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Kate and Danny each have $10. Together, they flip a fair coi

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Kate and Danny each have $10. Together, they flip a fair coi

by BTGmoderatorAT » Tue Dec 12, 2017 5:43 am
Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than $10 but less than $15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

I'm confused how to set up the formulas here. Can any experts help?
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by GMATGuruNY » Tue Dec 12, 2017 5:49 am
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by regor60 » Tue Dec 12, 2017 6:24 am
ardz24 wrote:Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than $10 but less than $15?

(A) 5/16
(B) 1/2
(C) 12/30
(D) 15/32
(E) 3/8

I'm confused how to set up the formulas here. Can any experts help?
You can start by identifying the combinations of tails and heads and the total Kate receives from each:

0 T, 5H = $5
1T, 4H = 7
2T,3H=9
3T,2H=11
4T,1H=13
5T,0H=15

So, only the 3T/2H and 4T/1H combinations satisfy the problem statement.

The tricky part is recognizing that there are several ways to achieve each of the above patterns, for example, 3T/2H:

THTHT or TTTHH or HHTTT, etc.

Each one of these has a probability of (1/2)^5 or 1/32. So, need to count them.

3T/2H = 5!/3!*2! = 10 ways of achieving

4T/1H = 5!/4!*1! = 5 ways of achieving

Total ways 15. Total probability: 15*1/32 = [spoiler]15/32, D[/spoiler]
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