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 B1 Let the random variable X denote the number of even numbers. P(getting x even numbers) =P(X=x)nCxpxq10−x,x=0,1,2,...,n Given n=10 ∴P(X=x)=10Cxpxq10−x,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=q3⇒3p=5q=5(1−p)
⇒8p=5⇒p=58
∴q=1−p=1−58=38 ∴P(getting x even numbers) 10Cx(58)x(38)10−x,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)