JEE Main 2023 — Statistics Question with Solution

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	$$\frac{{}^7{C}_4}{{}^{10}{C}_4}=\frac{35}{210}$$	0	0
1	$$\frac{{}^3{C}_1\times {}^7{C}_3}{{}^{10}{C}_4}=\frac{105}{210}$$	$$\frac{105}{210}$$	$$\frac{105}{210}$$
2	$$\frac{{}^3{C}_2\times {}^7{C}_2}{{}^{10}{C}_4}=\frac{63}{210}$$	$$\frac{126}{210}$$	$$\frac{252}{210}$$
3	$$\frac{{}^3{C}_3\times {}^7{C}_1}{{}^{10}{C}_4}=\frac{7}{210}$$	$$\frac{21}{210}$$	$$\frac{63}{210}$$

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

$$p_i$$ = Probability of rotten apple.

We know,

Mean $$(\mu )=\sum p_i{x}_i$$

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

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

Also,

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

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

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

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

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

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

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