\displaystyle \textbf{Question 1: }~\text{Find the mean and standard deviation of each of the following} \\ \text{probability distributions:}
\displaystyle \text{(i)}~\begin{array}{c|ccc} x_i & 2 & 3 & 4\\ \hline p_i & 0.2 & 0.5 & 0.3 \end{array}
\displaystyle \text{(ii)}~\begin{array}{c|cccc} x_i & 1 & 3 & 4 & 5\\ \hline p_i & 0.4 & 0.1 & 0.2 & 0.3 \end{array}
\displaystyle \text{(iii)}~\begin{array}{c|cccc} x_i & -5 & -4 & 1 & 2\\ \hline p_i & \frac{1}{4} & \frac{1}{8} & \frac{1}{2} & \frac{1}{8} \end{array}
\displaystyle \text{(iv)}~\begin{array}{c|ccccc} x_i & -1 & 0 & 1 & 2 & 3\\ \hline p_i & 0.3 & 0.1 & 0.1 & 0.3 & 0.2 \end{array}
\displaystyle \text{(v)}~\begin{array}{c|cccc} x_i & 1 & 2 & 3 & 4\\ \hline p_i & 0.4 & 0.3 & 0.2 & 0.1 \end{array}
\displaystyle \text{(vi)}~\begin{array}{c|cccc} x_i & 0 & 1 & 3 & 5\\ \hline p_i & 0.2 & 0.5 & 0.2 & 0.1 \end{array}
\displaystyle \text{(vii)}~\begin{array}{c|ccccc} x_i & -2 & -1 & 0 & 1 & 2\\ \hline p_i & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \end{array}
\displaystyle \text{(viii)}~\begin{array}{c|ccccc} x_i & -3 & -1 & 0 & 1 & 3\\ \hline p_i & 0.05 & 0.45 & 0.20 & 0.25 & 0.05 \end{array}
\displaystyle \text{(ix)}~\begin{array}{c|cccccc} x_i & 0 & 1 & 2 & 3 & 4 & 5\\ \hline p_i & \frac{1}{6} & \frac{5}{18} & \frac{2}{9} & \frac{1}{6} & \frac{1}{9} & \frac{1}{18} \end{array} \displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 2 & 0.2 & 0.4 & 0.8\\ 3 & 0.5 & 1.5 & 4.5\\ 4 & 0.3 & 1.2 & 4.8\\ \hline & & \sum p_i x_i = 3.1 & \sum p_i x_i^2 = 10.1 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 3.1
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 10.1 - (3.1)^2
\displaystyle = 10.1 - 9.61
\displaystyle = 0.49
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{0.49}
\displaystyle = 0.7

\displaystyle \textbf{(ii)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 1 & 0.4 & 0.4 & 0.4\\ 3 & 0.1 & 0.3 & 0.9\\ 4 & 0.2 & 0.8 & 3.2\\ 5 & 0.3 & 1.5 & 7.5\\ \hline & & \sum p_i x_i = 3 & \sum p_i x_i^2 = 12 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 3
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 12 - 3^2
\displaystyle = 12 - 9
\displaystyle = 3
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{3}
\displaystyle \approx 1.732

\displaystyle \textbf{(iii)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline -5 & \frac{1}{4} & -\frac{5}{4} & \frac{25}{4}\\ -4 & \frac{1}{8} & -\frac{4}{8} & \frac{16}{8}\\ 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\ 2 & \frac{1}{8} & \frac{2}{8} & \frac{4}{8}\\ \hline & & \sum p_i x_i = -1 & \sum p_i x_i^2 = \frac{74}{8} \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = -1
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = \frac{74}{8} - (-1)^2
\displaystyle = \frac{74}{8} - 1
\displaystyle = \frac{66}{8}
\displaystyle = 8.25
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{8.25}
\displaystyle \approx 2.872

\displaystyle \textbf{(iv)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline -1 & 0.3 & -0.3 & 0.3\\ 0 & 0.1 & 0 & 0\\ 1 & 0.1 & 0.1 & 0.1\\ 2 & 0.3 & 0.6 & 1.2\\ 3 & 0.2 & 0.6 & 1.8\\ \hline & & \sum p_i x_i = 1 & \sum p_i x_i^2 = 3.4 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 1
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 3.4 - 1
\displaystyle = 2.4
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{2.4}
\displaystyle \approx 1.549

\displaystyle \textbf{(v)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 1 & 0.4 & 0.4 & 0.4\\ 2 & 0.3 & 0.6 & 1.2\\ 3 & 0.2 & 0.6 & 1.8\\ 4 & 0.1 & 0.4 & 1.6\\ \hline & & \sum p_i x_i = 2 & \sum p_i x_i^2 = 5 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 2
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 5 - 2^2
\displaystyle = 5 - 4
\displaystyle = 1
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{1}
\displaystyle = 1

\displaystyle \textbf{(vi)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 0 & 0.2 & 0 & 0\\ 1 & 0.5 & 0.5 & 0.5\\ 3 & 0.2 & 0.6 & 1.8\\ 5 & 0.1 & 0.5 & 2.5\\ \hline & & \sum p_i x_i = 1.6 & \sum p_i x_i^2 = 4.8 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 1.6
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 4.8 - 1.6^2
\displaystyle = 4.8 - 2.56
\displaystyle = 2.24
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{2.24}
\displaystyle \approx 1.497

\displaystyle \textbf{(vii)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline -2 & 0.1 & -0.2 & 0.4\\ -1 & 0.2 & -0.2 & 0.2\\ 0 & 0.4 & 0 & 0\\ 1 & 0.2 & 0.2 & 0.2\\ 2 & 0.1 & 0.2 & 0.4\\ \hline & & \sum p_i x_i = 0 & \sum p_i x_i^2 = 1.2 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 0
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 1.2 - 0^2
\displaystyle = 1.2
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{1.2}
\displaystyle \approx 1.095

