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

Question

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

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

chetan2u · Accepted Answer

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

nikhiljd · Answer

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}

Alex75PAris · Answer

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}

Bunuel · Answer

This is a copy of the following question: kate-and-david-each-have-10-together-they-flip-a-coin-97177.html

Dinesh8y · Answer

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

sxsd · Answer

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 -

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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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.

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