| Given, a coin is tossed thrice, so its sample space is |
| S = {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} |
| Let X be a random variable that denotes the number of tails. Then, X can take values 0, 1, 2 and 3. |
| Now, P(X = 0) = P(no tail occurs) \[=\frac{1}{8}\] |
| P(X = 1) = P (only one tail occurs) \[=\frac{3}{8}\] |
| P(X = 2) = P (exactly two tails occur) \[=\frac{3}{8}\] |
| and P (X = 3) = P (three tails occur) \[=\frac{1}{8}\] |
| |
| \[\therefore \] The probability distribution of number of tails is as follows |
| |
| X | 0 | 1 | 2 | 3 | | P(X) | \[\frac{1}{8}\] | \[\frac{3}{8}\] | \[\frac{3}{8}\] | \[\frac{1}{8}\] | |
| Now, we find mean and variance using the following table: |
| |
| X | P(X) | X. P(X) | \[{{\mathbf{X}}^{\mathbf{2}}}\cdot \mathbf{P(X)}\] | | 0 | \[\frac{1}{8}\] | 0 | 0 | | 1 | \[\frac{3}{8}\] | \[\frac{3}{8}\] | \[\frac{3}{8}\] | | 2 | \[\frac{3}{8}\] | \[\frac{6}{8}\] | \[\frac{12}{8}\] | | 3 | \[\frac{1}{8}\] | \[\frac{3}{8}\] | \[\frac{9}{8}\] | | | | \[\Sigma \,X\cdot P(X)=\frac{12}{8}=\frac{3}{2}\] | \[\Sigma \,{{X}^{2}}\cdot P(X)=\frac{24}{8}=3\] | |
| Clearly, mean \[\Sigma \,X\cdot P(X)=\frac{3}{2}\] |
| and variance \[\Sigma \,{{X}^{2}}\cdot P(X)-{{[\Sigma \,X\cdot P(X)]}^{2}}\] |
| \[=3-{{\left( \frac{3}{2} \right)}^{2}}=3-\frac{9}{4}=\frac{12-9}{4}=\frac{3}{4}\] |
| OR |
| Let us define the following events |
| A: Number on the drawn card is more than 3 and B: Number on the drawn card is an even number. |
| Then, A = {4, 5, 6, 7, 8, 9, 10, 11, 12} |
| and B = {2, 4, 6, 8, 10, 12} |
| \[\Rightarrow \] \[A\cap B=\left\{ 4,\,\,6,\,\,8,\,\,10,\,\,12 \right\}~\] |
| Now, \[P(A)=\frac{n(A)}{n(S)}=\frac{9}{12}=\frac{3}{4}\] |
| \[P(B)=\frac{n(B)}{n(S)}=\frac{6}{12}=\frac{1}{2}\] |
| and \[P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{5}{12}\] |
| Hence, required probability |
| \[=P(B\,/A)=\frac{n(B\,\cap A)}{n(A)}=\frac{\frac{5}{12}}{\frac{3}{4}}=\frac{5}{12}\times \frac{4}{3}=\frac{5}{9}\] |
