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13 Jul 2010, 09:43
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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. $$\frac{11}{16}$$ [Reveal] Spoiler: OA Last edited by Bunuel on 13 Apr 2012, 13:52, edited 2 times in total. Edited the question and added the OA Kudos [?]: 235 [0], given: 1 Math Expert Joined: 02 Sep 2009 Posts: 41892 Kudos [?]: 129150 [0], given: 12194 Re: probability [#permalink] ### Show Tags 13 Jul 2010, 11:04 Expert's post 3 This post was BOOKMARKED bibha wrote: 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?".

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

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 $$\frac{5!}{3!2!}$$.

Hence $$P=\frac{5!}{3!2!}*(\frac{1}{2})^5$$.

Check this links for similar problems:
viewtopic.php?f=140&t=56812&hilit=+probability+occurring+times
viewtopic.php?f=140&t=88069&hilit=+probability+occurring+times
viewtopic.php?f=140&t=87673&hilit=+probability+occurring+times

Also you can check Probability chapter of Math Book for more (link in my signature).

Hope it helps.
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04 Jan 2011, 21:32
Thanks for the explanation

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Re: coin probability [#permalink]

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09 Mar 2011, 22:45

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Re: coin probability [#permalink]

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10 Mar 2011, 00:49
Harriet has won, or that 3 <= # of head <=4 because only in those cases Harriet will have Money > 10 and < 15

So Prpb = 5C3(1/2)^3* (1/2)^2 + 5C4(1/2)^4* (1/2)^1

= (1/2)^5{(5 * 4)/2 + 5} = 15 * (1/2)^5 = 15/32

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10 Jan 2015, 00:28
Two options for Kate to get >10 and <15 dollars

Lose (times): 1, 2

Win (times): 4, 3

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

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

we have 5!/3!*2!=10 such cases. So, (1/32)*10=10/32

5/32+10/32=15/32

B

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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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Re: Kate and David each have $10. Together they flip a coin 5 [#permalink] 27 Aug 2017, 00:28 Display posts from previous: Sort by # Kate and David each have$10. Together they flip a coin 5

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