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) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

= ⁶C_x (1/2)^6-x(1/2)^x

= ⁶C_x (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