There are 13 hearts in a full deck of 52 cards. In a certain

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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

The OA is B

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by [email protected] » Thu Feb 07, 2019 10:12 am
Hi All,

The probability of pulling a heart out of a full deck of cards is 13/52 = 1/4, so the probability of NOT pulling a heart out of a full deck = 1 - 1/4 = 3/4.

The probability of pulling a heart on the THIRD try, and NOT pulling a heart on the first two tries is:

(3/4)(3/4)(1/4) = 9/64

HOWEVER, the question asked for the probability of pulling the first heart on the third draw OR LATER. This is really asking for the probability of NOT pulling a heart on the first two pulls, since a heart on the third or fourth or fifth or sixth, etc. draw satisfies what we're looking for.

So, NO HEART on the first two draws = (3/4)(3/4) = 9/16

Whatever happens next is irrelevant; you would eventually pull a heart (either on the third draw or later).

Final Answer: B

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by Scott@TargetTestPrep » Sun Feb 10, 2019 7:37 am
swerve wrote: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

The OA is B

Source: Magoosh
Since we want the probability of drawing the first heart on the third draw or later, we can do the "opposite" by determining the probability of drawing the heart on the first draw and that of drawing the first heart on the second draw. After we determine these probabilities, we will subtract the sum of their probabilities from 1. In other words, we will use the following:

1 = P(heart on the first draw) + P(heart on the second draw) + P(heart on the third or a later draw)

Let H = heart and N = non-heart, so P(H) = ¼ and P(NH) = ¾ x ¼ = 3/16. Thus, we know that the probability of a heart on the first draw is 1/4, and the probability of drawing a heart on the second draw is 3/16.

Therefore, the probability of drawing the first heart on the third draw or later is:

1 - 1/4 - 3/16 = 16/16 - 4/16 - 3/16 = 9/16

Answer: B

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Re:

by geneing » Thu Dec 07, 2023 6:23 pm
Scott@TargetTestPrep wrote:
Sun Feb 10, 2019 7:37 am
swerve wrote: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

The OA is B

Source: Magoosh
Since we want the probability of drawing the first heart on the third draw or later, we can do the "opposite" by determining the probability of drawing the heart on the first draw and that of drawing the first heart on the second draw. After we determine these probabilities, we will subtract the sum of their probabilities from 1. In other words, we will use the following:

1 = P(heart on the first draw) + P(heart on the second draw) + P(heart on the third or a later draw)

Let H = heart and N = non-heart, so P(H) = ¼ and P(NH) = ¾ x ¼ = 3/16. Thus, we know that the probability of a heart on the first draw is 1/4, and the probability of drawing a heart on the second draw is 3/16.

Therefore, the probability of drawing the first heart on the third draw or later is:

1 - 1/4 - 3/16 = 16/16 - 4/16 - 3/16 = 9/16

Answer: B
Yes, I agree with you that the probability that one will pick the first heart on the third draw or later is 0.3125.
Therefore, the answer is (B). 9/16.

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swerve wrote:
Thu Feb 07, 2019 10:01 am
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

The OA is B

Source: Magoosh
To calculate the probability of picking the first heart on the third draw or later, we can use the complement rule. This means we find the probability of not picking a heart on the first two draws and then subtracting that from 1.
The probability of not picking a heart on the first draw is 39/52 (since there are 39 non-heart cards out of 52). After not picking a heart on the first draw, there are 38 non-heart cards left out of 51, so the probability of not picking a heart on the second draw is 38/51.
Therefore, the probability of not picking a heart on the first two draws is (39/52) * (38/51) which equals approximately 0.444 or 4/9.

Subtracting this from 1 gives us the probability of picking the first heart on the third draw or later, which is 1 - 4/9 = 5/9.
Converting 5/9 to a fraction with a denominator of 16, we get 9/16, which matches option B.