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CUET Mathematics 2025 PYQs3 June Shift 2. Free, no login required.

Algebra

past year questions, CUET Mathematics 2025 3 June Shift 2

Probability

past year questions, Algebra, CUET Mathematics 2025 3 June Shift 2

Medium

applied

A random variable X has the following probability distribution:

X012
P(X)1/41/21/4

then, which of the following is correct?

Solution

✅ Correct Option: 3

A probability distribution table shows:

  • X = 0 with probability 14\frac{1}{4}
  • X = 1 with probability 12\frac{1}{2}
  • X = 2 with probability 14\frac{1}{4}

The mean is the expected value of X.

Mean =(0×14)+(1×12)+(2×14)= (0 \times \frac{1}{4}) + (1 \times \frac{1}{2}) + (2 \times \frac{1}{4})

Mean =0+12+12= 0 + \frac{1}{2} + \frac{1}{2}

Mean =1= 1

To find variance, first calculate E(X2)E(X^2):

E(X2)=(02×14)+(12×12)+(22×14)E(X^2) = (0^2 \times \frac{1}{4}) + (1^2 \times \frac{1}{2}) + (2^2 \times \frac{1}{4})

E(X2)=0+12+1E(X^2) = 0 + \frac{1}{2} + 1

E(X2)=32E(X^2) = \frac{3}{2}

Variance =E(X2)−(Mean)2= E(X^2) - (\text{Mean})^2

Variance =32−(1)2= \frac{3}{2} - (1)^2

Variance =32−1= \frac{3}{2} - 1

Variance =12= \frac{1}{2}

Testing the options with Mean =1= 1 and Variance =12= \frac{1}{2}:

Option 1: mean = variance → 1=121 = \frac{1}{2} (False)

Option 2: 2 mean = variance → 2(1)=122(1) = \frac{1}{2} → 2=122 = \frac{1}{2} (False)

Option 3: mean = 2 variance → 1=2(12)1 = 2(\frac{1}{2}) → 1=11 = 1 (True)

Option 4: 3 mean = 2 variance → 3(1)=2(12)3(1) = 2(\frac{1}{2}) → 3=13 = 1 (False)

Therefore, Option 3 is correct: mean =2= 2 variance

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Match List - I with List - II.

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(B) If P(F)≠0P(F) \neq 0, then P(E′/F)P(E'/F) is equal to(II) P(E)=P(F)P(E) = P(F)
(C) If E and F are independent events, then(III) P(E∩F′)=P(E)⋅P(F′)P(E \cap F') = P(E) \cdot P(F')
(D) If P(E∩F)≠0P(E \cap F) \neq 0 and P(E/F)=P(F/E)P(E/F) = P(F/E), then(IV) [1−P(E)][1−P(F)][1-P(E)][1-P(F)]

Choose the correct answer from the options given below :

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