Question

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{3}{8}$
• D. $\frac{7}{8}$

Answer 1

The correct option is D $\frac{7}{8}$

Total no of probabilities= 2^3 =8
for 2/1/0 heads probability is =7(only 1 left HHH)
So we cannot include HHH, because it contains 3 heads.

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}.

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{8}$
Answer D $\frac{7}{8}$