MHT CET · Maths · Probability
If three fair coins are tossed, then variance of number of heads obtained, is
- A 0.25
- B 3
- C 0.75
- D 1.5
Answer & Solution
Correct Answer
(C) 0.75
Step-by-step Solution
Detailed explanation
Let r.v X denotes the no. of heads obtained \(\therefore \quad\) Probability distribution of X is given as
\(\mathrm{E}(x) =0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}=\frac{3}{2} \)
\( \mathrm{E}\left(x^2\right) =0 \times \frac{1}{8}+1 \times \frac{3}{8}+4 \times \frac{3}{8}+9 \times \frac{1}{8}=3 \)
\( \therefore \mathrm{~V}(x) =\mathrm{E}\left(x^2\right)-[\mathrm{E}(x)]^2 \)
\( =3-\frac{9}{4} \)
\( =\frac{3}{4} \)
\( =0.75\)
\(\mathrm{E}(x) =0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}=\frac{3}{2} \)
\( \mathrm{E}\left(x^2\right) =0 \times \frac{1}{8}+1 \times \frac{3}{8}+4 \times \frac{3}{8}+9 \times \frac{1}{8}=3 \)
\( \therefore \mathrm{~V}(x) =\mathrm{E}\left(x^2\right)-[\mathrm{E}(x)]^2 \)
\( =3-\frac{9}{4} \)
\( =\frac{3}{4} \)
\( =0.75\)
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