question_answer

A clever student used a biased coin so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find probability distribution and mean of numbers of tails.   

Answer:

Clearly, the probability of getting a tail is \[\frac{1}{4}\] and probability of getting a head is \[\frac{3}{4}.\]             Let X denotes the number of tails in two tosses of a coin, then it can take values 0, 1 and 2.        Clearly, X follows binomial with n = 2, \[p=\frac{1}{4}\] and \[q=\frac{3}{4}\] \[\therefore \] P(X = 0) = P(getting 0 tail) \[{{=}^{2}}{{C}_{0}}{{\left( \frac{3}{4} \right)}^{2}}\] \[[\because P(X=r){{=}^{n}}{{C}_{r}}\,\,{{p}^{r}}\,\,{{q}^{n\,-\,r}}]\] \[=1\times \frac{9}{16}=\frac{9}{16}\] P(X = 1) = P(getting 1 tail)             \[{{=}^{2}}{{C}_{1}}{{\left( \frac{1}{4} \right)}^{1}}{{\left( \frac{3}{4} \right)}^{1}}=2\times \frac{3}{16}=\frac{6}{16}\]      P(X = 2) = P(getting 2 tail)             \[{{=}^{2}}{{C}_{2}}{{\left( \frac{1}{4} \right)}^{2}}{{\left( \frac{3}{4} \right)}^{0}}=1\times \frac{1}{16}=\frac{1}{16}\] \[\therefore \] Required probability distribution is

X	0	1	2
P (X)	\[\frac{9}{16}\]	\[\frac{6}{16}\]	\[\frac{1}{16}\]

Now, mean \[=\Sigma {{x}_{i}}\,P({{x}_{i}})=0\times \frac{9}{16}+1\times \frac{6}{16}+2\times \frac{1}{16}\]             \[=0+\frac{6}{16}+\frac{2}{16}=\frac{8}{16}=\frac{1}{2}\]