BTGmoderatorDC wrote: ↑

Thu May 28, 2020 6:49 pm

In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure how many draws did it take before the person picked a heart and won. What is the probability that there will be at least three draws involved in a win, i.e. someone picking her first heart on the third draw or later?

(A) 1/2
(B) 9/16
(C) 11/16
(D) 13/16
(E) 15/16

OA B

Source: Magoosh

Since the person must win in his 3rd/4th/5th/6th or the nth attempt, it's better that we calculate the probability of his win in the 1st and the 2nd attempt, and thereafter, we would deduct the result from 1 to get the answer.

Probability of winning in the 1st attempt = 13/52 = 1/4; there are 13 heart cards in a deck;

Probability of winning in the 2nd attempt
= (Prob. of losing in the 1st attempt) * (Probability of winning in the 1st attempt)
= (1 – 1/4)*(1/4) = 3/4*1/4 = 3/16

=> Probability of winning in the 1st or the 2nd attempts = 1/4 + 3/16 = 7/16

=> Probability that there will be at least three draws involved in a win = 1 – 7/16 = 9/16

The correct answer: B

Hope this helps!

-Jay
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