Question

If three coins are tossed simultaneously, then the probability of getting at least two heads, is

• A. $\frac{1}{4}$
• B. $\frac{3}{8}$
• C. $\frac{1}{2}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $\frac{1}{2}$

Possible cases $=\left\{\right(H,H,H),(H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H),(T,T,T\left)\right\}$
So, total number of possible cases $=8$
Cases with at least two heads: $=\left\{\right(H,H,H),(H,H,T),(H,T,H),(T,H,H\left)\right\}$

Total number of favourable cases $=4$
$P($ Atleast two head$)=$$\frac{4}{8}=\frac{1}{2}$