question_answer

A coin is biased such that a head is three times as likely to occur as a tail. When it tossed twice, then find the probability distribution of number of heads. Also, find the mean and variance.

Answer:

Let probability of getting a tail = p Then, probability of getting a head = 3p [\[\because \]head is three times as likely to occur than a tail]  \[\therefore \]     \[3p+p=1\] [total probability of all outcomes of an experiment is 1] \[\Rightarrow \]   \[4p=1\]\[\Rightarrow \] \[p=\frac{1}{4}\] \[\Rightarrow \]   P(getting a tail) \[=p=\frac{1}{4}\] and P(getting a head) \[=3p=\frac{3}{4}\] Now, let X = number of heads when the coin is tossed twice. So, X can take values 0, 1 and 2. Now, \[P\,(X=0)=P\] (no head occurs) \[=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}\] \[P(X=0)=P\] (one head occurs)             \[=\left( \frac{1}{4}\times \frac{3}{4} \right)+\left( \frac{3}{4}\times \frac{1}{4} \right)=\frac{1}{16}\] \[P(X=1)=P\] (two heads occur) \[=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}\] \[\therefore \] The required probability distribution is given below

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

Table for calculation of mean and variance is given below

\[{{\mathbf{x}}_{\mathbf{i}}}\]	\[{{\mathbf{p}}_{\mathbf{i}}}\]	\[{{\mathbf{x}}_{\mathbf{i}}}{{\mathbf{p}}_{\mathbf{i}}}\]	\[\mathbf{x}_{\mathbf{i}}^{\mathbf{2}}{{\mathbf{p}}_{\mathbf{i}}}\]
0	1/16	0	0
1	6/16	6/16	6/16
2	9/16	18/16	36/16
Total	\[\sum{{{p}_{i}}=1}\]	\[\sum{{{p}_{i}}{{x}_{i}}=\frac{24}{16}}\]	\[\sum{{{p}_{i}}x_{i}^{2}=\frac{42}{16}}\]

Now, mean \[=\sum\limits_{i\,\,=\,\,1}^{3}{{{x}_{i}}{{p}_{i}}=\frac{24}{16}=1.5}\] and variance \[=\sum\limits_{i\,\,=\,\,1}^{3}{x_{i}^{2}{{p}_{i}}}-{{\left( \sum\limits_{i\,\,=\,\,1}^{3}{{{x}_{i}}{{p}_{i}}} \right)}^{2}}\] \[=\frac{42}{16}-{{(1.5)}^{2}}=2.625-2.25=0.375\]Hence, the mean and variance are 1.5 and 0.375 respectively.