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Question

An urn contains 4 white and 3 red balls.

Three balls are drawn with replacement from this urn.

Then, the standard deviation of the number of red balls drawn is


A

67

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B

3649

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C

57

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D

2549

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Solution

The correct option is A

67


Explanation for correct option:

Step1. Finding the probability of red ball

Given, An urn contains 4 white and 3 red balls.

Three balls are drawn with replacement from this urn.

Let X represent the number of red balls.

P be the probability of getting red balls =37

Hence, the probability of not getting red balls,q=47

P(X=0)=C03×(37)0×(47)3

=64343

P(X=1)=C13×(37)1×(47)2

=144343

P(X=2)=C23×(37)2×(47)1

=108343

P(X=3)=C33×(37)3×(47)0

=27343

Step2. Finding variance

We know that, variancei=03pixi2-[pixi]2i=03

X=red ballP(X)XP(X)X2P(X)
06434300
1144343144343144343
2108343216343432343
32734381343243343

i=03pixi2=0+144343+432343+243343=819343

[pixi]i=03=[0+144343+216343+81343]=[441343]

Put all the value, we get

Variance =819343-(441343)2

=3649

Standard deviation =variance

Therefore standard deviation =67

Hence, correct option is (A)


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