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?

Answer 1

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`