\displaystyle \textbf{Question 1: }~\text{Can the mean of a binomial distribution be less than its variance?}
\displaystyle \text{Answer:}
\displaystyle \text{The mean of a binomial distribution is }np\text{ and the variance is }npq.
\displaystyle \text{If the mean is less than the variance, then }np<npq.
\displaystyle \text{Since }n>0\text{ and }p>0,\text{ divide both sides by }np.
\displaystyle \text{We obtain }1<q.
\displaystyle \text{But }q=1-p,\text{ so }q<1\text{ under all circumstances.}
\displaystyle \text{Hence, the mean of a binomial distribution cannot be less than its variance.}

\displaystyle \textbf{Question 2: }~\text{Determine the binomial distribution whose mean is }9\text{ and variance is }\frac{9}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{It is given that the mean, i.e. }np=9,\text{ and the variance, i.e. }npq=\frac{9}{4}.
\displaystyle \therefore \frac{npq}{np}=\frac{9}{4}\times\frac{1}{9}=\frac{1}{4}.
\displaystyle \text{Hence, }q=\frac{1}{4}.
\displaystyle \text{And }p=1-q=\frac{3}{4}.
\displaystyle \text{When }p=\frac{3}{4},
\displaystyle np=\frac{3n}{4}.
\displaystyle 9=\frac{3n}{4}.
\displaystyle \Rightarrow n=12.
\displaystyle \mathrm{P}(X=r)={}^{12}C_{r}\left(\frac{3}{4}\right)^{r}\left(\frac{1}{4}\right)^{12-r},\; r=0,1,2,3,4,5,\ldots,12.

\displaystyle \textbf{Question 3: }~\text{If the mean and variance of a binomial distribution are respectively }9 \\ \text{ and }6,\text{ find the distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: Mean }=9\text{ and variance }=6.
\displaystyle \therefore np=9\ldots(1)
\displaystyle npq=6\ldots(2)
\displaystyle \text{Dividing eq.\,(2) by eq.\,(1), we get }q=\frac{6}{9}=\frac{2}{3}\text{ and }p=1-q=\frac{1}{3}.
\displaystyle \text{As }np=9,\text{ substituting the value of }p,\text{ we get }\frac{n}{3}=9.
\displaystyle \text{Hence, }n=27.
\displaystyle \mathrm{P}(X=r)={}^{27}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{27-r},\; r=0,1,2,\ldots,27.

\displaystyle \textbf{Question 4: }~\text{Find the binomial distribution when the sum of its mean and variance} \\ \text{for }5\text{ trials is }4.8.
\displaystyle \text{Answer:}
\displaystyle \text{Number of trials in the binomial distribution is }5.
\displaystyle \text{If }p\text{ is the probability of success, then }np+npq=4.8.
\displaystyle \text{Or }5p+5p(1-p)=4.8.
\displaystyle \Rightarrow 10p-5p^{2}=4.8.
\displaystyle \text{Or }p^{2}-2p+0.96=0.
\displaystyle \text{By factorising, we get }(p-0.8)(p-1.2)=0.
\displaystyle \text{As }p\text{ cannot exceed }1,
\displaystyle p=0.8=\frac{4}{5}.
\displaystyle \text{And }q=1-p=\frac{1}{5}.
\displaystyle \therefore \mathrm{P}(X=r)={}^{5}C_{r}\left(\frac{4}{5}\right)^{r}\left(\frac{1}{5}\right)^{5-r},\; r=0,1,2,\ldots,5.

\displaystyle \textbf{Question 5: }~\text{Determine the binomial distribution whose mean is }20\text{ and variance }16.
\displaystyle \text{Answer:}
\displaystyle \text{Mean, i.e. }np=20\ldots(1)
\displaystyle \text{Variance, i.e. }npq=16\ldots(2)
\displaystyle \text{Dividing eq.\,(2) by eq.\,(1), we get }\frac{npq}{np}=\frac{16}{20}.
\displaystyle \Rightarrow q=\frac{4}{5}.
\displaystyle \Rightarrow p=1-q.
\displaystyle \therefore p=\frac{1}{5}.
\displaystyle \text{As }np=20,
\displaystyle \Rightarrow n=100.
\displaystyle \therefore \mathrm{P}(X=r)={}^{100}C_{r}\left(\frac{1}{5}\right)^{r}\left(\frac{4}{5}\right)^{100-r},\; r=0,1,2,\ldots,100.

