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

Total apples ($$N$$ ) = 10
Rotten apples ($$M$$ ) = 3
Good apples ($$N-M$$ ) = 7
Apples drawn ($$n$$ ) = 4 

Using the Distribution formula $$P(X=k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}}$$ :The total ways to draw 4 apples is $$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$ .

• $$P(X=0) = \frac{\binom{3}{0} \binom{7}{4}}{210} = \frac{1 \times 35}{210} = \frac{35}{210}$$
• $$P(X=1) = \frac{\binom{3}{1} \binom{7}{3}}{210} = \frac{3 \times 35}{210} = \frac{105}{210}$$
• $$P(X=2) = \frac{\binom{3}{2} \binom{7}{2}}{210} = \frac{3 \times 21}{210} = \frac{63}{210}$$
• $$P(X=3) = \frac{\binom{3}{3} \binom{7}{1}}{210} = \frac{1 \times 7}{210} = \frac{7}{210}$$

For the any distribution, we know that $$\sigma^2 = E[X^2] - \mu^2$$, so $$10(\mu^2 + \sigma^2) = 10E[X^2]$$ 

Now we calculate $$E[X^2]$$ :

$$E[X^2] = \sum x_i^2 P(x_i)$$ $$E[X^2] = 0^2\left(\frac{35}{210}\right) + 1^2\left(\frac{105}{210}\right) + 2^2\left(\frac{63}{210}\right) + 3^2\left(\frac{7}{210}\right)$$

$$E[X^2] = \frac{105 + 252 + 63}{210} = \frac{420}{210} = 2.0 $$

$$10(\mu^2 + \sigma^2) = 10 \times E[X^2] = 10 \times 2.0 = 20$$ 

The final value is 20.