MHT CET · Maths · Probability
In a game, 3 coins are tossed. A person is paid ₹ \(100\), if he gets all heads or all tails; and he is supposed to pay ₹ 40 , if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is
- A 10 loss
- B 5 loss
- C 5 gain
- D 10 gain
Answer & Solution
Correct Answer
(B) 5 loss
Step-by-step Solution
Detailed explanation
In a game, 3 coins are tossed.
\(\mathrm{P}(\) getting all heads or all tails \()=\frac{2}{8}=\frac{1}{4}\)
\(\mathrm{P}(\) getting one head or two heads \()=\frac{6}{8}=\frac{3}{4}\)
Let \(x\) : number of rupees the person gets.
\(\begin{aligned}
& P(X=100)=\frac{1}{4} \\
& P(X=-40)=\frac{3}{4}
\end{aligned}\)
\(\therefore \quad\) The amount he can expect to win \(=\) mean
\(\begin{aligned}
& =\sum \mathrm{n}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} \\
& =100 \times \frac{1}{4}+(-40) \times \frac{3}{4} \\
& =\frac{100}{4}-\frac{120}{4} \\
& =-5 \\
& =5 \mathrm{loss}
\end{aligned}\)
\(\mathrm{P}(\) getting all heads or all tails \()=\frac{2}{8}=\frac{1}{4}\)
\(\mathrm{P}(\) getting one head or two heads \()=\frac{6}{8}=\frac{3}{4}\)
Let \(x\) : number of rupees the person gets.
\(\begin{aligned}
& P(X=100)=\frac{1}{4} \\
& P(X=-40)=\frac{3}{4}
\end{aligned}\)
\(\therefore \quad\) The amount he can expect to win \(=\) mean
\(\begin{aligned}
& =\sum \mathrm{n}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} \\
& =100 \times \frac{1}{4}+(-40) \times \frac{3}{4} \\
& =\frac{100}{4}-\frac{120}{4} \\
& =-5 \\
& =5 \mathrm{loss}
\end{aligned}\)
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