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Question
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes?
(ii) at least 5 successes?
(iii) at most 5 successes?
Solution
The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.
Probability of getting an odd number in a single throw of a die is p = 3/6 =1/2
`:. q = 1 - p=1/2`
X has a binomial distribution.
Therefore, P (X = x) = `""^n"C"_(n-x) q^(n-x) p^x, "where" n = 0,1,2 ...n`
= `""^6"C"_x (1/2)^(6-x) .(1/2)^x`
= `""^6"C"_x(1/2)^6`
(i) P (5 successes) = P (X = 5)
= `""^6"C"_5(1/2)^6`
= `6·1/64`
= `3/32`
(ii) P(at least 5 successes) = P(X ≥ 5)
= P(x =5)+P(x=6)
= `""^6"C"_5 (1/2)^6 + ""^6"C"_6(1/2)^6`
= `6·1/64+1·1/64`
= `7/64`
(iii) P (at most 5 successes) = P(X ≤ 5)
= l - P(X>5)
= l - P (X =6)
= `l - ""^6"C"_6 (1/2)^6`
= `l - 1/64`
= `63/64`
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