\displaystyle \textbf{(viii)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline -3 & 0.05 & -0.15 & 0.45\\ -1 & 0.45 & -0.45 & 0.45\\ 0 & 0.20 & 0 & 0\\ 1 & 0.25 & 0.25 & 0.25\\ 3 & 0.05 & 0.15 & 0.45\\ \hline & & \sum p_i x_i = -0.2 & \sum p_i x_i^2 = 1.6 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = -0.2
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 1.6 - (-0.2)^2
\displaystyle = 1.6 - 0.04
\displaystyle = 1.56
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{1.56}
\displaystyle \approx 1.249

\displaystyle \textbf{(ix)  }
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 0 & \frac{1}{6} & 0 & 0\\ 1 & \frac{5}{18} & \frac{5}{18} & \frac{5}{18}\\ 2 & \frac{2}{9} & \frac{4}{9} & \frac{8}{9}\\ 3 & \frac{1}{6} & \frac{1}{2} & \frac{3}{2}\\ 4 & \frac{1}{9} & \frac{4}{9} & \frac{16}{9}\\ 5 & \frac{1}{18} & \frac{5}{18} & \frac{25}{18}\\ \hline & & \sum p_i x_i = \frac{35}{18} & \sum p_i x_i^2 = \frac{35}{6} \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = \frac{35}{18}
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = \frac{35}{6} - \left(\frac{35}{18}\right)^2
\displaystyle = \frac{35}{6} - \frac{1225}{324}
\displaystyle = \frac{1890 - 1225}{324}
\displaystyle = \frac{665}{324}
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{\frac{665}{324}}
\displaystyle = \frac{\sqrt{665}}{18}

\displaystyle \textbf{Question 2: }~\text{A discrete random variable }X\text{ has the probability distribution given below: }
\displaystyle \begin{array}{c|cccc} X & 0.5 & 1 & 1.5 & 2\\ \hline P(X) & k & k^2 & 2k^2 & k \end{array}
\displaystyle \text{(i) Find the value of }k.\ \text{(ii) Determine the mean of the distribution.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)  }
\displaystyle \text{The probability distribution of } X \text{ is given as:}
\displaystyle \begin{array}{c|cccc} X & 0.5 & 1 & 1.5 & 2\\ \hline P(X) & k & k^2 & 2k^2 & k \end{array}
\displaystyle \text{As, } \sum p_i = 1
\displaystyle \Rightarrow k + k^2 + 2k^2 + k = 1
\displaystyle \Rightarrow 3k^2 + 2k - 1 = 0
\displaystyle \Rightarrow 3k^2 + 3k - k - 1 = 0
\displaystyle \Rightarrow 3k(k+1) - 1(k+1) = 0
\displaystyle \Rightarrow (3k-1)(k+1) = 0
\displaystyle \Rightarrow k = \frac{1}{3} \text{ or } k = -1
\displaystyle \text{But } k \text{ cannot be negative}
\displaystyle \therefore k = \frac{1}{3}

\displaystyle \textbf{(ii)  }
\displaystyle \text{The probability distribution of } X \text{ is given as:}
\displaystyle \begin{array}{c|cccc} X & 0.5 & 1 & 1.5 & 2\\ \hline P(X) & k & k^2 & 2k^2 & k \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 0.5k + 1\cdot k^2 + 1.5\cdot 2k^2 + 2k
\displaystyle = 0.5\cdot \frac{1}{3} + 1\cdot \left(\frac{1}{3}\right)^2 + 1.5\cdot 2\left(\frac{1}{3}\right)^2 + 2\cdot \frac{1}{3}
\displaystyle = \frac{0.5}{3} + \frac{1}{9} + \frac{3}{9} + \frac{2}{3}
\displaystyle = \frac{1.5 + 1 + 3 + 6}{9}
\displaystyle = \frac{11.5}{9}
\displaystyle = \frac{115}{90}
\displaystyle = \frac{23}{18}

\displaystyle \textbf{Question 3: }~\text{Find the mean, variance and standard deviation of the following} \\ \text{probability distribution, where }p+q=1.
\displaystyle \begin{array}{c|cc} x_i & a & b\\ \hline p_i & p & q \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline a & p & ap & a^2 p\\ b & q & bq & b^2 q\\ \hline & & \sum p_i x_i = ap + bq & \sum p_i x_i^2 = a^2 p + b^2 q \end{array}
\displaystyle \text{Now,}
\displaystyle \text{Mean} = \sum p_i x_i = ap + bq
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2 = a^2 p + b^2 q - (ap + bq)^2
\displaystyle = a^2 p + b^2 q - a^2 p^2 - b^2 q^2 - 2abpq
\displaystyle = a^2 p - a^2 p^2 + b^2 q - b^2 q^2 - 2abpq
\displaystyle = a^2 p(1-p) + b^2 q(1-q) - 2abpq
\displaystyle = a^2 pq + b^2 qp - 2abpq \text{ since } p+q=1
\displaystyle = pq(a^2 + b^2 - 2ab)
\displaystyle = pq(a-b)^2
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{pq(a-b)^2}
\displaystyle = |a-b|\sqrt{pq}

\displaystyle \textbf{Question 4: }~\text{Find the mean and variance of the number of tails in three tosses} \\ \text{of a coin.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of tails in three tosses of a coin.}
\displaystyle \text{Then, } X \text{ can take the values } 0,1,2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\mathrm{HHH})=\frac{1}{8}
\displaystyle P(X=1)=P(\mathrm{THH}\text{ or }\mathrm{HHT}\text{ or }\mathrm{HTH})=\frac{3}{8}
\displaystyle P(X=2)=P(\mathrm{TTH}\text{ or }\mathrm{THT}\text{ or }\mathrm{HTT})=\frac{3}{8}
\displaystyle P(X=3)=P(\mathrm{TTT})=\frac{1}{8}
\displaystyle \text{Thus, the probability distribution of } X \text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 0 & \frac{1}{8}\\ 1 & \frac{3}{8}\\ 2 & \frac{3}{8}\\ 3 & \frac{1}{8} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 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}\\ \hline & & \sum p_i x_i = \frac{3}{2} & \sum p_i x_i^2 = 3 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = \frac{3}{2}
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 3 - \left(\frac{3}{2}\right)^2
\displaystyle = 3 - \frac{9}{4}
\displaystyle = \frac{3}{4}

