Q61 / 180 · KCET 2022 · Mathematics
Find the mean number of heads in three tosses of a fair coin.
- Correct answerA1.5
- B4.5
- C2.5
- D3.5
Solution
Given three coins are tossed.
Therefore, sample space,
$S=\{T T T, T T H, T H T, H T T, H H T$,
HTH, THH, HHHH
$\therefore n(S)=8$
Let $X$ represents 'number of heads'
$\because \quad X=0,1,2$ or 3 .
Probability distribution of $X$ is
$$ \begin{aligned} \text { Required mean } & =\Sigma X_i P_i \\ & =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right) \\ & =0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=1.5 \end{aligned} $$
$$ \begin{aligned} \text { Required mean } & =\Sigma X_i P_i \\ & =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right) \\ & =0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=1.5 \end{aligned} $$
Hence, the correct answer is option: 'A'.