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Question

An irregular six faced die is thrown and the expectation that in 10 throws it will give 5 even numbers is twice the expectation that it will give 4 even numbers. The number of times in 10000 sets of 10 throws you expect to give no even numbers is

A
6
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B
1
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C
2
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D
9
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Solution

The correct option is B 1
Let the random variable X denote the number of even numbers.
P(getting x even numbers)
=P(X=x) nCxpxq10x,x=0,1,2,...,n
Given n=10
P(X=x)=10Cxpxq10x,x=0,1,2,...10
P(getting 5 even numbers) =2P(getting 4 even numbers)
P(X=5)=2P(X=4)=10C5p5q5

=2(10C4p4q6)

p4=q33p=5q=5(1p)

8p=5p=58

q=1p=158=38
P(getting x even numbers)
10Cx(58)x(38)10x,x=0,1,2,...n
the required number of times that in 10000 sets of 10 throws each we get no even number.
=1000×P(X=0)=1000×10C0(58)0(38)10=1 (approx)

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