# A die is thrown 6

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  Rate this question :

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