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Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable \mathrm{X} denote the number of rotten apples. If \mathrm{\mu \: \: and\: \: \sigma ^{2}} represent mean and variance of \mathrm{X}, respectively, then \mathrm{10(\mu ^{2}+\sigma ^{2})} is equal to

Option: 1

250


Option: 2

25


Option: 3

30


Option: 4

20


Answers (1)

best_answer

Total Apple = 10, Rotten apple = 3, good apple = 7
Prob. of rotten apple \mathrm{\left ( P \right )=\frac{3}{10}}
Prob. of good apple \mathrm{\left ( q \right )=\frac{7}{10}}

\mathrm{X\rightarrow }Number of rotten apples

here \mathrm{x = 0, 1, 2, 3}

\begin{aligned} & \mathrm{p}(\mathrm{x}=0)={ }^4 \mathrm{C}_0\left(\frac{3}{10}\right)^0 \times \frac{7}{10} \times \frac{6}{9} \times \frac{5}{8} \times \frac{4}{7}=\frac{1}{6} \\ & p(x=1)={ }^4 \mathrm{C}_1\left(\frac{3}{10}\right) \times \frac{7}{9} \times \frac{6}{8} \times \frac{5}{7}=\frac{1}{2} \\ & p(x=2)={ }^4 \mathrm{C}_2\left(\frac{3}{10} \times \frac{2}{9}\right) \times \frac{7}{8} \times \frac{6}{7}=\frac{3}{10} \\ & p(x=3)={ }^4 C_3\left(\frac{3}{10} \times \frac{2}{9} \times \frac{1}{8}\right) \times \frac{7}{7}=\frac{1}{30} \end{aligned}

\mathrm{x_{i}} 0 1 2 3
\mathrm{p_{i}} \frac{35}{210} \frac{105}{210} \frac{3}{10} \frac{1}{30}

Now,

\begin{aligned} & \mu=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{1}{6} \times 0+\frac{1}{2} \times 1+2 \times \frac{3}{10}+3 \times \frac{1}{30}=\frac{6}{5} \\ & \text { and } \sigma^2=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\mu^2=\frac{1}{2}+\frac{3}{10} \times 4+\frac{1}{30} \times 9-\frac{36}{25}=\frac{14}{25} \\ & \therefore 10\left(\mu^2+\sigma^2\right)=10\left(\frac{36}{25}+\frac{14}{25}\right) \\ & =10 \times\left(\frac{50}{25}\right)=10 \times 2 \\ \end{aligned}

                  =20

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Pankaj

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