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Provide solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise 31.1 question 20 maths textbook solution

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Answer:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{969}{2530}$	$\frac{114}{253}$	$\frac{38}{253}$	$\frac{4}{253}$	$\frac{1}{2530}$

Given: from a lot containing 25 items, 5 of which are defective, 4 are chosen at random.Hint: use probability formula and combination formula

Let X be the number of defectives found obtains the probability distribution of x if the items are chose without replacement

Solution: X represents the number of defective items drawn.

∴x can take value 0,1,2,3,4

∴ there are total 25 items ( 20 good + 5 defective) items

N(s) = total possible ways of selling 5 items 25 $c_{4}$

P (x=0)= P (selecting no defective item)

$\frac{20 c_{4}}{25 c_{4}}=\frac{969}{2530}$

P (x=1)= P (selecting 1 defective item and 3 good item)

$=\frac{5 c_{2} \times 20 c_{2}}{25 c_{4}}=\frac{38}{253}$

P (x=2) = P ( selecting 2 defective item and 2 good item)

$=\frac{5 c_{2} \times 20 c_{2}}{25 c_{4}}=\frac{38}{253}$

P (x=3) = P ( selecting 3 defective item and 1 good item)

$=\frac{5 c_{3} \times 20 c_{1}}{25 c_{4}}=\frac{4}{253}$

P (x=4) =P (selecting 4 defective items and no good item)

$=\frac{5 c_{4}}{25 c_{4}}=\frac{1}{2530}$

Now we have $P_{1}$ and $X_{1}$

The following table gives probability Distribution of x:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{969}{2530}$	$\frac{114}{253}$	$\frac{38}{253}$	$\frac{4}{253}$	$\frac{1}{2530}$

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