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
Table for calculation of mean and variance is given below X 0 1 2 P(X) \[\frac{1}{16}\] \[\frac{6}{16}\] \[\frac{9}{16}\]
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. \[{{\mathbf{x}}_{\mathbf{i}}}\] \[{{\mathbf{p}}_{\mathbf{i}}}\] \[{{\mathbf{x}}_{\mathbf{i}}}{{\mathbf{p}}_{\mathbf{i}}}\] \[\mathbf{x}_{\mathbf{i}}^{\mathbf{2}}{{\mathbf{p}}_{\mathbf{i}}}\] 0 1/16 0 0 1 6/16 6/16 6/16 2 9/16 18/16 36/16 Total \[\sum{{{p}_{i}}=1}\] \[\sum{{{p}_{i}}{{x}_{i}}=\frac{24}{16}}\] \[\sum{{{p}_{i}}x_{i}^{2}=\frac{42}{16}}\]
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