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




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