Out of 6 tosses, exactly 3 heads implies exactly 3 tails.
You can get 3 heads and 3 tails in various ways: HHHTTT, HHTHTT, TTTHHH etc
How many such ways are there? We just need to find the number of ways of arranging 3 Hs and 3 Ts in a row. This can be done in 6!/3!*3! = 20 ways (the 3 Hs are identical and the 3 Ts are identical)

Probability of obtaining HHHTTT = (0.5)*(0.5)*(0.5)*(0.5)*(0.5)*(0.5)

Probability of obtaining 3Hs and 3 Ts in any order = (0.5)^6 * 20 = 5/16

Check out this post for a discussion on binomial probability concept:
http://www.veritasprep.com/blog/2012/02 ... obability/
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When a certain coin is flipped, the probability of heads is 0.5. If the coin is flipped 6 times, what is the probability that there are exactly 3 heads?

A. 1/4
B. 1/3
C. 5/16
D. 31/64
E. 1/2

We need the proability of HHHTTT (any combination of this): \(P=\frac{6!}{3!3!}*(\frac{1}{2})^6=\frac{20}{64}=\frac{5}{16}\). We are multiplying by \(\frac{6!}{3!3!}\) since HHHTTT scenario can occur in several ways: HHHTT, THHHTT, TTHHHT, ... basically the number of permutations of 6 letters HHHTTT, out of which 3 H's and T's are identical (\(\frac{6!}{3!3!}\));

Answer: C.

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Probability chapter of out Math Book:
math-probability-87244.html

Hope it helps.
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