Practicing Success
The mean number of heads in two tosses of a coin is : |
2 $\frac{1}{2}$ 1 $\frac{3}{2}$ |
1 |
The correct answer is Option (3) → 1 X → no of heads in two tosses $X=0⇒P(X=0)={^2C}_0(\frac{1}{2})^0(\frac{1}{2})^2=\frac{1}{4}$ $X=1⇒P(X=1)={^2C}_1(\frac{1}{2})(\frac{1}{2})=\frac{2}{4}=\frac{1}{2}$ $X=2⇒P(X=2)={^2C}_2(\frac{1}{2})^2(\frac{1}{2})^0=\frac{1}{4}$ Mean = $∑P(X_i)=1$ |