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Bunuel
Kate and David each start with $10. They flip a coin 5 times. For each flip that results in heads, Kate gives David $1, and for each flip that results in tails, David gives Kate $1. After flipping the coin 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. \(\frac{11}{16}\)
­10$ < Kate < 15$

If she wins 2 times then Kate will have : 12-3 = 9$
If Kate wins 1 time then , Kate will have : 11-4 = 7$
If she wins 5 times then , Kate will have 15$

So, Kate can get between 10$ to 15$ , if she wins 3 out of 5 times or 4 out of 5 times. 

Probability of getting Head =\( \frac{ 1 }{ (2^5) } = \frac{ 1}{32 } \)

Probability of getting 3 Heads out of 5 flips =
      \( 5 C 3 \) * (\( \frac{ 1 }{ 32 } \) ) = \( \frac{ 10 }{ 32 } \) 

Probability of getting 4 Heads out of 5 flips =  
     \( 5 C 4 \) * ( \( \frac{ 1 }{ 32 } \)) = \( \frac{ 5 }{ 32 } \) 

Combined =\( \frac{ 15}{32 }\)­
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I think this is a high-quality question and I agree with the explanation.
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I like the solution - it’s helpful.
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