NIMCET 2015 Question Paper with solutions

Question : 9 of 120

Marks: +1, -0

If a fair dice is rolled successively, then the probability that 1 appears in an even numbered throw is;

Solution:

Let A be the event 1 appear on the die

∴P(A)=1/6,P(\ov{A})=1-1/6=5/6

Now, 1 appears in an even numbered throw.
Required probability

=P(\ov{A})⋅P(A)+P(\ov{A})⋅P(\ov{A})⋅P(\ov{A})⋅P(\ov{A})+

={5}/{6}×{1}/{6}+{5}/{6}×{5}/{6}×{5}/{6}×{1}/{6}+. . .

={5}/{6}×{1}/{6}[1+{25}/{36}+({25}/{36})^{2}...]={5}/{36}({1}/{1-{25}/{36}})={5}/{11}

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