\displaystyle \textbf{Question 6: }~\text{In a binomial distribution the sum and product of the mean and the} \\ \text{variance are }\frac{25}{3}\text{ and }\frac{50}{3}\text{ respectively. Find the distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle \text{Sum of the mean and variance }=\frac{25}{3}.
\displaystyle \Rightarrow np+npq=\frac{25}{3}.
\displaystyle \Rightarrow np(1+q)=\frac{25}{3}\ldots(1)
\displaystyle \text{Product of the mean and variance }=\frac{50}{3}.
\displaystyle \Rightarrow np(npq)=\frac{50}{3}\ldots(2)
\displaystyle \text{Dividing eq.\,(2) by eq.\,(1), we get }\frac{np(npq)}{np(1+q)}=\frac{50}{3}\times\frac{3}{25}.
\displaystyle \Rightarrow \frac{npq}{1+q}=2.
\displaystyle \Rightarrow npq=2(1+q).
\displaystyle \Rightarrow np(1-p)=2(2-p).
\displaystyle \Rightarrow np=\frac{2(2-p)}{1-p}.
\displaystyle \text{Substituting this value in }np+npq=\frac{25}{3},\text{ we get }
\displaystyle \frac{2(2-p)}{1-p}(2-p)=\frac{25}{3}.
\displaystyle \Rightarrow 6(4-4p+p^{2})=25-25p.
\displaystyle \Rightarrow 6p^{2}+p-1=0.
\displaystyle \Rightarrow (3p-1)(2p+1)=0.
\displaystyle \Rightarrow p=\frac{1}{3}\text{ or }-\frac{1}{2}.
\displaystyle \text{As }p\text{ cannot be negative, the only answer is }p=\frac{1}{3}.
\displaystyle q=1-p=\frac{2}{3}.
\displaystyle \Rightarrow np+npq=\frac{25}{3}.
\displaystyle \Rightarrow n\left(\frac{1}{3}\right)\left(1+\frac{2}{3}\right)=\frac{25}{3}.
\displaystyle \Rightarrow n=15.
\displaystyle \therefore \mathrm{P}(X=r)={}^{15}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{15-r},\; r=0,1,2,\ldots,15.

\displaystyle \textbf{Question 7: }~\text{The mean of a binomial distribution is }20,\text{ and the standard deviation }4. \\ \text{ Calculate the parameters of the binomial distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Given that the mean, i.e. }np=20\ldots(1)
\displaystyle \text{and the standard deviation, i.e. }\sqrt{npq}=4.
\displaystyle \Rightarrow npq=16\ldots(2)
\displaystyle \text{Dividing eq.\,(2) by eq.\,(1), we get }q=\frac{16}{20}=\frac{4}{5}.
\displaystyle \text{And }p=1-q=\frac{1}{5}.
\displaystyle \therefore n=\frac{\text{Mean}}{p}=100.
\displaystyle \mathrm{P}(X=r)={}^{100}C_{r}\left(\frac{1}{5}\right)^{r}\left(\frac{4}{5}\right)^{100-r},\; r=0,1,2,\ldots,100.
\displaystyle \text{Therefore, the parameters are }n=100\text{ and }p=\frac{1}{5}.

\displaystyle \textbf{Question 8: }~\text{If the probability of a defective bolt is }0.1,\text{ find the (i) mean and} \\ \text{(ii) standard deviation for the distribution of bolts in a total of }400\text{ bolts.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of bolts }(n)=400\text{ and }p=\text{probability of a defective bolt }=0.1.
\displaystyle \text{(i) Mean }=np=400(0.1)=40.
\displaystyle \text{(ii) Variance }=npq=400(0.1)(1-0.1)=36.
\displaystyle \text{So, the standard deviation }=\sqrt{36}=6.

\displaystyle \textbf{Question 9: }~\text{Find the binomial distribution whose mean is }5\text{ and variance }\frac{10}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Mean of the binomial distribution, i.e. }np=5.
\displaystyle \text{Variance, i.e. }npq=\frac{10}{3}.
\displaystyle q=\frac{\text{Variance}}{\text{Mean}}=\frac{\frac{10}{3}}{5}=\frac{2}{3}.
\displaystyle \text{And }p=1-q=\frac{1}{3}.
\displaystyle np=5.
\displaystyle \Rightarrow n=15.
\displaystyle \therefore \mathrm{P}(X=r)={}^{15}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{15-r},\; r=0,1,2,\ldots,15.

\displaystyle \textbf{Question 10: }~\text{If on an average }9\text{ ships out of }10\text{ arrive safely to ports, find} \\ \text{the mean and S.D. of ships returning safely out of a total of }500\text{ ships.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ships }(n)=500.
\displaystyle \text{Let }X\text{ denote the number of ships returning safely to the ports.}
\displaystyle p=\frac{9}{10}\text{ and }q=1-p=\frac{1}{10}.
\displaystyle \text{Mean }=np=500\left(\frac{9}{10}\right)=450.
\displaystyle \text{Variance }=npq=500\left(\frac{9}{10}\right)\left(\frac{1}{10}\right)=45.
\displaystyle \text{Standard deviation }=\sqrt{45}\approx6.71.

\displaystyle \textbf{Question 11: }~\text{The mean and variance of a binomial variate with parameters }n \\ \text{ and }p\text{ are }16\text{ and }8\text{ respectively. Find }P(X=0),~P(X=1)\text{ and }P(X\ge 2).
\displaystyle \text{Answer:}
\displaystyle \text{Given: mean }=16\text{ and variance }=8.
\displaystyle \text{Let }n\text{ and }p\text{ be the parameters of the distribution.}
\displaystyle \text{That is, }np=16\text{ and }npq=8.
\displaystyle q=\frac{npq}{np}=\frac{8}{16}=\frac{1}{2}.
\displaystyle \text{And }p=1-q=\frac{1}{2}.
\displaystyle np=16.
\displaystyle \Rightarrow n=32.
\displaystyle \therefore \mathrm{P}(X=r)={}^{32}C_{r}\left(\frac{1}{2}\right)^{r}\left(\frac{1}{2}\right)^{32-r},\; r=0,1,2,\ldots,32.
\displaystyle \Rightarrow \mathrm{P}(X=0)=\left(\frac{1}{2}\right)^{32}.
\displaystyle \mathrm{P}(X=1)={}^{32}C_{1}\left(\frac{1}{2}\right)^{32}=\frac{32}{2^{32}}.
\displaystyle \mathrm{P}(X\geq2)=1-\mathrm{P}(X=0)-\mathrm{P}(X=1).
\displaystyle =1-\left(\frac{1}{2}\right)^{32}-\frac{32}{2^{32}}.
\displaystyle =1-\frac{1+32}{2^{32}}.
\displaystyle =1-\frac{33}{2^{32}}.

\displaystyle \textbf{Question 12: }~\text{In eight throws of a die }5\text{ or }6\text{ is considered a success, find} \\ \text{the mean number of successes and the standard deviation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the number of successes in }8\text{ throws.}
\displaystyle n=8.
\displaystyle p=\text{probability of getting }5\text{ or }6=\frac{2}{6}=\frac{1}{3}\text{ and }q=\frac{2}{3}.
\displaystyle \text{Mean }(np)=\frac{8}{3}.
\displaystyle \text{Variance }(npq)=\frac{16}{9}.
\displaystyle \text{Standard deviation }=\sqrt{\text{Variance}}=\frac{4}{3}.
\displaystyle \text{So, mean }\approx2.67\text{ and standard deviation }\approx1.33.

