There are 13 hearts in a full deck of 52 cards. 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 it took before the person picked a heart and, thereby, won. What is the probability that one will pick the first heart on the third draw or later?

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

Hi sunita123,

Here, we're told that 13 of the 52 cards are "hearts" - this is 1/4 of the deck. We're also told that after drawing a card, the card is PUT BACK in to the deck (so there will always be 52 cards to draw from).

To draw a heart for the first time on the 3rd draw (or later), the FIRST TWO draws MUST NOT be hearts....

Probability of NOT drawing a heart on the first draw = 3/4
Probability of NOT drawing a heart on the second draw = 3/4

(3/4)(3/4) = 9/16

From this point, you will eventually draw a heart (which is what the question is asking for), so the probability is 9/16.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi adkikani

Generally the definition of "favorable" in Probability is the event that you are looking for. So by definition 39 is your unfavorable outcome. Your calculation is perfectly ok here.

First two events are unfavorable and you are multiplying those events to get the final answer.

Hello guys! I need help form those of you who are great at probabilities.

In this one I thought "Oh, ok. So I could calculate the complement of the probability of reaching the third trial and NOT draw a heart: 1-(3/4)^3 = 37/64.

What was wrong with my train of thought?

There are 13 hearts in a full deck of 52 cards. 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 it took before the person picked a heart and, thereby, won. What is the probability that one will pick the first heart on the third draw or later?

Probability of selecting heart in first draw = \(\frac{13}{52}\) = \(\frac{1}{4}\)

Probability of selecting heard in second draw = \(\frac{3}{4}\) * \(\frac{1}{4}\) =\(\frac{3}{16}\)

Probability of selecting heart in 1st and second draw = \(\frac{1}{4}\) + \(\frac{3}{16}\) = \(\frac{7}{16}\)

Probability of selecting heart in 3rd draw = 1-\(\frac{7}{16}\) = \(\frac{9}{16}\)

Ans: B
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