\displaystyle \textbf{Question 1:}\ \ \text{The mean and variance of a binomial distribution are }4\text{ and }\\ \frac{4}{3}\text{ respectively, find }P(X\ge1).\qquad [\text{CBSE 2004, 2005}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ be a binomial variate with parameters }n\text{ and }p.\text{ Then,}
\displaystyle \text{Mean }=np\ \text{and Variance }=npq\qquad [\because\ \text{Mean }=4,\ \mathrm{Var}(X)=\frac{4}{3}\ (\text{Given})]
\displaystyle \Rightarrow np=4\ \text{and }\ npq=\frac{4}{3}
\displaystyle \Rightarrow \frac{npq}{np}=\frac{4/3}{4}\Rightarrow q=\frac{1}{3}\Rightarrow p=1-\frac{1}{3}=\frac{2}{3}\qquad [\because\ p=1-q]
\displaystyle \text{Putting }p=\frac{2}{3}\text{ in }np=4,\text{ we get}
\displaystyle n\times\frac{2}{3}=4\Rightarrow n=6
\displaystyle \text{Thus, we have }n=6,\ p=\frac{2}{3}\text{ and }q=\frac{1}{3}
\displaystyle \therefore\ P(X=r)={}^{n}C_{r}p^{r}q^{n-r}
\displaystyle \Rightarrow P(X=r)={}^{6}C_{r}\left(\frac{2}{3}\right)^{r}\left(\frac{1}{3}\right)^{6-r},\ r=0,1,2,\ldots,6
\displaystyle \text{Now, }\ P(X\ge1)=1-P(X<1)=1-P(X=0)=1-{}^{6}C_{0}\left(\frac{2}{3}\right)^{0}\left(\frac{1}{3}\right)^{6}=1-\left(\frac{1}{3}\right)^{6}=1-\frac{1}{729}=\frac{728}{729}

\displaystyle \textbf{Question 2:}\ \ \text{If the sum of the mean and variance of a binomial distribution for 5} \\ \text{trials is }1.8,\text{ find the distribution.}\qquad [\text{CBSE 2004}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }n\text{ and }p\text{ be the parameters of the distribution. Then,}
\displaystyle \text{Mean }=np\ \text{and Variance }=npq
\displaystyle \text{It is given that}
\displaystyle n=5\ \text{and, Mean+Variance }=1.8
\displaystyle \text{Now, Mean+Variance }=1.8
\displaystyle \Rightarrow np+npq=1.8
\displaystyle \Rightarrow 5p+5pq=1.8
\displaystyle \Rightarrow 5p+5p(1-p)=1.8
\displaystyle \Rightarrow 5p^{2}-10p+1.8=0
\displaystyle \Rightarrow p^{2}-2p+0.36=0
\displaystyle \Rightarrow (p-0.2)(p-1.8)=0\Rightarrow p=0.2\qquad [\because\ p\le1]
\displaystyle \text{Thus, we have}
\displaystyle n=5,\ p=0.2\ \text{and }\ q=0.8
\displaystyle \text{Therefore, if }X\text{ denotes the binomial variate, then}
\displaystyle P(X=r)={}^{5}C_{r}(0.2)^{r}(0.8)^{5-r},\ r=0,1,2,3,4,5
\displaystyle \text{This is the required binomial distribution.}

\displaystyle \textbf{Question 3:}\ \ \text{If two dice are rolled 12 times, obtain the mean and the variance of the} \\ \text{distribution of successes, if getting a total greater than 4 is considered a success.} \\ \qquad [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }X\text{ denote the number of successes in 12 trials. Then, }X\text{ follows binomial} \\ \text{distribution with parameters }n=12\text{ and }p.
\displaystyle \text{We have,}
\displaystyle p=\text{Probability of getting a total greater than 4 in a single throw of a pair of dice.}
\displaystyle \Rightarrow p=1-\text{Probability of getting a total less than or equal to }4
\displaystyle \Rightarrow p=1-\frac{6}{36}=\frac{5}{6}
\displaystyle \therefore\ q=1-p=1-\frac{5}{6}=\frac{1}{6}
\displaystyle \text{Mean }=np=\frac{5}{6}\times12=10\ \text{and, Variance }=npq=12\times\frac{5}{6}\times\frac{1}{6}=\frac{5}{3}

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