If X is a random variable which can assume values 0, 1, 2, 3 or 4 such that P(X = 1) = P(X = 2) and 3P(X = 3) = 4P(X = 4) = P(X = 0) = 1/8, then P(X > 0) is: Option 1: 1/8 Option 2: 5/16 Option 3: 19/96 Option 4: 7/8

Answer: 7/8. Solution: To find P(X > 0), we can use the fundamental property that the sum of all probabilities in a probability distribution must equal 1. * P(X = 0) = 1/8 * 4P(X = 4) = 1/8 P(X = 4) = 1/32 * 3P(X = 3) = 1/8 P(X = 3) = 1/24 The event X > 0 includes the outcomes X=1, X=2, X=3, and X=4. The complement of this event is X = 0. The formula for the complement is: P(X > 0) = 1 - P(X = 0) Substituting the known value for P(X = 0): P(X > 0) = 1 - 1/8 P(X > 0) = 8/8 - 1/8 P(X > 0) = 7/8

Algebra

Probability

Medium

common

If $X$ is a random variable which can assume values $0, 1, 2, 3$ or $4$ such that $P(X = 1) = P(X = 2)$ and $3P(X = 3) = 4P(X = 4) = P(X = 0) = \frac{1}{8}$ , then $P(X > 0)$ is:

$\frac{1}{8}$

$\frac{5}{16}$

$\frac{19}{96}$

$\frac{7}{8}$

✅ Correct Option: 4

To find $P(X > 0)$ , we can use the fundamental property that the sum of all probabilities in a probability distribution must equal $1$ .

$P(X = 0) = \frac{1}{8}$
$4P(X = 4) = \frac{1}{8} \implies P(X = 4) = \frac{1}{32}$
$3P(X = 3) = \frac{1}{8} \implies P(X = 3) = \frac{1}{24}$

The event $X > 0$ includes the outcomes $X=1, X=2, X=3,$ and $X=4$ . The complement of this event is $X = 0$ .

The formula for the complement is:

$P(X > 0) = 1 - P(X = 0)$

Substituting the known value for $P(X = 0)$ :

$P(X > 0) = 1 - \frac{1}{8}$

$P(X > 0) = \frac{8}{8} - \frac{1}{8}$

$P(X > 0) = \frac{7}{8}$

2025: 15 May Shift 2

A player participates in 3 matches against three teams T₁, T₂ and T₃.The probability of winning a match against teams T₁, T₂ and T₃ are 0.2, 0.3 and 0.9 respectively. If 'wins' can be regarded as independent events, then the probability that he

(A) wins all the 3 matches is 0.054

(B) wins no match is 0.054

(D) wins exactly one match is 0.542

Choose the correct answer from the options given below:

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Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. The probability that it was drawn from Bag II is :

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The probability that it will rain on any particular day is 50%. The probability that it rains only on the first 4 days of the week is:

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In Binomial distribution with parameters $n = 100$ and p, Variance of distribution is maximum when p is equal to :

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Two persons A and B throw a die alternately till one of them gets a 'three' and wins the game. The probability of A's winning if A starts first 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 XXX is a random variable which can assume values 0,1,2,30, 1, 2, 30,1,2,3 or 444 such that P(X=1)=P(X=2)P(X = 1) = P(X = 2)P(X=1)=P(X=2) and 3P(X=3)=4P(X=4)=P(X=0)=183P(X = 3) = 4P(X = 4) = P(X = 0) = \frac{1}{8}3P(X=3)=4P(X=4)=P(X=0)=81​, then P(X>0)P(X > 0)P(X>0) is:

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CUET Mathematics Past Year Questions (Topic-Wise)

If $X$ is a random variable which can assume values $0, 1, 2, 3$ or $4$ such that $P(X = 1) = P(X = 2)$ and $3P(X = 3) = 4P(X = 4) = P(X = 0) = \frac{1}{8}$ , then $P(X > 0)$ is: