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?
We know that the repeated tosses of a dice are known as Bernouli trials.
Let the number of successes of getting an odd number in an experiment of 6 trials be x.
Probability of getting an odd number in a single throw of a dice(p) 
Thus,
Now, here x has a binomial distribution.
Thus, P(X = x) = nCxqn-xpx , where x = 0, 1, 2, …n
= 6Cx (1/2)6-x(1/2)x
= 6Cx (1/2)6
(i) Probability of getting 5 successes = P(X = 5)
= 6C5(1/2)6
(ii) Probability of getting at least 5 successes = P(X ≥ 5)
= P(X = 5) + P(X = 6)
= 6C5(1/2)6 + 6C5 (1/2)6
(iii) Probability of getting at most 5 successes = P(X ≤ 5)
We can also write it as: 1 – P(X>5)
= 1 – P(X = 6)
= 1 – 6C6 (1/2)6
