InterviewSolution
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InterviewSolution
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A dice is thrown `(2n+1)` times. The probability that faces with even numbers show odd number of times isA. `(1)/(2)`B. `lt(1)/(2)`C. `gt(1)/(2)`D. none of these |
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Answer» Correct Answer - A We,have p= Probability of getting an even number in a throw `rArrp=(3)/(6)=(1)/(2)` `therefore q=1-p=(1)/(2)` Let X denote the number of times an even number is shown in `(2n+1)` throws of a dice. Then, X follows binomial distribution such that `P(X=r)=.^2n+1C_(r)((1)/(2))^2n+1-r((1)/(2))^r=.^2n+1C_(r) ((1)/(2))^2n+1`. `therefore` Required Probability `=sum_(r=1)^(2n+1)P(X=r)=sum _(r=1)^(2n+1)2n+1.^C_r((1)/(2))^2n+1` `=((1)/(2))^2n+1{.^(2n+1)C_1+.^(2n+1)C_3+.....+.^(2n+1)C_(2n+1)}` `=(1)/(2^(2n+1))xx2^2n=(1)/(2)`. |
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