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Re: In order, Anna, Beth, and Carrie take turns flipping the [#permalink]
I added answer choices...
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Re: In order, Anna, Beth, and Carrie take turns flipping the [#permalink]
Dan wrote:
In order, Anna, Beth, and Carrie take turns flipping the same fair coin. The first one to toss a head wins. What is the probability that Beth wins?

A. 1/4
B. 1/2
C. 1/3
D. 1/8
E. 2/7


prob(Beth wins) = 1/4 + 1/32 +...+ 1/4*(8^(n-1))
n goes to infinity

E, D, A are out since they are less than 1/4,

1/3? or 1/2 I have no idea how sum it up.
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Re: In order, Anna, Beth, and Carrie take turns flipping the [#permalink]
For Beth to win, she has to be ths first one to toss heads, for every possible round (ad inifinitum)

1/4 + 1/8(1/4)+ (1/8)^2*1/4+....

This is an infinite geometric series, a+ar+ar^2...

The sum is a/1-r = 1/4/(1-1/8) = (1/4)/(7/8) = 2/7

Answer is E

Sparky, 2/7 > 1/4
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Re: In order, Anna, Beth, and Carrie take turns flipping the [#permalink]
well..what can I say?!

E is correct.



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