\displaystyle \textbf{Question 5: }~\text{Two cards are drawn simultaneously from a pack of }52\text{ cards.} \\ \text{Compute the mean and standard deviation of the number of kings. [CBSE\ 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of kings in a sample of }2\text{ cards drawn from a well-} \\ \text{shuffled pack of }52\text{ playing cards.}
\displaystyle \text{Then, } X \text{ can take the values }0,1\text{ and }2.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{no king})
\displaystyle = \frac{{}^{48}C_2}{{}^{52}C_2} = \frac{1128}{1326} = \frac{188}{221}
\displaystyle P(X=1)=P(\text{1 king})
\displaystyle = \frac{{}^{4}C_1 \times {}^{48}C_1}{{}^{52}C_2} = \frac{192}{1326} = \frac{32}{221}
\displaystyle P(X=2)=P(\text{2 kings})
\displaystyle = \frac{{}^{4}C_2}{{}^{52}C_2} = \frac{6}{1326} = \frac{1}{221}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 0 & \frac{188}{221}\\ 1 & \frac{32}{221}\\ 2 & \frac{1}{221} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 0 & \frac{188}{221} & 0 & 0\\ 1 & \frac{32}{221} & \frac{32}{221} & \frac{32}{221}\\ 2 & \frac{1}{221} & \frac{2}{221} & \frac{4}{221}\\ \hline & & \sum p_i x_i = \frac{34}{221} & \sum p_i x_i^2 = \frac{36}{221} \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = \frac{34}{221}
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = \frac{36}{221} - \left(\frac{34}{221}\right)^2
\displaystyle = \frac{7956 - 1156}{48841}
\displaystyle = \frac{6800}{48841}
\displaystyle = \frac{400}{2873}

\displaystyle \textbf{Question 6: }~\text{Find the mean, variance and standard deviation of the number of tails} \\ \text{in three tosses of a coin.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of tails in three tosses of a coin.}
\displaystyle \text{Then, } X \text{ can take the values }0,1,2 \text{ and }3.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\mathrm{HHH})=\frac{1}{8}
\displaystyle P(X=1)=P(\mathrm{THH}\text{ or }\mathrm{HHT}\text{ or }\mathrm{HTH})=\frac{3}{8}
\displaystyle P(X=2)=P(\mathrm{TTH}\text{ or }\mathrm{THT}\text{ or }\mathrm{HTT})=\frac{3}{8}
\displaystyle P(X=3)=P(\mathrm{TTT})=\frac{1}{8}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 0 & \frac{1}{8}\\ 1 & \frac{3}{8}\\ 2 & \frac{3}{8}\\ 3 & \frac{1}{8} \end{array}
\displaystyle \text{Computation of mean and standard deviation}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 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}\\ \hline & & \sum p_i x_i = \frac{3}{2} & \sum p_i x_i^2 = 3 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = \frac{3}{2}
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 3 - \left(\frac{3}{2}\right)^2
\displaystyle = 3 - \frac{9}{4}
\displaystyle = \frac{3}{4}
\displaystyle \text{Standard Deviation} = \sqrt{\text{Variance}}
\displaystyle = \sqrt{\frac{3}{4}}
\displaystyle \approx 0.87

\displaystyle \textbf{Question 7: }~\text{Two bad eggs are accidently mixed up with ten good ones. Three eggs are} \\ \text{drawn at random with replacement from this lot. Compute the mean for the number} \\ \text{of bad eggs drawn.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of bad eggs in a sample of }3\text{ eggs drawn from a lot} \\ \text{containing }2\text{ bad eggs and }10\text{ good eggs.}
\displaystyle \text{Then, } X \text{ can take the values }0,1 \text{ and }2.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{no bad egg})
\displaystyle = \frac{{}^{10}C_3}{{}^{12}C_3}
\displaystyle = \frac{120}{220}
\displaystyle = \frac{6}{11}
\displaystyle P(X=1)=P(\text{1 bad egg})
\displaystyle = \frac{{}^{2}C_1 \times {}^{10}C_2}{{}^{12}C_3}
\displaystyle = \frac{90}{220}
\displaystyle = \frac{9}{22}
\displaystyle P(X=2)=P(\text{2 bad eggs})
\displaystyle = \frac{{}^{2}C_2 \times {}^{10}C_1}{{}^{12}C_3}
\displaystyle = \frac{10}{220}
\displaystyle = \frac{1}{22}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 0 & \frac{6}{11}\\ 1 & \frac{9}{22}\\ 2 & \frac{1}{22} \end{array}
\displaystyle \text{Computation of mean}
\displaystyle \begin{array}{c|c|c} x_i & p_i & p_i x_i\\ \hline 0 & \frac{6}{11} & 0\\ 1 & \frac{9}{22} & \frac{9}{22}\\ 2 & \frac{1}{22} & \frac{2}{22}\\ \hline & & \sum p_i x_i = \frac{11}{22} \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 0 + \frac{9}{22} + \frac{2}{22}
\displaystyle = \frac{11}{22}
\displaystyle = \frac{1}{2}

