Probability

[outty]

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

[outty]

I managed to solve this a different way, the explanation given by Magoosh seems a little more convoluted. I'm just wondering if my method is correct or just lucky.

P(win on 3rd or more) = P(not heart) * P(not heart) * P(heart)

P(win on 3rd or more) = 3/4 * 3/4 * 1/4

Ans = [spoiler]B, 9/16[/spoiler]

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Hi outty,

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).

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

[masoom j negi]

Here, we need to find the probability of not drawing a heart in 1st and 2nd attempt.
We won't consider the probability of drawing a heart in 3rd attempt because we are supposed to calculate the probability of drawing atleast in the 3rd draw and not surely in 3rd draw.
So, probability of not drawing in 1st attempt = 3/4
Probability of not drawing in 2nd attempt = 3/4
Probability of not drawing in 1st and 2nd draw = 9/16.

[Scott@TargetTestPrep]

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

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