\displaystyle \textbf{Question 13: }~\text{Find the expected number of boys in a family with }8\text{ children,} \\ \text{assuming the sex distribution to be equally probable.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n=8.
\displaystyle \text{Let }p\text{ be the probability of a boy in the family.}
\displaystyle p=\frac{1}{2},\; q=\frac{1}{2}.
\displaystyle \text{Expected number of boys }=\text{ mean}.
\displaystyle \Rightarrow np=4.

\displaystyle \textbf{Question 14: }~\text{The probability is }0.02\text{ that an item produced by a factory is defective.} \\ \text{A shipment of }10{,}000\text{ items is sent to its warehouse. Find the expected number of} \\ \text{defective items and the standard deviation.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n=10000.
\displaystyle \text{Let }p\text{ (the probability of getting a defective item)}=0.02.
\displaystyle q=1-0.02=0.98.
\displaystyle \text{Mean }=\text{ expected number of defective items.}
\displaystyle \Rightarrow np=10000(0.02)=200.
\displaystyle \text{Variance }(npq)=200(0.98)=196.
\displaystyle \text{Standard deviation }=\sqrt{\text{Variance}}=14.
\displaystyle \text{So, mean }=200\text{ and standard deviation }=14.

\displaystyle \textbf{Question 15: }~\text{A die is thrown thrice. A success is }1\text{ or }6\text{ in a throw. Find the} \\ \text{mean and variance of the number of successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }n=3.
\displaystyle p=\text{probability of getting }1\text{ or }6=\frac{1}{3}.
\displaystyle \text{And }q=1-\frac{1}{3}=\frac{2}{3}.
\displaystyle \text{Mean }=np=1.
\displaystyle \text{Variance }=npq=\frac{2}{3}.

\displaystyle \textbf{Question 16: }~\text{If a random variable }X\text{ follows binomial distribution with mean }\\ 3\text{ and variance }\frac{3}{2},\text{ find }P(X\le 5).\text{ \ [CBSE\ 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{Mean }(np)=3\text{ and variance }(npq)=\frac{3}{2}.
\displaystyle \therefore q=\frac{npq}{np}=\frac{\frac{3}{2}}{3}=\frac{1}{2}.
\displaystyle \text{And }p=1-\frac{1}{2}=\frac{1}{2}.
\displaystyle n=\frac{\text{Mean}}{p}.
\displaystyle \Rightarrow n=6.
\displaystyle \text{Hence, the distribution is given by }
\displaystyle \mathrm{P}(X=r)={}^{6}C_{r}\left(\frac{1}{2}\right)^{r}\left(\frac{1}{2}\right)^{6-r},\; r=0,1,2,\ldots,6.
\displaystyle =\frac{{}^{6}C_{r}}{2^{6}}.
\displaystyle \therefore \mathrm{P}(X\leq5)=1-\mathrm{P}(X=6).
\displaystyle =1-\frac{1}{64}.
\displaystyle =\frac{63}{64}.

\displaystyle \textbf{Question 17: }~\text{If }X\text{ follows binomial distribution with mean }4\text{ and variance }2,\text{ find} \\ P(X\ge 5).\text{ \ [CBSE\ 2001\ C]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, mean }(np)=4\text{ and variance }(npq)=2.
\displaystyle \therefore q=\frac{npq}{np}=\frac{2}{4}=\frac{1}{2}\text{ and }p=\frac{1}{2}.
\displaystyle n=\frac{\text{Mean}}{p}.
\displaystyle =4\times2.
\displaystyle =8.
\displaystyle \text{Hence, the distribution is given by }
\displaystyle \mathrm{P}(X=r)={}^{8}C_{r}\left(\frac{1}{2}\right)^{r}\left(\frac{1}{2}\right)^{8-r},\; r=0,1,2,\ldots,8.
\displaystyle \mathrm{P}(X\geq5)=\mathrm{P}(X=5)+\mathrm{P}(X=6)+\mathrm{P}(X=7)+\mathrm{P}(X=8).
\displaystyle =\left(\frac{1}{2}\right)^{8}\left[{}^{8}C_{5}+{}^{8}C_{6}+{}^{8}C_{7}+{}^{8}C_{8}\right].
\displaystyle =\frac{56+28+8+1}{2^{8}}.
\displaystyle =\frac{93}{256}.

