KCET · Maths · Probability
Find the mean number of heads in three tosses of a fair coin.
- A 1.5
- B 4.5
- C 2.5
- D 3.5
Answer & Solution
Correct Answer
(A) 1.5
Step-by-step Solution
Detailed explanation
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
\( \begin{aligned} \text { Required mean } & =\Sigma X_i P_i \\ & =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right) \\ & =0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=1.5 \end{aligned} \)
\( \begin{aligned} \text { Required mean } & =\Sigma X_i P_i \\ & =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right) \\ & =0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=1.5 \end{aligned} \)
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