If three unbiased coins are tossed, find the probability that they are all heads
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{9}\)
\(\frac{1}{8}\)
Correct answer is D
P(H) = \(\frac{1}{2}\) and P(T) = \(\frac{1}{2}\)
Using the binomial prob. distribution,
(H + T)3 = H3 + 3H2T1 + 3HT2 + T3
Hence the probability that three heads show in a toss of the three coins is H3
= (\(\frac{1}{2}\))3
= \(\frac{1}{8}\)