The binomial distribution for which the mean is 5 and variance 4, is Option 1: P(X = r) = ^25C_r (4/5)^r (1/5)^(25-r), r = 0, 1, 2, 3, ..., 25 Option 2: P(X = r) = ^25C_r (1/5)^r (4/5)^(25-r), r = 0, 1, 2, 3, ..., 25 Option 3: P(X = r) = ^25C_r (1/5)^25 (4/5)^(25-r), r = 0, 1, 2, 3, ..., 25 Option 4: P(X = r) = ^25C_r (1/5)^(25-r) (4/5)^25, r = 0, 1, 2, 3, ..., 25

Answer: P(X = r) = ^25C_r (1/5)^r (4/5)^(25-r), r = 0, 1, 2, 3, ..., 25. Solution: For a binomial distribution with mean = 5 and variance = 4, the parameters n (number of trials) and p (probability of success) need to be determined. For binomial distribution: Mean = n x p = 5 Variance = n x p x (1 - p) = 4 Dividing variance by mean: (np(1-p))/np = 4/5 1 - p = 4/5 p = 1 - 4/5 p = 1/5 Using the mean formula with p = 1/5: n x 1/5 = 5 n = 25 The binomial distribution formula with n = 25, p = 1/5, and (1-p) = 4/5: P(X = r) = ^25C_r x (1/5)^r x (4/5)^(25-r) where r = 0, 1, 2, 3, ..., 25

Algebra

Probability

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The binomial distribution for which the mean is 5 and variance 4, is

$P(X = r) = ^{25}C_r (\frac{4}{5})^r (\frac{1}{5})^{25-r}$ , $r = 0, 1, 2, 3, ..., 25$

$P(X = r) = ^{25}C_r (\frac{1}{5})^r (\frac{4}{5})^{25-r}$ , $r = 0, 1, 2, 3, ..., 25$

$P(X = r) = ^{25}C_r (\frac{1}{5})^{25} (\frac{4}{5})^{25-r}$ , $r = 0, 1, 2, 3, ..., 25$

$P(X = r) = ^{25}C_r (\frac{1}{5})^{25-r} (\frac{4}{5})^{25}$ , $r = 0, 1, 2, 3, ..., 25$

✅ Correct Option: 2

For a binomial distribution with mean = 5 and variance = 4, the parameters n (number of trials) and p (probability of success) need to be determined.

For binomial distribution:

Mean = $n \times p = 5$

Variance = $n \times p \times (1 - p) = 4$

Dividing variance by mean:

$\dfrac{np(1-p)}{np} = \dfrac{4}{5}$

$1 - p = \dfrac{4}{5}$

$p = 1 - \dfrac{4}{5}$

$p = \dfrac{1}{5}$

Using the mean formula with $p = \dfrac{1}{5}$ :

$n \times \dfrac{1}{5} = 5$

$n = 25$

The binomial distribution formula with $n = 25$ , $p = \dfrac{1}{5}$ , and $(1-p) = \dfrac{4}{5}$ :

$P(X = r) = ^{25}C_r \times \left(\dfrac{1}{5}\right)^r \times \left(\dfrac{4}{5}\right)^{25-r}$

where $r = 0, 1, 2, 3, ..., 25$

2023: 23 May Shift 3

If a fair coin is tossed 10 times the probability of atleast 6 heads is:

2026: 26 May Shift 2

A random variable $X$ has the following probability distribution:

X	0	1	2	3	4
P(X)	0	$\dfrac{k}{2}$	$\dfrac{3k}{2}$	$\dfrac{k}{3}$	$2k$

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	Value of $k$	I.	$\dfrac{7}{13}$
B.	Value of $P(X \leq 3)$	II.	$\dfrac{3}{26}$
C.	Value of $P(X > 1)$	III.	$\dfrac{23}{26}$
D.	Value of $P(X < 2)$	IV.	$\dfrac{3}{13}$

Choose the correct answer from the options given below:

2026: 12 May Shift 2

Suppose that A and B are independent events with $P(A)=\dfrac{1}{7}$ and $P(B)=\dfrac{3}{7}$ , then $P(A\cap B)$ is

2025: 15 May Shift 1

If $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$ , then $P(A \cap B)$ is

2025: 19 May Shift 1

Bag A contains 2 unbiased and 3 biased coins whereas Bag B contains 3 unbiased and 2 biased coins. A bag is selected at random and 2 coins are taken out simultaneously. The probability, that both coins are unbiased is:

2025: 13 May Shift 2

Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice.

Match List-I with List-II

List-I	List-II
X	Probability, P(X)
(A) 4	(I) $\frac{1}{6}$
(B) 5	(II) $\frac{5}{36}$
(C) 6	(III) $\frac{1}{12}$
(D) 7	(IV) $\frac{1}{9}$

Choose the correct answer from the options given below:

The binomial distribution for which the mean is 5 and variance 4, is

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