JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 3)
Explanation
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 $$({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}$$
$$ = {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}$$
$$ = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}$$
$$\therefore$$ $$10(\mu^2 + {\sigma ^2})$$
$$ = 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)$$
$$ = 10\left( {{{36 + 14} \over {25}}} \right)$$
$$ = 10 \times {{50} \over {25}} = 20$$
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