Question

A coin is tossed twice. Events E and F are defined as follows :E=heads on first toss, F= heads on second toss.Find the probability of $E\cup F.$

• A. Hence p $(E\cup F)=\frac{1}{4}$
• B. Hence p $(E\cup F)=\frac{3}{4}$
• C. Hence p $(E\cup F)=\frac{1}{2}$
• D. none of these

Answer 1

Answer: (B)

Hence p $(E\cup F)=\frac{3}{4}$

Let H denotes head and T denotes tail .
Sample space $S=\left\{\right(H,H),(H,T),(T,H),(T,T\left)\right\}$
Total number of sample points $n\left(S\right)=4$
Now, $E=\left\{\right(H,H),(H,T\left)\right\}$
and $F=\left\{\right(H,H),(T,H\left)\right\}$
$E\cup F=\left\{\right(H,H),(H,T),(T,H\left)\right\}$
So $n(E\cup F)=3$
Hence $P(E\cup F)=\frac{n(E\cup F)}{n\left(S\right)}=\frac{3}{4}$