\displaystyle \textbf{Question 8: }~\text{A pair of fair dice is thrown. Let }X\text{ be the random variable which} \\ \text{denotes the minimum of the two numbers which appear. Find the probability} \\ \text{distribution, mean and variance of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the event of getting twice the number.}
\displaystyle \text{Then, } X \text{ can take the values }1,2,3,4,5 \text{ and }6.
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 1 & \frac{11}{36}\\ 2 & \frac{9}{36}\\ 3 & \frac{7}{36}\\ 4 & \frac{5}{36}\\ 5 & \frac{3}{36}\\ 6 & \frac{1}{36} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 1 & \frac{11}{36} & \frac{11}{36} & \frac{11}{36}\\ 2 & \frac{9}{36} & \frac{18}{36} & \frac{36}{36}\\ 3 & \frac{7}{36} & \frac{21}{36} & \frac{63}{36}\\ 4 & \frac{5}{36} & \frac{20}{36} & \frac{80}{36}\\ 5 & \frac{3}{36} & \frac{15}{36} & \frac{75}{36}\\ 6 & \frac{1}{36} & \frac{6}{36} & \frac{36}{36}\\ \hline & & \sum p_i x_i = \frac{91}{36} = 2.5 & \sum p_i x_i^2 = \frac{301}{36} \approx 8.4 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 2.5
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 8.4 - 6.25
\displaystyle = 2.15

\displaystyle \textbf{Question 9: }~\text{A fair coin is tossed four times. Let }X\text{ denote the number of heads} \\ \text{occurring. Find the probability distribution, mean and variance of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{If a coin is tossed }4\text{ times, then the possible outcomes are } \mathrm{HHHH}, \mathrm{HHHT}, \mathrm{HHTT}, \mathrm{HTTT}, \mathrm{THHH}, \ldots
\displaystyle \text{For the longest string of heads, } X \text{ can take the values }0,1,2,3 \text{ and }4.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{0 head})=\frac{1}{16}
\displaystyle P(X=1)=P(\text{1 head})=\frac{4}{16}
\displaystyle P(X=2)=P(\text{2 heads})=\frac{6}{16}
\displaystyle P(X=3)=P(\text{3 heads})=\frac{4}{16}
\displaystyle P(X=4)=P(\text{4 heads})=\frac{1}{16}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 0 & \frac{1}{16}\\ 1 & \frac{4}{16}\\ 2 & \frac{6}{16}\\ 3 & \frac{4}{16}\\ 4 & \frac{1}{16} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 0 & \frac{1}{16} & 0 & 0\\ 1 & \frac{4}{16} & \frac{4}{16} & \frac{4}{16}\\ 2 & \frac{6}{16} & \frac{12}{16} & \frac{24}{16}\\ 3 & \frac{4}{16} & \frac{12}{16} & \frac{36}{16}\\ 4 & \frac{1}{16} & \frac{4}{16} & \frac{16}{16}\\ \hline & & \sum p_i x_i = 2 & \sum p_i x_i^2 = 5 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 2
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 5 - 4
\displaystyle = 1

\displaystyle \textbf{Question 10: }~\text{A fair die is tossed. Let }X\text{ denote twice the number appearing.} \\ \text{Find the probability distribution, mean and variance of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the event of getting twice the number.}
\displaystyle \text{Then, } X \text{ can take the values }2,4,6,8,10 \text{ and }12.
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 2 & \frac{1}{6}\\ 4 & \frac{1}{6}\\ 6 & \frac{1}{6}\\ 8 & \frac{1}{6}\\ 10 & \frac{1}{6}\\ 12 & \frac{1}{6} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 2 & \frac{1}{6} & \frac{2}{6} & \frac{4}{6}\\ 4 & \frac{1}{6} & \frac{4}{6} & \frac{16}{6}\\ 6 & \frac{1}{6} & \frac{6}{6} & \frac{36}{6}\\ 8 & \frac{1}{6} & \frac{8}{6} & \frac{64}{6}\\ 10 & \frac{1}{6} & \frac{10}{6} & \frac{100}{6}\\ 12 & \frac{1}{6} & \frac{12}{6} & \frac{144}{6}\\ \hline & & \sum p_i x_i = 7 & \sum p_i x_i^2 = \frac{364}{6} \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 7
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = \frac{364}{6} - 49
\displaystyle = \frac{364 - 294}{6}
\displaystyle = \frac{70}{6}
\displaystyle = \frac{35}{3}

\displaystyle \textbf{Question 11: }~\text{A fair die is tossed. Let }X\text{ denote }1\text{ or }3\text{ according as an} \\ \text{odd or an even number appears. Find the probability distribution, mean and variance of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X=1 \text{ for the appearance of odd numbers }1,3 \text{ or }5\text{ on the die.}
\displaystyle \text{Then, } P(X=1)=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{Let } X=3 \text{ for the appearance of even numbers }2,4 \text{ or }6\text{ on the die.}
\displaystyle \text{Then, } P(X=3)=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 1 & \frac{1}{2}\\ 3 & \frac{1}{2} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\ 3 & \frac{1}{2} & \frac{3}{2} & \frac{9}{2}\\ \hline & & \sum p_i x_i = 2 & \sum p_i x_i^2 = 5 \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = 2
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = 5 - 4
\displaystyle = 1

