Practicing Success
The mean number of heads in three tosses of a fair coin is : |
$\frac{1}{2}$ 1 $\frac{3}{2}$ $\frac{1}{4}$ |
$\frac{3}{2}$ |
Let X be no. of heads in three tosses of a fair coin Using binomial probability distribution For X = 0 we get P(X = 0) = ${ }^3 C_0\left(\frac{1}{2}\right)^3$ For X = 1 we get P(X = 1) = ${ }^3 C_1\left(\frac{1}{2}\right)^3$ For X = 2 we get P(X = 2) = ${ }^3 C_2\left(\frac{1}{2}\right)^3$ For X = 3 we get P(X = 3) = ${}^3 C_3\left(\frac{1}{2}\right)^3$ So
for Mean y = $\sum P(X) X$ $=0 \times { }^3 C_0\left(\frac{1}{2}\right)^3+{ }^3 C_1\left(\frac{1}{2}\right)^3 \times 1+ { }^3 C_2\left(\frac{1}{2}\right)^3 \times 2+{}^3 C_3\left(\frac{1}{2}\right)^3 \times 3$ $=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2}$ |