MHT CET · Maths · Probability
Three fair coins with faces numbered 1 and 0 are tossed simultaneously. Then variance \((\mathrm{X})\) of the probability distribution of random variable \(\mathrm{X}\), where \(\mathrm{X}\) is the sum of numbers on the upper most faces, is
- A \(0.7\)
- B \(0.75\)
- C \(0.65\)
- D \(0.6\)
Answer & Solution
Correct Answer
(B) \(0.75\)
Step-by-step Solution
Detailed explanation
Possible value of \(\mathrm{X}\) are \(0,1,2,3\) Here,
\(\mathrm{S}=\{000,001,010,1,00,111,110,101,011\}\)
\(\mathrm{n}(\mathrm{S})=8\)
\(\therefore \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8} \)
\( \mathrm{E}\left(\mathrm{X}^2\right)= \)
\( \therefore \sum x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}^2=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8}=\frac{24}{8} \)
\(=3-\left(\frac{3}{2}\right)^2 \)
\( =3-\frac{9}{4} \)
\( =\frac{3}{4} \)
\( =0.75\)
\(\mathrm{S}=\{000,001,010,1,00,111,110,101,011\}\)
\(\mathrm{n}(\mathrm{S})=8\)
\(\therefore \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}=0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8} \)
\( \mathrm{E}\left(\mathrm{X}^2\right)= \)
\( \therefore \sum x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}^2=0+\frac{3}{8}+\frac{12}{8}+\frac{9}{8}=\frac{24}{8} \)
\(=3-\left(\frac{3}{2}\right)^2 \)
\( =3-\frac{9}{4} \)
\( =\frac{3}{4} \)
\( =0.75\)
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