probability and counting

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probability and counting

by Amrabdelnaby » Tue Nov 17, 2015 6:01 am
Dear all,

although i solved the below probability problem in 2 different ways using probability and combinations, i was wondering if there is any way easier and shorter than mine because each time i solved it took me more than 2 minutes doing many calculations. please let me know if there is a shortcut so solving it without doing many calculations.

7. 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 did it take before the person picked a heart and won? What is the probability that one will have at least three draws before one picks a heart?

A. 1/2

B. 9/16

C. 11/16

D. 13/16

E. 15/16

Here is how I did it:

Probability:

number of different possibilities: 52 x 52 x 52 = 52^3
break rule: 39/52 x 39/52 x 39/52 = 39^3/52^3 --> here it took me so long to do calculations
1 - break: 52^3/52^3 - 39^3/52^3 --> also here it took a lot of time to do calculations to arrive to the right answer choice


Combinations:
Ignore rule: 52 x 52 x 52 = 52^3
break rule: 39 x 39 x 39
(ignore - break)/ignore = (52^3 - 39^3)/52^3

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by regor60 » Tue Nov 17, 2015 6:31 am
Probability of taking 3 or more draws to get a heart equals 1 - (Probability of getting a heart in 2 or fewer draws).

Probability of heart in one draw: 13/52 = 1/4

Probability of not getting heart in first draw but in second: 39/52 x 13/52 = 3/16

Therefore probability of getting a heart in 2 or fewer draws equals sum of above or 7/16

Therefore probability of drawing heart in 3 or more equals 1 - 7/16, or 9/16

Answer is B.

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by Amrabdelnaby » Tue Nov 17, 2015 7:51 am
Hi Regor,

Why can't we just do 1- (3^3/4^3) which is equal to 37/64 which is very close to 9/16