If in a binomial distribution n = 4, P(X=0) = 16/81, then P(X=4) equals : Option 1: 1/16 Option 2: 1/81 Option 3: 1/27 Option 4: 1/8

Answer: 1/81. Solution: P(X=0) = q^4 = 16/81 q = 2/3, so p = 1/3. P(X=4) = p^4 = (1/3)^4 = 1/81.

Algebra

Probability

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core

If in a binomial distribution $n = 4$ , $P(X=0) = \frac{16}{81}$ , then $P(X=4)$ equals :

$\frac{1}{16}$

$\frac{1}{81}$

$\frac{1}{27}$

$\frac{1}{8}$

✅ Correct Option: 2

$P(X=0) = q^4 = \frac{16}{81} \Rightarrow q = \frac{2}{3}$ , so $p = \frac{1}{3}$ .

$P(X=4) = p^4 = \left(\frac{1}{3}\right)^4 = \frac{1}{81}$ .

2025: 22 May Shift 1

Consider two independent events A and B such that $P(A) = 0.3$ , $P(B) = 0.6$ .

Match List-I with List-II

List-I	List-II
(A) $P(A$ and $B)$	(I) 0.28
(B) $P(A$ and not $B)$	(II) 0.18
(C) $P(A$ or $B)$	(III) 0.12
(D) $P$ (neither A nor B)	(IV) 0.72

Choose the correct answer from the options given below:

2026: 18 May Shift 2

A farmer has a tubewell which works by connecting two switches $S_1$ and $S_2$ . The probability that $S_1$ works is 9/10 and the probability that $S_2$ works is 8/10. The farmer connects them in series in 30% cases and in parallel in 70% cases. If it is known that the tubewell was working, what is the probability that it was a series connection.

2025: 30 May Shift 2

The probability of a shooter of hitting the target is $\frac{1}{4}$ . The minimum number of fire needed so that the probability of hitting the target atleast once is greater than $\frac{7}{16}$ is:

2026: 30 May Shift 2

A pair of dice is thrown, then the probability of getting a sum of 7, is

2022: 6 Aug Shift 2

If the mean and variance of a binomially distributed random variable X are 4 and 2 respectively, then $P(X = 2)$ is equal to

2026: 12 May Shift 2

From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let X denotes the number of defective items in the sample. If the sample is drawn randomly, then Match List-I with List-II (P(X) denotes the probability of occurence of X)

List-I	List-II
(A) P (X = 0)	(I) $\dfrac{^3C_3\times{}^7C_1}{^{10}C_4}$
(B) P (X = 1)	(II) $\dfrac{^7C_4\times{}^3C_0}{^{10}C_4}$
(C) P (X = 2)	(III) $\dfrac{^3C_1\times{}^7C_3}{^{10}C_4}$
(D) P (X = 3)	(IV) $\dfrac{^3C_2\times{}^7C_2}{^{10}C_4}$

Choose the correct answer from the options given below:

If in a binomial distribution n=4n = 4n=4, P(X=0)=1681P(X=0) = \frac{16}{81}P(X=0)=8116​, then P(X=4)P(X=4)P(X=4) equals :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If in a binomial distribution $n = 4$ , $P(X=0) = \frac{16}{81}$ , then $P(X=4)$ equals :