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Josh and Dan have a 12 apples each. Together they flip a coin 6 times.

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Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 03 Jun 2016, 05:03
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Josh and Dan have a 12 apples each. Together they flip a coin 6 times. For every heads, Josh receives an apple from Dan, and for every tails Dan receives an apple from Josh. After the coin has been flipped 6 times, what is the probability that Josh has more than 12 apples but fewer than 18?

(A) 1/64
(B) 15/64
(C) 21/64
(D) 21/32
(E) 5/6

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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 03 Jun 2016, 05:49
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snorkeler wrote:
Josh and Dan have a 12 apples each. Together they flip a coin 6 times. For every heads, Josh receives an apple from Dan, and for every tails Dan receives an apple from Josh. After the coin has been flipped 6 times, what is the probability that Josh has more than 12 apples but fewer than 18?

(A)1/64
(B)15/64
(C)21/64
(D)21/32
(E)5/6



Good and slightly tough for 600-700 level..

1) proper method
Josh and Dan have 6 apples..
In what circumstances, will Josh have more than 12 - Only when he has more number of heads , BUT all 6 will make the # of apples to be 18 with J..
so # of heads have to be 4 or 5..

1) # of heads 4 = \(6C4* \frac{1}{2}^4*\frac{1}{2}^2 =\frac{6*5}{2}*\frac{1}{2}^6 = \frac{15}{2^6} = \frac{15}{64}.\).

2) # of heads 5 = \(6C5 *\frac{1}{2}^5*\frac{1}{2}^1 = 6*\frac{1}{2}^6 = \frac{6}{2^6} = \frac{6}{64}.\).

Total = \(\frac{15}{64}+\frac{6}{64}=\frac{21}{64}\)

C

(2) approximation-

now we have to have heads as 4 or 5 to ensure that the # of apples with J is >12 and <18...
now # of heads can be any of 6 numbers = 1,2,3,4,5,6.. BUT only 5 and 6 is what we are looking for..

Probability of having heads 4 or 5 times out of 6 is 2/6 = 1/3 ..
Our answer must be CLOSE to 1/3..
ONLY 21 /64 is close to 1/3, which is 21/63

C
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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 04 Jun 2016, 02:51
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minimum number of times to win is 4 to have more than 12 apples .
coz: if he wins 3 times => he gets 3 apples and give 3 apples ; no change
so he needs to win 4 times OR 5 times
Not 6 times because ,
if he wins 6 times => he gets 6 apples and he will have 18 apples
Now number of ways to win 4 times out of 6 times 6C4
Now number of ways to win 5 times out of 6 times 6C5
Total number of ways \(2^6\) ways
Probability =(6C4) + (6C5)/2^6
=\({21}/{64}\)
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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 04 Jun 2016, 03:35
Let's write J = Number of Josh's apples
Let's write D = Number of Dan's apples

For the 6 flips, we can have (H = head, T= Tail) :
HH HH HH -> J + 6 apples ; D - 6 apples
HH HH HT -> J + 4 apples ; D - 4 apples
HH HT HT -> J + 2 apples ; D - 2 apples
HT HT HT -> J + 0 apples ; D - 0 apples
etc...

First, we have to notice that Josh cannot have an odd value of apples.
Given that, Josh can have : 13, 14, 15, 16, or 17 apples = 14 or 16.

We want to calculate the probability that Josh has more than 12 apples but fewer than 18 : P = probability (J14 or J16)
These cases are mutually exclusive, so P = p(J14) + p(J16)

p(J14) = \(\frac{Number.of.outcomes.for.which.Josh.has.14.apples}{Total.number.of.outcomes}\)
Number of outcomes for which Josh has 14 apples = HH HT HT = \(\frac{6!}{4!*2}\) = 15 (mississippi rule)
Total number of outcomes = \(2^6\) = 64
So, p(J14) = \(\frac{15}{64}\)

Similarly, p(J16) = \(\frac{6}{64}\)

Finally, P = p(J14) + p(J16) = \(\frac{15}{64}\) + \(\frac{6}{64}\) = \(\frac{21}{64}\)
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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 05 Jun 2016, 04:33
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snorkeler wrote:
Josh and Dan have a 12 apples each. Together they flip a coin 6 times. For every heads, Josh receives an apple from Dan, and for every tails Dan receives an apple from Josh. After the coin has been flipped 6 times, what is the probability that Josh has more than 12 apples but fewer than 18?

(A) 1/64
(B) 15/64
(C) 21/64
(D) 21/32
(E) 5/6


This is a copy of the following question: kate-and-david-each-have-10-together-they-flip-a-coin-97177.html
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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 01 Feb 2018, 00:27
Only possible if number of heads>number of tails
6 Heads not possible as the number of apples will increase to n18.
Only possibilities are 5 Heads and 1 Tail OR 4 Heads and 2 Tails
6C5/64+6C4/64 = 21/24
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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times.  [#permalink]

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New post 04 Feb 2018, 11:31
snorkeler wrote:
Josh and Dan have a 12 apples each. Together they flip a coin 6 times. For every heads, Josh receives an apple from Dan, and for every tails Dan receives an apple from Josh. After the coin has been flipped 6 times, what is the probability that Josh has more than 12 apples but fewer than 18?

(A) 1/64
(B) 15/64
(C) 21/64
(D) 21/32
(E) 5/6



Let us start with counting the number of apples Josh can have that satisfies the condition that "Josh has more than 12 apples but fewer than 18"
13 - Not possible. Since he gets one apple for every heads and gives one away for every heads. There is no scenario where his "net gain" would be +1 apple if the coin is only being tossed 6 times.
14 - Possible for 4 Heads and 2 Tails OR for 4 Tails and 2 Heads
15 - Not possible. Same as 13.
16 - Possible for 5 Heads and 1 Tail OR 5 Tails and 1 Head
17 - Not possible.

So the 4 cases where this scenario is possible are - [4H2T + 4T2H + 5H1T + 5T1H]

No. of ways to get 4H and 2T in 6 tosses = HHHHTT, HHTTHH, and so on, so to calculate this - 6!/2!4! = 15 and prob. of getting either head or tails = 1/2, so 15*(1/2) = 15/2
No. of ways to get 4T and 2H = 15 x Prob of H or T = 15*(1/2) = 15/2
No. of ways to get 5H and 1T = 6!/5! x 1/2
No. of ways to get 5T and 1 H x Prob of getting head or tails = 6!/5! x 1/2

Adding all 4 we get 21

Total possibilities of H or T with 6 tosses = 2x2x2x2x2x2 = 64

Therefore, 21/64
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Re: Josh and Dan have a 12 apples each. Together they flip a coin 6 times. &nbs [#permalink] 04 Feb 2018, 11:31
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