A fair coin is tossed three times. Find the probability of getting two heads and one tail.
\(\frac{1}{2}\)
\(\frac{3}{8}\)
\(\frac{1}{4}\)
\(\frac{1}{8}\)
Correct answer is B
Pr(head) = \(\frac{1}{2}\), Pr(tail) = \(\frac{1}{2}\):Pr(2 heads)
= \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\)
Pr(2 heads and tail) 3 times
= (\(\frac{1}{4}\) x \(\frac{1}{2}\)) x 3
= \(\frac{3}{8}\)