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

EMPOWERgmatRichC niks18 Bunuel amanvermagmat chetan2u



Can I also say, out of 42 cards, 13 are hearts so the probability of NOT choosing a heart is
52-13 ie 39 (Favorable outcomes) divided by all possible outcomes ie 52. 39/52 = 3/4
Same for the second draw: 3/4
Total Probability = 9/16

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.

Hi adkikani,

Yes - that way of writing out the math is the exact same calculation that I did (I just reduced the fraction immediately, then subtracted - whereas you subtracted first, then reduced the fraction).

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

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?

Hi JohnAHD,

Your calculation is based on the idea that the FIRST 3 cards are NOT hearts. However, the prompt asks for the probability of pulling a heart on the 3rd card or later.

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

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

P(h)=13/52=1/4
the person replaces the card to the deck, so first draw heart, P(A) and second draw heart, P(B), will have equal chance.

We wish to only draw a heart once, so we can eliminate the times when a heart is drawn both times
P(A)+P(B)-P(AB)
1/4+1/4-1/16=4/16+4/16-1/16=7/16

We will use 1-P to find when a heart is not drawn once in the first two tries

1-7/16=9/16

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