Question

Three unbiased cins are tossed. What is the probability of getting atmost two heads?

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

Answer 1

Answer: (D)

$\frac{7}{8}$

All posible outcomes are:

$\{H,H,H\},\{H,H,T\},\{H,T,H\},\{T,H,H\},\{H,T,T\},\{T,H,T\},\{T,T,H\},\{T,T,T\}$

Total number of outcomes$=8$ (a)

Number of outcomes with atmost two heads (all outcomes except one with three heads)$=7$ (b)

Probability of getting atmost two heads$=\frac{\left(b\right)}{\left(a\right)}=\frac{7}{8}$

Hence the correct option is (D).