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Three rotten apples were accidentally mixed with seven good apples, and the four were pulled out one by one without replacement. Let X be a random variable the number of bad apples.If   $\mu$ and $\sigma^2$ denote the mean and variance of X respectively then $10\left(\mu^2+\sigma^2\right)$ then equal to:

$X_i$ $p_i$ $p_ix_i$ $p_i\left(x_i\right)$

0 $\frac{_{}^{7}\textrm{C}_{4}}{_{}^{10}\textrm{C}_{4}}=\frac{35}{100}$ $0$ $0$

1 $\frac{_{}^{3}\textrm{C}_{1}\times _{}^{7}\textrm{C}_{3}}{_{}^{10}\textrm{C}_{4}}=\frac{105}{210}$ $\frac{105}{210}$ $\frac{105}{210}$

2 $\frac{_{}^{3}\textrm{C}_{2}\times _{}^{7}\textrm{C}_{2}}{_{}^{10}\textrm{C}_{4}}=\frac{63}{210}$ $\frac{126}{210}$ $\frac{126}{210}$

3 $\frac{_{}^{3}\textrm{C}_{3}\times _{}^{7}\textrm{C}_{1}}{_{}^{10}\textrm{C}_{4}}=\frac{7}{210}$ $\frac{21}{210}$ $\frac{63}{210}$

Here, $X_i$   is the number of rotten apples.

And $p_i$   is the probability of a rotten apple.

We know that:

Mean, $\mu=\sum p_iX_i$

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

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

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

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

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

Or,      $10\left(\mu^2+\sigma^2\right)$

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

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

Option: 1
$20$

Option: 2
$30$

Option: 3
$250$

Option: 4
$25$

Answers (1)

best_answer

$X_i$	$p_i$	$p_ix_i$	$p_i\left(x_i\right)^2$
0
1
2
3

Posted by

Kshitij

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