The mean number of heads in two tosses of a coin is :
Answer & explanation
Correct answer: option 3
In two tosses of a coin
No. of heads = 0, 1, 2
|
X = |
0 |
1 |
2 |
|
P(x) = |
${}^{2}C_{0}\frac{1}{2^{2}}$ |
${}^{2}C_{1}\frac{1}{2^{2}}$ |
${}^{2}C_{2}\frac{1}{2^{2}}$ |
So mean = $\sum x_i P\left(x_i\right)$
$=0 \times{ }^2 C_0 \times \frac{1}{2^2}+2 C_1 \times \frac{1}{2^2}+2 \times{ }^2 C_2 \times \frac{1}{2^2}$
$=\frac{2}{4}+\frac{2}{4}=1$