Three unbiased coins are tossed. Then, the probability of getting at most two heads is
Three unbiased coins are tossed. Then, the probability of getting at most two heads is
$3 / 4$
$1 / 4$
$3 / 8$
$7 / 8$
Solution
3 unbiased coins are tossed.
Sample space,
$S=\{$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT $\}$
$n(S)=8$
Let $E$ be the event of getting at most two heads.
$\Rightarrow E^{\prime}=\{\mathrm{HHH}\}$
$P(E)=1-P\left(E^{\prime}\right)=1-\frac{n\left(E^{\prime}\right)}{n(S)}=1-\frac{1}{8}=\frac{7}{8}$