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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Probability

Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let XX denote the random variable of number of aces obtained in the two drawn cards. Then, P(X=1)+P(X=2)P(X=1) + P(X=2) equals
  1. A.
    25169\dfrac{25}{169}
  2. B.
    52169\dfrac{52}{169}
  3. C.
    49169\dfrac{49}{169}
  4. D.
    24169\dfrac{24}{169}
Show answer and solution

Correct option: A

Solution

P(ace) = 4/52 = 1/13, P(no ace) = 12/13.

P(X=1) = 2·(1/13)·(12/13) = 24/169.
P(X=2) = (1/13)² = 1/169.

Sum = 24/169 + 1/169 = 25/169.

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