Solution (a) Let S be the number of crash-free days,which is a binomial random variable with n=50 and p =0.95, so that E[S]=50·0.95=47.5 ·And 0s=V50.0.95.0.05=1.54Solution (a) • Let 𝑆 be the number of crash-free days, which is a binomial random variable with 𝑛 = 50 and 𝑝 = 0.95, • so that 𝐄[𝑆] = 50 · 0.95 = 47.5 • And 𝜎𝑠 = 50 · 0.95 · 0.05 = 1.54