\displaystyle \textbf{Question 12: }~\text{A fair coin is tossed four times. Let }X\text{ denote the longest string of} \\ \text{heads occurring. Find the probability distribution, mean and variance of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{If a coin is tossed }4\text{ times, then the possible outcomes are } \mathrm{HHHH}, \mathrm{HHHT}, \\ \mathrm{HHTH}, \mathrm{HHTT}, \mathrm{HTHH}, \mathrm{HTHT}, \mathrm{HTTH}, \mathrm{HTTT}, \mathrm{THHH}, \\ \mathrm{THHT}, \mathrm{THTH}, \mathrm{THTT}, \mathrm{TTHH}, \mathrm{TTHT}, \mathrm{TTTH}, \mathrm{TTTT}.
\displaystyle \text{For the longest string of heads, } X \text{ can take the values }0,1,2,3 \text{ and }4.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{0 head})=\frac{1}{16}
\displaystyle P(X=1)=P(\text{1 head})=\frac{7}{16}
\displaystyle P(X=2)=P(\text{2 heads})=\frac{5}{16}
\displaystyle P(X=3)=P(\text{3 heads})=\frac{2}{16}
\displaystyle P(X=4)=P(\text{4 heads})=\frac{1}{16}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 0 & \frac{1}{16}\\ 1 & \frac{7}{16}\\ 2 & \frac{5}{16}\\ 3 & \frac{2}{16}\\ 4 & \frac{1}{16} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 0 & \frac{1}{16} & 0 & 0\\ 1 & \frac{7}{16} & \frac{7}{16} & \frac{7}{16}\\ 2 & \frac{5}{16} & \frac{10}{16} & \frac{20}{16}\\ 3 & \frac{2}{16} & \frac{6}{16} & \frac{18}{16}\\ 4 & \frac{1}{16} & \frac{4}{16} & \frac{16}{16}\\ \hline & & \sum p_i x_i = \frac{27}{16} & \sum p_i x_i^2 = \frac{61}{16} \end{array}
\displaystyle \text{Mean} = \sum p_i x_i = \frac{27}{16}
\displaystyle \text{Variance} = \sum p_i x_i^2 - (\text{Mean})^2
\displaystyle = \frac{61}{16} - \left(\frac{27}{16}\right)^2
\displaystyle = \frac{976 - 729}{256}
\displaystyle = \frac{247}{256}
\displaystyle \approx 0.9

\displaystyle \textbf{Question 13: }~\text{Two cards are selected at random from a box which contains five cards} \\ \text{numbered }1,1,2,2,\text{ and }3\text{. Let }X\text{ denote the sum and }Y\text{ the maximum of the two} \\ \text{numbers drawn. Find the probability distribution, mean and variance of }\\ X\text{ and }Y\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }5\text{ cards numbered }1,1,2,2 \text{ and }3.
\displaystyle \text{Let } X=\text{sum of two numbers on cards }=2,3,4,5.
\displaystyle \text{Let } Y=\text{maximum of two numbers }=1,2,3.
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} x & P(X)\\ \hline 2 & \frac{1}{10}\\ 3 & \frac{4}{10}\\ 4 & \frac{3}{10}\\ 5 & \frac{2}{10} \end{array}
\displaystyle \text{Total number of cards }=2\text{ cards with }1+2\text{ cards with }2+1\text{ card with }3=5.
\displaystyle \text{Total number of possible choices in drawing two cards from }5= {}^{5}C_2= \frac{5!}{2!3!} = 10
\displaystyle P(11)=\frac{{}^{2}C_2\times {}^{2}C_0\times {}^{1}C_0}{{}^{5}C_2} = \frac{1\times1\times1}{10}= \frac{1}{10}
\displaystyle P(12)=\frac{{}^{2}C_1\times {}^{2}C_1\times {}^{1}C_0}{{}^{5}C_2}= \frac{2\times2\times1}{10}= \frac{4}{10}= \frac{2}{5}
\displaystyle P(13)=\frac{{}^{2}C_1\times {}^{2}C_0\times {}^{1}C_1}{{}^{5}C_2} = \frac{2\times1\times1}{10}= \frac{2}{10}= \frac{1}{5}
\displaystyle \begin{array}{c|c|c|c|c} & \text{``1''s} & \text{``2''s} & \text{``3''s} & \text{Total}\\ \hline \text{Available} & 2 & 2 & 1 & 5\\ \text{To Choose} & 1 & 0 & 1 & 2\\ \text{Choices} & {}^{2}C_1 & {}^{2}C_0 & {}^{1}C_1 & {}^{5}C_2 \end{array}
\displaystyle = \frac{{}^{2}C_1 \times {}^{2}C_0 \times {}^{1}C_1}{{}^{5}C_2} = \frac{2\times1\times1}{10} = \frac{2}{10} = \frac{1}{5}
\displaystyle \text{``2'' and ``2'' on them}
\displaystyle P(22)=\frac{{}^{2}C_0 \times {}^{2}C_2 \times {}^{1}C_0}{{}^{5}C_2} = \frac{1\times1\times1}{10} = \frac{1}{10}
\displaystyle \begin{array}{c|c|c|c|c} & \text{``1''s} & \text{``2''s} & \text{``3''s} & \text{Total}\\ \hline \text{Available} & 2 & 2 & 1 & 5\\ \text{To Choose} & 0 & 1 & 1 & 2\\ \text{Choices} & {}^{2}C_0 & {}^{2}C_1 & {}^{1}C_1 & {}^{5}C_2 \end{array}
\displaystyle = \frac{{}^{2}C_0 \times {}^{2}C_1 \times {}^{1}C_1}{{}^{5}C_2} = \frac{1\times2\times1}{10} = \frac{2}{10}= \frac{1}{5}
\displaystyle \text{Probability for the sum of the numbers on the cards drawn to be}
\displaystyle 2 \Rightarrow p(x=2)=p(11) = \frac{1}{10}
\displaystyle 3 \Rightarrow p(x=3)=p(12)= \frac{4}{10}
\displaystyle 4 \Rightarrow p(x=4)=p(13\text{ or }22) = p(13)+p(22) = \frac{2}{10}+\frac{1}{10} = \frac{3}{10}
\displaystyle 5 \Rightarrow p(x=5)=p(23) = \frac{2}{10}
\displaystyle \text{The probability distribution of }x\text{ would be}
\displaystyle \begin{array}{c|cccc} x & 2 & 3 & 4 & 5\\ \hline P(X=x) & \frac{1}{10} & \frac{4}{10} & \frac{3}{10} & \frac{2}{10} \end{array}
\displaystyle \text{Calculations for mean and standard deviation}
\displaystyle \begin{array}{c|c|c|c|c} x & P(X=x) & px & x^2 & px^2\\ \hline 2 & \frac{1}{10} & \frac{2}{10} & 4 & \frac{4}{10}\\ 3 & \frac{4}{10} & \frac{12}{10} & 9 & \frac{36}{10}\\ 4 & \frac{3}{10} & \frac{12}{10} & 16 & \frac{48}{10}\\ 5 & \frac{2}{10} & \frac{10}{10} & 25 & \frac{50}{10}\\ \hline \text{Total} & 1 & \frac{36}{10}=3.6 & & \frac{138}{10}=13.8 \end{array}
\displaystyle \text{The expected value of the sum}
\displaystyle \Rightarrow E(x)=\sum px
\displaystyle = 3.6
\displaystyle \text{Variance of the sum of the numbers on the cards}
\displaystyle \Rightarrow \mathrm{var}(x)=E(x^2)-(E(x))^2
\displaystyle = 13.8-(3.6)^2
\displaystyle = 13.8-12.96
\displaystyle = 0.84
\displaystyle \text{Standard deviation of the sum of the numbers on the cards}
\displaystyle \Rightarrow \sqrt{\mathrm{var}(x)}
\displaystyle = \sqrt{0.84}
\displaystyle \approx 0.917
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 2 & \frac{1}{10} & \frac{2}{10} & \frac{4}{10}\\ 3 & \frac{4}{10} & \frac{12}{10} & \frac{36}{10}\\ 4 & \frac{3}{10} & \frac{12}{10} & \frac{48}{10}\\ 5 & \frac{2}{10} & \frac{10}{10} & \frac{50}{10}\\ \hline & & \sum p_i x_i=\frac{36}{10}=3.6 & \sum p_i x_i^2=\frac{138}{10}=13.8 \end{array}
\displaystyle \text{Mean}=\sum p_i x_i=3.6
\displaystyle \text{Variance}=\sum p_i x_i^2-(\text{Mean})^2
\displaystyle =13.8-12.96
\displaystyle =0.84
\displaystyle \text{Thus, the probability distribution of }Y\text{ is given by}
\displaystyle \begin{array}{c|c} Y & P(Y)\\ \hline 1 & 0.1\\ 2 & 0.5\\ 3 & 0.4 \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} y_i & p_i & p_i y_i & p_i y_i^2\\ \hline 1 & 0.1 & 0.1 & 0.1\\ 2 & 0.5 & 1 & 2\\ 3 & 0.4 & 1.2 & 3.6\\ \hline & & \sum p_i y_i = 2.3 & \sum p_i y_i^2 = 5.7 \end{array}
\displaystyle \text{Mean} = \sum p_i y_i = 2.3
\displaystyle \text{Variance} = \sum p_i y_i^2 - (\text{Mean})^2
\displaystyle = 5.7 - 5.29
\displaystyle = 0.41