\displaystyle \textbf{Question 18: }~\text{The mean and variance of a binomial distribution are }\frac{4}{3} \text{ and }\frac{8}{9} \\ \text{respectively. Find }P(X\ge 1).\text{ \ [CBSE\ 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Mean }(np)=\frac{4}{3}\text{ and variance }(npq)=\frac{8}{9}.
\displaystyle \therefore q=\frac{npq}{np}=\frac{\frac{8}{9}}{\frac{4}{3}}=\frac{2}{3}.
\displaystyle \text{And }p=1-\frac{2}{3}=\frac{1}{3}.
\displaystyle \text{Therefore, }n=\frac{\text{Mean}}{p}=4.
\displaystyle \text{Hence, the distribution is given by }
\displaystyle \mathrm{P}(X=r)={}^{4}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{4-r},\; r=0,1,2,3,4.
\displaystyle \mathrm{P}(X\geq1)=1-\mathrm{P}(X=0).
\displaystyle =1-\left(\frac{2}{3}\right)^{4}.
\displaystyle =\frac{81-16}{81}.
\displaystyle =\frac{65}{81}.

\displaystyle \textbf{Question 19: }~\text{If the sum of the mean and variance of a binomial distribution for } \\ 6\text{ trials is }\frac{10}{3},\text{ find the distribution. \ [CBSE\ 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Given that }n=6.
\displaystyle \text{The sum of the mean and variance of a binomial distribution for }6\text{ trials is }\frac{10}{3}.
\displaystyle \Rightarrow 6p+6pq=\frac{10}{3}.
\displaystyle \Rightarrow 18p+18p(1-p)=10.
\displaystyle \Rightarrow 18p^{2}-36p+10=0.
\displaystyle \Rightarrow (3p-1)(6p-10)=0.
\displaystyle \Rightarrow p=\frac{1}{3}\text{ or }\frac{5}{3}.
\displaystyle p=\frac{5}{3}\text{ is neglected as it is greater than }1.
\displaystyle \therefore p=\frac{1}{3}.
\displaystyle \Rightarrow q=1-p=\frac{2}{3}.
\displaystyle \text{Hence, the distribution is given by }
\displaystyle \mathrm{P}(X=r)={}^{6}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{6-r},\; r=0,1,2,\ldots,6.

\displaystyle \textbf{Question 20: }~\text{A pair of dice is thrown }4\text{ times. If getting a doublet is considered} \\ \text{a success, find the probability distribution of number of successes and hence find} \\ \text{its mean. \ [CBSE\ 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the number of times a doublet is obtained in four throws.}
\displaystyle \text{Then }p=\text{probability of success in one throw of a pair of dice }=\frac{6}{36}=\frac{1}{6}.
\displaystyle \text{And }q=\frac{5}{6},\; n=4.
\displaystyle \mathrm{P}(X=r)={}^{4}C_{r}\left(\frac{1}{6}\right)^{r}\left(\frac{5}{6}\right)^{4-r},\; r=0,1,2,3,4.
\displaystyle \text{As }n=4\text{ and }p=\frac{1}{6},
\displaystyle \text{mean }=np=\frac{4}{6}=\frac{2}{3}.
\displaystyle \therefore \mathrm{P}(X=r)={}^{4}C_{r}\left(\frac{1}{6}\right)^{r}\left(\frac{5}{6}\right)^{4-r},\; r=0,1,2,3,4.
\displaystyle \text{The distribution is as follows:}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline \mathrm{P}(X) & \frac{5^{4}}{6^{4}} & \frac{4\times5^{3}}{6^{4}} & \frac{6\times5^{2}}{6^{4}} & \frac{4\times5}{6^{4}} & \frac{1}{6^{4}} \end{array}

