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Kate and David each have $10. Together they flip a coin 5 [#permalink]

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13 Jul 2010, 09:43

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A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

66% (01:43) correct
34% (01:25) wrong based on 172 sessions

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Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?

A. \(\frac{5}{16}\) B. \(\frac{15}{32}\) C. \(\frac{1}{2}\) D. \(\frac{21}{32}\) E. \(\frac{11}{16}\)

Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15? a. 5/16 b. 15/32 c. 1/2 d. 21/32 e. 11/16

After 5 tries Kate to have more than initial sum of 10$ and less than 15$ must win 3 or 4 times (if she wins 2 or less times she'll have less than 10$ and if she wins 5 times she'll have 15$).

So the question becomes "what is the probability of getting 3 or 4 tails in 5 tries?".

\(P(t=3 \ or \ t=4)=P(t=3)+P(t=4)=C^3_5*(\frac{1}{2})^5+C^4_5*(\frac{1}{2})^5=\frac{15}{32}\)

Answer: B.

To elaborate more:

If the probability of a certain event is \(p\), then the probability of it occurring \(k\) times in \(n\)-time sequence is: \(P = C^k_n*p^k*(1-p)^{n-k}\)

For example for the case of getting 3 tails in 5 tries: \(n=5\) (5 tries); \(k=3\) (we want 3 tail); \(p=\frac{1}{2}\) (probability of tail is 1/2).

So, \(P = C^k_n*p^k*(1-p)^{n-k}=C^3_5*(\frac{1}{2})^3*(1-\frac{1}{2})^{(5-3)}=C^3_5*(\frac{1}{2})^5\)

OR: probability of scenario t-t-t-h-h is \((\frac{1}{2})^3*(\frac{1}{2})^2\), but t-t-t-h-h can occur in different ways:

t-t-t-h-h - first three tails and fourth and fifth heads; h-h-t-t-t - first two heads and last three tails; t-h-h-t-t - first tail, then two heads, then two tails; ...

Certain # of combinations. How many combinations are there? Basically we are looking at # of permutations of five letters t-t-t-h-h, which is \(\frac{5!}{3!2!}\).

Re: Kate and David each have $10. Together they flip a coin 5 [#permalink]

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05 Jan 2015, 12:57

Hello from the GMAT Club BumpBot!

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Re: Kate and David each have $10. Together they flip a coin 5 [#permalink]

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06 Jun 2016, 15:34

Hello from the GMAT Club BumpBot!

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Re: Kate and David each have $10. Together they flip a coin 5 [#permalink]

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07 Jun 2016, 09:04

Temurkhon wrote:

First option: 1/2 (win)*1/2 (lose)*1/2 (lose)*1/2 (lose)*1/2 (lose)=1/32.

we have 5!/4!*1!=5 such cases. So, (1/32)*5=5/32

B

Could anyone please shed me some light on this part? Why is it that we have to count only the number of cases we can arrange winnings and not just 5! ? Kindly thanks!
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Please kindly +Kudos if my posts or questions help you!

First option: 1/2 (win)*1/2 (lose)*1/2 (lose)*1/2 (lose)*1/2 (lose)=1/32.

we have 5!/4!*1!=5 such cases. So, (1/32)*5=5/32

B

Could anyone please shed me some light on this part? Why is it that we have to count only the number of cases we can arrange winnings and not just 5! ? Kindly thanks!

We are arranging 1 win and 4 loses (WLLL). WLLL can be arranged in 5!/4! number of ways.
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Re: Kate and David each have $10. Together they flip a coin 5 [#permalink]

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27 Aug 2017, 00:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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