\displaystyle \textbf{Question 14: }~\text{A die is tossed twice. A `success' is getting an odd number on a toss.} \\ \text{Find the variance of the number of successes.}
\displaystyle \text{Answer:}
\displaystyle \text{It is given that ``success'' denotes the event of getting the numbers }1,3 \text{ or }5.
\displaystyle P(\text{success})=\frac{1}{2}
\displaystyle \text{Also, ``failure'' denotes the event of getting the numbers }2,4 \text{ or }6.
\displaystyle P(\text{failure})=\frac{1}{2}
\displaystyle \text{Let }X\text{ denote the event of getting success. Then, }X\text{ can take the values }0,1 \text{ and }2.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{no success})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}
\displaystyle P(X=1)=P(\text{1 success})=\left(\frac{1}{2}\times\frac{1}{2}\right)+\left(\frac{1}{2}\times\frac{1}{2}\right)=\frac{1}{2}
\displaystyle P(X=2)=P(\text{2 successes})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} X & P(X)\\ \hline 0 & \frac{1}{4}\\ 1 & \frac{1}{2}\\ 2 & \frac{1}{4} \end{array}
\displaystyle \text{Computation of mean and variance}
\displaystyle \begin{array}{c|c|c|c} x_i & p_i & p_i x_i & p_i x_i^2\\ \hline 0 & \frac{1}{4} & 0 & 0\\ 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\ 2 & \frac{1}{4} & \frac{1}{2} & 1\\ \hline & & \sum p_i x_i = 1 & \sum p_i x_i^2 = \frac{3}{2} \end{array}
\displaystyle \text{Mean}=\sum p_i x_i=1
\displaystyle \text{Variance}=\sum p_i x_i^2-(\text{Mean})^2
\displaystyle =\frac{3}{2}-1
\displaystyle =\frac{1}{2}

\displaystyle \textbf{Question 15: }~\text{A box contains }13\text{ bulbs, out of which }5\text{ are defective. } 3\text{ bulbs are} \\ \text{randomly drawn, one by one without replacement, from the box. Find the probability} \\ \text{distribution of the number of defective bulbs. \ [CBSE\ 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of defective bulbs in a sample of }3\text{ bulbs drawn from a} \\ \text{bag containing }13\text{ bulbs. }5\text{ bulbs in the bag turn out to be defective.}
\displaystyle \text{Then, }X\text{ can take the values }0,1,2 \text{ and }3.
\displaystyle P(X=0)=P(\text{no defective bulb})=\frac{{}^{8}C_3}{{}^{13}C_3}=\frac{56}{286}=\frac{28}{143}
\displaystyle P(X=1)=P(\text{1 defective bulb})=\frac{{}^{5}C_1\times{}^{8}C_2}{{}^{13}C_3}=\frac{140}{286}=\frac{70}{143}
\displaystyle P(X=2)=P(\text{2 defective bulbs})=\frac{{}^{5}C_2\times{}^{8}C_1}{{}^{13}C_3}=\frac{80}{286}=\frac{40}{143}
\displaystyle P(X=3)=P(\text{3 defective bulbs})=\frac{{}^{5}C_3}{{}^{13}C_3}=\frac{10}{286}=\frac{5}{143}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} X & P(X)\\ \hline 0 & \frac{28}{143}\\ 1 & \frac{70}{143}\\ 2 & \frac{40}{143}\\ 3 & \frac{5}{143} \end{array}