\displaystyle \textbf{Question 21: }~\text{Find the probability distribution of the number of doublets in three} \\ \text{throws of a pair of dice and hence find its mean. \ [CBSE\ 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of outcomes when two dice are thrown }=6\times6=36.
\displaystyle \text{Let }X\text{ be the number of doublets in three throws of a pair of dice.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=3.
\displaystyle p=\mathrm{P}(\text{getting a doublet in one throw})=\frac{6}{36}=\frac{1}{6}\text{ and }q=\frac{5}{6}.
\displaystyle \therefore \mathrm{P}(X=r)={}^{3}C_{r}\left(\frac{1}{6}\right)^{r}\left(\frac{5}{6}\right)^{3-r},\; r=0,1,2,3.
\displaystyle \text{Mean }(np)=3\left(\frac{1}{6}\right)=\frac{1}{2}.
\displaystyle \text{The distribution is as follows:}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3\\ \hline \mathrm{P}(X) & \frac{125}{216} & \frac{75}{216} & \frac{15}{216} & \frac{1}{216} \end{array}

\displaystyle \textbf{Question 22: }~\text{From a lot of }15\text{ bulbs which include }5\text{ defective, a sample of } 4\text{ bulbs} \\ \text{is drawn one by one with replacement. Find the probability distribution of} \\ \text{number of defective bulbs. Hence, find the mean of the distribution. \ [CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let getting a defective bulb in a trial be a success.}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of getting a defective bulb}=\frac{5}{15}=\frac{1}{3}.
\displaystyle q=\text{probability of getting a non-defective bulb}=1-p=1-\frac{1}{3}=\frac{2}{3}.
\displaystyle \text{Let }X\text{ denote the number of successes in a sample of }4\text{ trials.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with parameters }n=4\text{ and }p=\frac{1}{3}.
\displaystyle \therefore \mathrm{P}(X=r)={}^{4}C_{r}p^{r}q^{4-r}={}^{4}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{4-r}=\frac{{}^{4}C_{r}2^{4-r}}{3^{4}},\; r=0,1,2,3,4.
\displaystyle \text{i.e.}
\displaystyle \mathrm{P}(X=0)=\frac{{}^{4}C_{0}2^{4}}{3^{4}}=\frac{16}{81}.
\displaystyle \mathrm{P}(X=1)=\frac{{}^{4}C_{1}2^{3}}{3^{4}}=\frac{32}{81}.
\displaystyle \mathrm{P}(X=2)=\frac{{}^{4}C_{2}2^{2}}{3^{4}}=\frac{24}{81}.
\displaystyle \mathrm{P}(X=3)=\frac{{}^{4}C_{3}2^{1}}{3^{4}}=\frac{8}{81}.
\displaystyle \mathrm{P}(X=4)=\frac{{}^{4}C_{4}2^{0}}{3^{4}}=\frac{1}{81}.
\displaystyle \text{So, the probability distribution of }X\text{ is given as follows:}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline \mathrm{P}(X) & \frac{16}{81} & \frac{32}{81} & \frac{24}{81} & \frac{8}{81} & \frac{1}{81} \end{array}
\displaystyle \text{Now,}
\displaystyle \text{Mean, }E(X)=0\times\frac{16}{81}+1\times\frac{32}{81}+2\times\frac{24}{81}+3\times\frac{8}{81}+4\times\frac{1}{81}.
\displaystyle =\frac{32+48+24+4}{81}.
\displaystyle =\frac{108}{81}.
\displaystyle =\frac{4}{3}.
\displaystyle \text{Note: We can also calculate the mean of the binomial distribution by}
\displaystyle \text{Mean, }E(X)=np=4\times\frac{1}{3}=\frac{4}{3}.

