An unbiased coin is tossed 4 times. Find the probability distribution…

The correct answer is Option (1) → Mean: 2, Variance: 1

When a fair coin is tossed, probability of getting a head = $p=\frac{1}{2}$

So, $q = 1 − p = 1-\frac{1}{2}=\frac{1}{2}$

Here number of trials = number of times the coin is tossed = 4.

Thus, we get a binomial distribution with $p=\frac{1}{2},q=\frac{1}{2}$ and $n=4$.

∴ P(r successes) = ${^4C}_r p^rq^{4- r} = {^4C}_r. (\frac{1}{2})^r(\frac{1}{2})^{4-r}={^4C}_r(\frac{1}{2})^4$.

Let X denote the number of heads obtained, then X can take values 0, 1, 2, 3, 4.

$P(0)= {^4C}_0 (\frac{1}{2})^4 =\frac{1}{16}$, $P(1)= {^4C}_1 (\frac{1}{2})^4 =\frac{}{16}$

$P(2)= {^4C}_2 (\frac{1}{2})^4 =\frac{6}{16}$, $P(3)= {^4C}_3 (\frac{1}{2})^4 =\frac{4}{16}$ and

$P(4)= {^4C}_4 (\frac{1}{2})^4 =\frac{1}{16}$

∴ Probability distribution is $\begin{pmatrix}0&1&2&3&4\\\frac{1}{16}&\frac{4}{16}&\frac{6}{16}&\frac{4}{16}&\frac{1}{16}\end{pmatrix}$

Mean of number of heads obtained = $Σp_ix_i$

$=\frac{1}{16}×0+\frac{4}{16}×1+\frac{6}{16}×2+\frac{4}{16}×3+\frac{1}{16}×4$

$=\frac{0+4+12+12+4}{16}=\frac{32}{16}=2$

Now $Σp_i{x_i}^2=\frac{1}{16}×0^2+\frac{4}{16}×1^2+\frac{6}{16}×2^2+\frac{4}{16}×3^2+\frac{1}{16}×4^2$

$=\frac{0+4+24+36+16}{16}=\frac{80}{16}=5$

∴ Variance = $Σp_i{x_i}^2-μ^2 = 5-2^2 = 1$.