Three unbiased coins are tossed. What is the probability of getting at most two heads?
$ \dfrac{3}{4} $
$ \dfrac{1}{4} $
$ \dfrac{3}{8} $
$ \dfrac{7}{8} $
Explanation:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
$\therefore P\left(E\right) $ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{7}{8} $.