\displaystyle \textbf{Question 23: }~\text{A die is thrown three times. Let }X\text{ be the number of twos seen.} \\ \text{Find the expectation of }X. 
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of getting the number two in a throw}=\frac{1}{6}.
\displaystyle \text{And }q=1-p=1-\frac{1}{6}=\frac{5}{6}.
\displaystyle \text{Let }X\text{ denote the number of twos seen.}
\displaystyle \text{So, }X\text{ follows a binomial distribution with parameters }n=3\text{ and }p=\frac{1}{6}.
\displaystyle \therefore E(X)=np=3\times\frac{1}{6}=\frac{1}{2}.

\displaystyle \textbf{Question 24: }~\text{A die is tossed twice. A `success' is getting an even number on a toss.} \\ \text{Find the variance of number of successes.  }
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of getting an even number on a toss}=\frac{3}{6}=\frac{1}{2}.
\displaystyle \text{And }q=1-p=1-\frac{1}{2}=\frac{1}{2}.
\displaystyle \text{Let }X\text{ denote a success of getting an even number on a toss.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with parameters }n=2\text{ and }p=\frac{1}{2}.
\displaystyle \therefore \mathrm{Var}(X)=npq=2\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{2}.

\displaystyle \textbf{Question 25: }~\text{Three cards are drawn successively with replacement from a well} \\ \text{shuffled pack of }52\text{ cards. Find the probability distribution of the number of spades.} \\ \text{Hence, find the mean of the distribution. \ [CBSE\ 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle p=\text{probability of getting a spade in a draw}=\frac{13}{52}=\frac{1}{4}.
\displaystyle \text{And }q=1-p=1-\frac{1}{4}=\frac{3}{4}.
\displaystyle \text{Let }X\text{ denote a success of getting a spade in a throw.}
\displaystyle \text{Then }X\text{ follows a binomial distribution with parameters }n=3\text{ and }p=\frac{1}{4}.
\displaystyle \therefore \mathrm{P}(X=r)={}^{3}C_{r}p^{r}q^{3-r}={}^{3}C_{r}\left(\frac{1}{4}\right)^{r}\left(\frac{3}{4}\right)^{3-r}=\frac{{}^{3}C_{r}3^{3-r}}{4^{3}},\; r=0,1,2,3.
\displaystyle \text{So, the probability distribution of }X\text{ is given by:}
\displaystyle \mathrm{P}(X=r)=\frac{27}{64}\left(\frac{{}^{3}C_{r}}{3^{r}}\right),\; r=0,1,2,3.
\displaystyle \text{Now,}
\displaystyle \text{Mean, }E(X)=np=3\times\frac{1}{4}=\frac{3}{4}.

\displaystyle \textbf{Question 26: }~\text{An urn contains }3\text{ white and }6\text{ red balls. Four balls are drawn one by} \\ \text{one with replacement from the urn. Find the probability distribution of the number of} \\ \text{red balls drawn. Also, find the mean and variance of the distribution. \ [CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ denote the total number of red balls when four balls are drawn one by one with replacement.}
\displaystyle \mathrm{P}(\text{getting a red ball in one draw})=\frac{2}{3}.
\displaystyle \mathrm{P}(\text{getting a white ball in one draw})=\frac{1}{3}.
\displaystyle \text{So, }X\sim\text{Binomial}\left(n=4,\;p=\frac{2}{3}\right).
\displaystyle \text{The probability distribution is:}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline \mathrm{P}(X) & \frac{1}{81} & \frac{8}{81} & \frac{24}{81} & \frac{32}{81} & \frac{16}{81} \end{array}
\displaystyle \text{Using the formula for mean, we have } \overline{X}=\sum P_{i}X_{i}.
\displaystyle \text{Mean }(\overline{X})=0\times\frac{1}{81}+1\times\frac{8}{81}+2\times\frac{24}{81}+3\times\frac{32}{81}+4\times\frac{16}{81}.
\displaystyle =\frac{1}{81}(8+48+96+64).
\displaystyle =\frac{216}{81}.
\displaystyle =\frac{8}{3}.
\displaystyle \text{Using the formula for variance, we have } \mathrm{Var}(X)=\sum P_{i}X_{i}^{2}-\left(\sum P_{i}X_{i}\right)^{2}.
\displaystyle \mathrm{Var}(X)=\left\{0\times\frac{1}{81}+1\times\frac{8}{81}+4\times\frac{24}{81}+9\times\frac{32}{81}+16\times\frac{16}{81}\right\}-\left(\frac{8}{3}\right)^{2}.
\displaystyle =\frac{648}{81}-\frac{64}{9}.
\displaystyle =\frac{8}{9}.
\displaystyle \text{Hence, the mean of the distribution is }\frac{8}{3}\text{ and the variance is }\frac{8}{9}.

