of their squares is 5, then the probability of success is

Solution

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