here are the outcomes that can happen:

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

of the eight outcomes there are 3 that have exactly two tails so the probability is 3/8

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another way. This method will be more general and useful

Let X be the number of tails in three tosses of a fair coin. X has the binomial distribution with n = 3 trials and success probability 1/2.

In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, …, n

P[X = x] = 0 for any other value of x.

P( X = 2 ) = 3! / (2! (3-2)!) * (1/2)^2 * (1-1/2)^(3-2)

= 3 * 1/2 * 1/2 * 1/2 = 3/8

another example, say the coin was not fair, say the probability of tails was 0.35. Using the binomial distributional you will find

P( X = 2 ) = 0.238875

Second, you haven’t told us where you’re stuck. Simply giving you the answer is cheating.

I suggest one of these things:

(1) Use the binomial expression, p=1/2, n=3, m=2, and calculate the answer.

(2) Go to the third row of Pascal’s triangle, look up the 2nd entry, and divide by 2^3.

(3) Explain where you’re stuck, and we’ll help you over the hard part.