Multiple choice

If a fair coin is tossed 15 times, what is the probability of getting head as many times in the first ten throws as in the last five?

  1. Hence the required probability=$\displaystyle \frac{3003}{32768.}$
  2. Hence the required probability=$\displaystyle \frac{1001}{32768.}$
  3. Hence the required probability=$\displaystyle \frac{3003}{65536.}$
  4. Hence the required probability=$\displaystyle \frac{1001}{65536.}$
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A Correct answer
Explanation

Let H1 be heads in first 10 tosses and H2 be heads in last 5. We want P(H1 = H2). This occurs if H1=0, H2=0; H1=1, H2=1; H1=2, H2=2; H1=3, H2=3; H1=4, H2=4; H1=5, H2=5. Summing the probabilities using combinations (10Ck * 5Ck) / 2^15 gives 3003 / 32768.

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