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A man takes a step forward with probability `0.4` and backward with probability `0.6`. The probability that at the end of eleven steps he is just one step away from the starting point, is A. `.^11(C)_(5)(0.4)^6(0.6)^5` B. `.^11(C)_(6)(0.4)^5(0.6)^6` C. `.^11(C)_(5)(0.4)^5(0.6)^6` D. `.^n(C)_(5)(0.4)^5(0.6)^5` Select the correct answer from above options

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Correct Answer - C Let p denote the probability that the mant eks a step foraward. Then, `p=0.4` `therefore q=1 -p=1-0.4=0.6` Let X denote the number of steps taken in the forward direction. Since the stpes are idependent of each other, therefore X is a binomial variate with parameters `n=11` and `p=0.4` such that `P(X=r)=.^11C_r(0.4)^r(0.6)^11 -r, r=0,1,2,.....,11 ..(i)` Since the man is one step away from the initial poin, he is either one step forward or one step backward from the initial point at the end of eleven steps. If he is one step forward, then he must have taken six steps, forward and five steps backward and if he is one step backward, then he must have taken five steps forward and six backward. Thus , either `X=6 or X=5`. `therefore` Required Probability `=P[(X=5)or (X=6)]` `P(X=5)+P(X=6)` `=.^11C_5(0.4)^5(0.6)^(11-5)+.^11C_6(0.6)^6(0.6)^(11-6)["Using"(i)]` `=.^11C_5(0.4)^5(0.6)^5[0.6_0.4]` `[because .^11C_5=.^11C_6]` `=.^C_5(0.4)^5(0.6)^5`

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