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CUET Mathematics 2025 PYQs13 May Shift 2. Free, no login required.

Algebra

past year questions, CUET Mathematics 2025 13 May Shift 2

Probability

past year questions, Algebra, CUET Mathematics 2025 13 May Shift 2

Medium

applied

A random variable X has the following probability distribution:

X0123
P(X)0.10.20.30.4

The variance of the X will be:

Solution

✅ Correct Option: 1

The variance of random variable X is calculated using the formula:

Variance =E(X2)−[E(X)]2= E(X^2) - [E(X)]^2

The expected value E(X)E(X) is:

E(X)=(0×0.1)+(1×0.2)+(2×0.3)+(3×0.4)E(X) = (0 \times 0.1) + (1 \times 0.2) + (2 \times 0.3) + (3 \times 0.4)

E(X)=0+0.2+0.6+1.2E(X) = 0 + 0.2 + 0.6 + 1.2

E(X)=2E(X) = 2

The expected value of X2X^2 is:

E(X2)=(02×0.1)+(12×0.2)+(22×0.3)+(32×0.4)E(X^2) = (0^2 \times 0.1) + (1^2 \times 0.2) + (2^2 \times 0.3) + (3^2 \times 0.4)

E(X2)=(0×0.1)+(1×0.2)+(4×0.3)+(9×0.4)E(X^2) = (0 \times 0.1) + (1 \times 0.2) + (4 \times 0.3) + (9 \times 0.4)

E(X2)=0+0.2+1.2+3.6E(X^2) = 0 + 0.2 + 1.2 + 3.6

E(X2)=5E(X^2) = 5

The variance is:

Variance =E(X2)−[E(X)]2= E(X^2) - [E(X)]^2

Variance =5−(2)2= 5 - (2)^2

Variance =5−4= 5 - 4

Variance =1= 1

Therefore, the variance of X is 11.

Related questions:

2022: 17 Aug Shift 2

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows :

For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive.

Based on the above informations, answer the question :

The probability of the person to be tested as COVID positive, given that he is actually not having COVID is :

2022: 17 Aug Shift 2

It is given that only 0.1% of a large population have COVID infection. In this population, the reliability of COVID RTPCR-test is specified as follows :

For persons having COVID, 90% of the test detects the disease but 10% goes undetected. For persons not having COVID, 99% of the test is judged COVID negative but 1% are diagnosed as COVID positive.

Based on the above informations, answer the question :

The probability of the person tested as COVID positive, given that he is actually having COVID is :

2022: 10 Aug Shift 1

Bag I contains 4 red and 5 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 black. Then the probability that it was drawn from Bag II, is

2026: 18 May Shift 2

If P(A)=0.4P(A) = 0.4, P(B)=XP(B) = X and P(Aˉ∩Bˉ)=0.4P(\bar{A} \cap \bar{B}) = 0.4, where A and B are independent event, then the value of X is:

2025: 27 May Shift 1

Let box I contains 3 black and 4 white balls, box II contains 2 black and 2 white balls, box III contains 4 black and 3 white balls. A box is selected at random and then a ball is randomly drawn from the selected box. If the color of the ball is black then the probability that the ball is drawn from box III, is:

2026: 15 May Shift 1

If E and F are two independent events such that P(E)=13,P(F)=35P(E)=\dfrac{1}{3}, P(F)=\dfrac{3}{5}, then which of the following is/are correct ?

A. P(E∩F)=15P(E \cap F) = \dfrac{1}{5}

B. P(E‾F)=23P\left(\dfrac{\overline{E}}{F}\right) = \dfrac{2}{3}

C. P(EF‾)=23P\left(\dfrac{E}{\overline{F}}\right) = \dfrac{2}{3}

D. P(EF)=23P\left(\dfrac{E}{F}\right) = \dfrac{2}{3}

Choose the correct answer from the options given below:

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