In a certain game, you pick a card from a standard deck of

Here is another method of solving the same:

P(hearts)=13/52

p(atleast three draws involved in win)= 1-(atmost two draws involved in win)

=1-((win in first draw)+((lose in first draw)*(win in second draw))

=1-((13/52)+((39/52)*(13/52 )))=9/16

In order to win on precisely the third draw, three things need to happen: you must lose on the first draw AND lose on the second draw AND win on the third draw. So we need to multiply the probability of those three events together: (3/4)(3/4)(1/4) = 9/64

If we want to know the probability the game lasts at least three draws, then we just need to lose on the first two draws, so the probability of that is (3/4)(3/4) = 9/16.

If we subtract that probability from 1, to get 7/16, we're finding the probability the game does NOT last at least three draws, so 7/16 is the probability the player wins on either the first or second draw. If that were the question, you could also solve in another way, by dividing the problem into two cases: the player wins 1/4 of the time on the first draw. 3/4 of the time the player loses on the first draw, but then wins on the second draw with a probability of 1/4, so the probability of winning precisely on the second draw is (3/4)(1/4) = 3/16. Adding the probabilities of winning on the first and second draws, we get 1/4 + 3/16 = 7/16

I'd add that the question in this thread is not a realistic GMAT problem (the math involved is realistic, but the setup is not). You don't need to know anything at all about decks of cards for the test. The wording is also a lot more complicated than what you'd see on the real test.

Hi Mike,

Could you please help where I am going wrong. Why haven't we multiplied by 1/4?

I did this question in the mentioned way:

The probability of not drawing a heart in the first draw : 3/4
The probability of not drawing a heart in the second draw : 3/4
The probability of drawing a heart in the third draw : 3/4

So probability of not drawing a heart in first and second draw but drawing a heart in the third draw is 3/4*3/4*1/4=9/64.


Thanks

Got it. Thanks Mike

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.

If the question were to ask what is the probability that one would win on the third draw, would the correct answer be 7/16?
1-(3/4)^2=7/16

However, since the question is asking what are the odds of having a minimum of 3 draws to gain a victory (we don't care about the the third draw), we just use (3/4)^2=9/16, correct?

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