MHT CET · Maths · Probability
A fair coin is tossed 2 times. A person receives \(₹ X^{3}\) if he gets \(X\) number of heads.
His expected gain is \(=\)
- A \(₹ 2.00\)
- B \(₹ 1.00\)
- C \(₹ 2.50\)
- D \(₹ 5.20\)
Answer & Solution
Correct Answer
(C) \(₹ 2.50\)
Step-by-step Solution
Detailed explanation
A fair coin is tossed 2 times. Possible outcomes are HH, HT, TH, TT \(\therefore X\) takes values \(0,1,2\)
\(\therefore P(X=0)=\frac{1}{4}, P(X=1)=\frac{2}{4}=\frac{1}{2}, P(X=2)\) \(=\frac{1}{4}\)
Given a person receives \(₹ X^{3}\) if we gets \(X\) no. of heads.
\( \therefore \text {Expected gain }=\left(\frac{1}{4} \times 0\right)+\left(\frac{1}{2} \times 1^{3}\right)~+\) \(\left(\frac{1}{4} \times 2^{3}\right) \)
\(=0+\frac{1}{2}+\frac{8}{4}=2.5\)
\(\therefore P(X=0)=\frac{1}{4}, P(X=1)=\frac{2}{4}=\frac{1}{2}, P(X=2)\) \(=\frac{1}{4}\)
Given a person receives \(₹ X^{3}\) if we gets \(X\) no. of heads.
\( \therefore \text {Expected gain }=\left(\frac{1}{4} \times 0\right)+\left(\frac{1}{2} \times 1^{3}\right)~+\) \(\left(\frac{1}{4} \times 2^{3}\right) \)
\(=0+\frac{1}{2}+\frac{8}{4}=2.5\)
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