\displaystyle \textbf{Question 27: }~\text{Five bad oranges are accidently mixed with }20\text{ good ones. If four} \\ \text{oranges are drawn one by one successively with replacement, then find the probability} \\ \text{distribution of number of bad oranges drawn. Hence, find the mean and variance of the} \\ \text{distribution. \ [CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }X\text{ be the random variable denoting the number of bad oranges drawn.}
\displaystyle \mathrm{P}(\text{getting a good orange})=\frac{20}{25}=\frac{4}{5}.
\displaystyle \mathrm{P}(\text{getting a bad orange})=\frac{5}{25}=\frac{1}{5}.
\displaystyle \text{The probability distribution of }X\text{ is given by}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline \mathrm{P}(X) & \frac{256}{625} & \frac{256}{625} & \frac{96}{625} & \frac{16}{625} & \frac{1}{625} \end{array}
\displaystyle \text{Mean of }X\text{ is given by } \overline{X}=\sum P_{i}X_{i}.
\displaystyle =0\times\frac{256}{625}+1\times\frac{256}{625}+2\times\frac{96}{625}+3\times\frac{16}{625}+4\times\frac{1}{625}.
\displaystyle =\frac{1}{625}(256+192+48+4).
\displaystyle =\frac{500}{625}.
\displaystyle =\frac{4}{5}.
\displaystyle \text{Variance of }X\text{ is given by } \mathrm{Var}(X)=\sum P_{i}X_{i}^{2}-\left(\sum P_{i}X_{i}\right)^{2}.
\displaystyle =\left\{0\times\frac{256}{625}+1\times\frac{256}{625}+4\times\frac{96}{625}+9\times\frac{16}{625}+16\times\frac{1}{625}\right\}-\left(\frac{4}{5}\right)^{2}.
\displaystyle =\frac{800}{625}-\frac{16}{25}.
\displaystyle =\frac{400}{625}.
\displaystyle =\frac{16}{25}.
\displaystyle \text{Thus, the mean and variance of the distribution are }\frac{4}{5}\text{ and }\frac{16}{25},\text{ respectively.}

\displaystyle \textbf{Question 28: }~\text{Three cards are drawn successively with replacement from a well shuffled} \\ \text{pack of }52\text{ cards. Find the mean and variance of number of red cards. \ [CBSE\ 2017]}
\displaystyle \text{Answer:}
\displaystyle \textbf{Question 28: }\text{Three cards are drawn successively with replacement from a well shuffled pack of }52\text{ cards. Find the mean and variance of number of red cards.}
\displaystyle \text{Let }X\text{ denote the number of red cards drawn in }3\text{ draws (with replacement).}
\displaystyle \text{Then }X\text{ follows a binomial distribution with }n=3\text{ and }p=\mathrm{P}(\text{red card})=\frac{26}{52}=\frac{1}{2}.
\displaystyle \text{So }q=1-p=\frac{1}{2}.
\displaystyle \text{Mean }=E(X)=np=3\times\frac{1}{2}=\frac{3}{2}.
\displaystyle \text{Variance }=\mathrm{Var}(X)=npq=3\times\frac{1}{2}\times\frac{1}{2}=\frac{3}{4}.
\displaystyle \text{Answer: Mean }=\frac{3}{2}\text{ and Variance }=\frac{3}{4}.

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