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Kate and David each have $10. Together they flip a coin 5 [#permalink]
 06 Mar 2008, 07:33
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
* \frac{5}{16}
* \frac{15}{32}
* \frac{1}{2}
* \frac{21}{32}
* \frac{11}{16}
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Re: Maths: Probability [#permalink]
06 Mar 2008, 08:11
This means Kate should win atleast once and atmost 4 times. There is only one way to win all the tosses, and only 1 way to loose all the matches.
Total Outcome = 2^5 = 32
Probable Outcome = 32 - 1 -1 = 30
Probability = 30/32 = 15/16
But I don't see this option in Answers.
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Re: Maths: Probability [#permalink]
06 Mar 2008, 08:11
7-p409567?t=57304&hilit=kate+coin#p409567
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Re: Maths: Probability [#permalink]
06 Mar 2008, 08:57
Kate has to get 3 tails or 4 tails to get the dollars in range 10-15 not inclusive.
Total cases = 2^5 =32
3 tails 2heads, ways = 10
4 tails 1head = 5
hence prob = 15/32
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Re: Maths: Probability [#permalink]
06 Mar 2008, 09:51
i used binomial theorem to get 15/32.
First, figure out cases that result in Kate have btwn 11 and 14 dollars. The only cases that work are her winning three times, or winning four times.
for the first case, I have: (5C3)*(1/2)^3*(1/2)^2 = 10/32
for the second case, I have : (5C4)*(1/2)^4*(1/2)^1 = 5/32
total prob is 15/32
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Re: Maths: Probability [#permalink]
06 Mar 2008, 10:25
I got 5/16
to go from $10 to $11 Kate has to get 1 tails (1/2)
to go from $10 to $12 Kate has to get 2 tails (1/2 and 1/2)
to go from $10 to $13 Kate has to get 3 tails (1/2; 1/2; 1/2)
to go from $10 to $14 Kate has to get 4 tails (1/2; 1/2; 1/2; 1/2)
Total number of times = 1+2+3+4 = 10
10/32 = 5/16...
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Re: Maths: Probability [#permalink]
06 Mar 2008, 21:32
vshaunak@gmail.com wrote: Kate has to get 3 tails or 4 tails to get the dollars in range 10-15 not inclusive.
Total cases = 2^5 =32
3 tails 2heads, ways = 10
4 tails 1head = 5
hence prob = 15/32 Shaunak are you including the scenarios where Kate can have 12, 13 dollars in the above calculation?
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Re: Maths: Probability [#permalink]
07 Mar 2008, 12:36
suntaurian 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
* \frac{5}{16}
* \frac{15}{32}
* \frac{1}{2}
* \frac{21}{32}
* \frac{11}{16} actually my answer is different. at a glance probability must be >1/2. it is more probable that kate win at least one match or that she lose one than otherwise. moreover, given that 2^5 is the total number of cases, we may have 1,2,3 or 4 wins for Kate out of 4 possible cases in which she can win. we use permutations since order matters, thus 4P1+4P2+4P3+4P4=21. Denominator is 32. 21/32 is my choice
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Re: Maths: Probability [#permalink]
09 Mar 2008, 08:59
This can be solved using binomial probability.... Results of 5 coin tosses: 5 Tails -> Kate = $15 4T, 1H -> Kate = $13 3T, 2H -> Kate = $11 2T, 3H -> Kate = $9 1T, 4H -> Kate = $7 5H -> Kate = $5 We need to find out the probability for 4T,1H & 3T,2H as total money has to be >$10 and <$15 C_5^4*(\frac{1}{2})^5 + C_5^3*(\frac{1}{2})^5 --> \frac{15}{32}
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Re: Maths: Probability [#permalink]
09 Mar 2008, 13:59
Thanks Neelesh. Now I get it.
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Re: Maths: Probability
[#permalink]
09 Mar 2008, 13:59
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