\displaystyle \textbf{Question 16: }~\text{In roulette, the wheel has }13\text{ numbers }0,1,2,\ldots,12\text{ marked} \\ \text{on equally spaced slots. A player bets } \text{Rs }10\text{ on a given number. He receives} \\ \text{Rs }100\text{ from the organiser of the game if the ball comes to rest in this slot;} \\ \text{otherwise he gets nothing. If }X\text{ denotes the player's net gain/loss, find }E(X)\text{.}


\displaystyle \text{Answer:}
\displaystyle \text{The wheel has }13\text{ numbers, i.e. }0,1,2,\ldots,12\text{ marked on equally spaced slots.}
\displaystyle \therefore \text{Probability of ball resting on any particular number }=\frac{1}{13}
\displaystyle \text{Let the player set Rs }10\text{ on a given number }k.
\displaystyle P(\text{player receives Rs }100)=P(\text{ball rests on it})=\frac{1}{13}
\displaystyle X\text{ denotes the player's net gain or loss. If he gets the required number, then his gain is Rs }90\,(100-10).
\displaystyle \text{If the ball does not rest on the number, then it rests on any of the other }12\text{ numbers. In that} \\ \text{case, the player's loss is Rs }10.
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \text{Computation of mean}
\displaystyle \begin{array}{c|c|c} x_i & p_i & p_i x_i\\ \hline 90 & \frac{1}{13} & \frac{90}{13}\\ -10 & \frac{12}{13} & -\frac{120}{13}\\ \hline & & \sum p_i x_i=-\frac{30}{13} \end{array}
\displaystyle \text{Mean}=E(x)=\sum p_i x_i=-\frac{30}{13}

\displaystyle \textbf{Question 17: }~\text{Three cards are drawn at random (without replacement) from a well} \\ \text{shuffled pack of }52\text{ cards. Find the probability distribution of number of red cards.} \\ \text{Hence find the mean of the distribution. \ [CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of red cards drawn.}
\displaystyle \text{Then, }X\text{ can take the values }0,1,2 \text{ or }3.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\mathrm{BBB})=\frac{26}{52}\times\frac{25}{51}\times\frac{24}{50}=\frac{2}{17}
\displaystyle P(X=1)=P(\mathrm{RBB}\text{ or }\mathrm{BRB}\text{ or }\mathrm{BBR})=3\times\frac{26}{52}\times\frac{26}{51}\times\frac{25}{50}=\frac{13}{34}
\displaystyle P(X=2)=P(\mathrm{RRB}\text{ or }\mathrm{RBR}\text{ or }\mathrm{BRR})=3\times\frac{26}{52}\times\frac{25}{51}\times\frac{26}{50}=\frac{13}{34}
\displaystyle P(X=3)=P(\mathrm{RRR})=\frac{26}{52}\times\frac{25}{51}\times\frac{24}{50}=\frac{2}{17}
\displaystyle \text{Thus, the probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|c} X & P(X)\\ \hline 0 & \frac{2}{17}\\ 1 & \frac{13}{34}\\ 2 & \frac{13}{34}\\ 3 & \frac{2}{17} \end{array}
\displaystyle \text{Mean}=\sum p_i x_i=0\times\frac{2}{17}+1\times\frac{13}{34}+2\times\frac{13}{34}+3\times\frac{2}{17}
\displaystyle =0+\frac{13}{34}+\frac{26}{34}+\frac{6}{17}
\displaystyle =\frac{51}{34}
\displaystyle =\frac{3}{2}

\displaystyle \textbf{Question 18: }~\text{An urn contains }5\text{ red }2\text{ black balls. Two balls are randomly} \text{drawn, without} \\ \text{replacement. Let }X\text{ represent the number of black balls drawn. What are the possible values} \\ \text{of }X\text{? Is }X\text{ a random variable? If yes, find the mean and variance of }X\text{. \ [CBSE\ 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{As, }X\text{ represents the number of black balls drawn.}
\displaystyle \text{So, it can take the values }0,1\text{ and }2.\text{ Yes, }X\text{ is a random variable.}
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\mathrm{RR})=\frac{5}{7}\times\frac{4}{6}=\frac{10}{21}
\displaystyle P(X=1)=P(\mathrm{RB}\text{ or }\mathrm{BR})=2\times\frac{5}{7}\times\frac{2}{6}=\frac{10}{21}
\displaystyle P(X=2)=P(\mathrm{BB})=\frac{2}{7}\times\frac{1}{6}=\frac{1}{21}
\displaystyle \text{Mean}=\sum p_i x_i=0\times\frac{10}{21}+1\times\frac{10}{21}+2\times\frac{1}{21}
\displaystyle =\frac{10}{21}+\frac{2}{21}
\displaystyle =\frac{12}{21}
\displaystyle =\frac{4}{7}
\displaystyle \text{Also, }\sum p_i x_i^2=0^2\times\frac{10}{21}+1^2\times\frac{10}{21}+2^2\times\frac{1}{21}
\displaystyle =\frac{10}{21}+\frac{4}{21}
\displaystyle =\frac{14}{21}
\displaystyle =\frac{2}{3}
\displaystyle \text{So, variance}=\sum p_i x_i^2-(\text{Mean})^2
\displaystyle =\frac{2}{3}-\left(\frac{4}{7}\right)^2
\displaystyle =\frac{2}{3}-\frac{16}{49}
\displaystyle =\frac{98-48}{147}
\displaystyle =\frac{50}{147}

