Three rotten apples are mixed accidently with seven good apples and four apples are drawn …

Question

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 :

Accepted Answer

3 rotten apples are mixed with 7 good apples.

$$\therefore$$ Total apples = 10

Among those 10 apples 4 are chosen randomly.

${x_i}$
    ${p_i}$

${p_i}{x_i}$
    ${p_i}{({x_i})^2}$

0
    $${{{}^7{C_4}} \over {{}^{10}{C_4}}} = {{35} \over {210}}$$
    0
    0

1
    $${{{}^3{C_1} \times {}^7{C_3}} \over {{}^{10}{C_4}}} = {{105} \over {210}}$$
    $${{105} \over {210}}$$
    $${{105} \over {210}}$$

2
    $${{{}^3{C_2} \times {}^7{C_2}} \over {{}^{10}{C_4}}} = {{63} \over {210}}$$
    $${{126} \over {210}}$$
    $${{252} \over {210}}$$

3
    $${{{}^3{C_3} \times {}^7{C_1}} \over {{}^{10}{C_4}}} = {7 \over {210}}$$
    $${{21} \over {210}}$$
    $${{63} \over {210}}$$

${x_i}$ = Number of rotten apples drawn.

${p_i}$ = Probability of rotten apple.

We know,

Mean $$(\mu ) = \sum {{p_i}{x_i}} $$

$$ = 0 + {{105} \over {210}} + {{126} \over {210}} + {{21} \over {210}}$$

$$ = {{252} \over {210}} = {6 \over 5}$$

Also,

Variance $$({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}$$

$$ = {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}$$

$$ = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}$$

$$\therefore$$ $$10(\mu^2  + {\sigma ^2})$$

$$ = 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)$$

$$ = 10\left( {{{36 + 14} \over {25}}} \right)$$

$$ = 10 \times {{50} \over {25}} = 20$$

Solutions · 1