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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 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 :

JEE · Math · previous-year question

  1. A.20correct
  2. B.30
  3. C.250
  4. D.25

Answer

A. 20

Explanation

3 rotten apples are mixed with 7 good apples. $$\therefore$$ Total apples = 10 Among those 10 apples 4 are chosen randomly. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} $${x_i}$$ $${p_i}$$ $${p_i}{x_i}$$ $${p_i}{({x_i})^2}$$ 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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