\displaystyle \textbf{Question 19: }~\text{Two numbers are selected at random (without replacement) from} \\ \text{positive integers }2,3,4,5,6\text{ and }7\text{. Let }X\text{ denote the larger of the two numbers obtained.} \\ \text{Find the mean and variance of the probability distribution of }X\text{. \ [CBSE\ 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{As, }X\text{ denotes the larger of the two numbers obtained.}
\displaystyle \text{So, }X\text{ can take the values }3,4,5,6 \text{ and }7.
\displaystyle \text{Now,}
\displaystyle P(X=3)=P\{(2,3)\text{ or }(3,2)\}=2\times\frac{1}{6}\times\frac{1}{5}=\frac{1}{15}
\displaystyle P(X=4)=P\{(2,4)\text{ or }(4,2)\text{ or }(3,4)\text{ or }(4,3)\}=4\times\frac{1}{6}\times\frac{1}{5}=\frac{2}{15}
\displaystyle P(X=5)=P\{(2,5)\text{ or }(5,2)\text{ or }(3,5)\text{ or }(5,3)\text{ or }(4,5)\text{ or }(5,4)\}=6\times\frac{1}{6}\times\frac{1}{5}=\frac{1}{5}
\displaystyle P(X=6)=P\{(2,6)\text{ or }(6,2)\text{ or }(3,6)\text{ or }(6,3)\text{ or }(4,6)\text{ or }(6,4)\text{ or }(5,6)\text{ or }(6,5)\}=8\times\frac{1}{6}\times\frac{1}{5}=\frac{4}{15}
\displaystyle P(X=7)=P\{(2,7)\text{ or }(7,2)\text{ or }(3,7)\text{ or }(7,3)\text{ or }(4,7)\text{ or }(7,4)\text{ or }(5,7)\text{ or }(7,5)\text{ or }(6,7)\text{ or }(7,6)\}=10\times\frac{1}{6}\times\frac{1}{5}=\frac{1}{3}
\displaystyle \text{The probability distribution of }X\text{ is as follows:}
\displaystyle \begin{array}{c|ccccc} X & 3 & 4 & 5 & 6 & 7\\ \hline P(X) & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{4}{15} & \frac{1}{3} \end{array}
\displaystyle \text{So, Mean, }E(X)=3\times\frac{1}{15}+4\times\frac{2}{15}+5\times\frac{1}{5}+6\times\frac{4}{15}+7\times\frac{1}{3}
\displaystyle =\frac{3}{15}+\frac{8}{15}+\frac{15}{15}+\frac{24}{15}+\frac{35}{15}
\displaystyle =\frac{85}{15}
\displaystyle =\frac{17}{3}
\displaystyle \text{Also, }E(X^2)=3^2\times\frac{1}{15}+4^2\times\frac{2}{15}+5^2\times\frac{1}{5}+6^2\times\frac{4}{15}+7^2\times\frac{1}{3}
\displaystyle =\frac{9}{15}+\frac{32}{15}+\frac{75}{15}+\frac{144}{15}+\frac{245}{15}
\displaystyle =\frac{505}{15}
\displaystyle =\frac{101}{3}
\displaystyle \mathrm{Var}(X)=E(X^2)-[E(X)]^2
\displaystyle =\frac{101}{3}-\left(\frac{17}{3}\right)^2
\displaystyle =\frac{101}{3}-\frac{289}{9}
\displaystyle =\frac{303-289}{9}
\displaystyle =\frac{14}{9}

\displaystyle \textbf{Question 20: }~\text{In a game, a man wins } \text{Rs }5\text{ for getting a number greater than } \\ 4\text{ and loses } \text{Rs }1\text{ otherwise, when a fair die is thrown. The man decided to throw} \\ \text{a die thrice but to quit as and when he gets a number greater than }4\text{. Find the } \\ \text{expected value of the amount he wins/lose. \ [CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{The man may get a number greater than }4\text{ in the first throw and then he quits the game.}
\displaystyle \text{He may get a number less than or equal to }4\text{ in the first throw and in the second throw} \\ \text{he may get a number greater than }4\text{ and quits the game.}
\displaystyle \text{In the first two throws he gets a number less than or equal to }4\text{ and in the third throw he} \\ \text{may get a number greater than }4.
\displaystyle \text{He may not get a number greater than }4\text{ in any one of the three throws.}
\displaystyle \text{Let }X\text{ be the amount he wins or loses.}
\displaystyle \text{Then, }X\text{ can take the values }-3,3,4,5\text{ such that}
\displaystyle P(X=5)=P(\text{getting number greater than }4\text{ in first throw})=\frac{1}{3}
\displaystyle P(X=4)=P(\text{getting number less than or equal to }4\text{ in first throw and number greater than } \\ 4\text{ in second throw})
\displaystyle =\frac{4}{6}\times\frac{2}{6}=\frac{2}{9}
\displaystyle P(X=3)=P(\text{getting number less than or equal to }4\text{ in the first two throws and number} \\ \text{greater than }4\text{ in third throw})
\displaystyle =\frac{4}{6}\times\frac{4}{6}\times\frac{2}{6}=\frac{4}{27}
\displaystyle P(X=-3)=P(\text{getting number less than or equal to }4\text{ in all three throws})
\displaystyle =\frac{4}{6}\times\frac{4}{6}\times\frac{4}{6}=\frac{8}{27}
\displaystyle \begin{array}{c|cccc} X & 5 & 4 & 3 & -3\\ \hline P(X) & \frac{1}{3} & \frac{2}{9} & \frac{4}{27} & \frac{8}{27} \end{array}
\displaystyle E(X)=5\times\frac{1}{3}+4\times\frac{2}{9}+3\times\frac{4}{27}-3\times\frac{8}{27}
\displaystyle =\frac{1}{27}\left(45+24+12-24\right)
\displaystyle =\frac{57}{27}
\displaystyle \text{Expected value of the amount he wins or loses is }\frac{57}{27}= \frac{19}{9}

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