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) = ^{n}C_{x}q^{n-x}p^{x} , where x = 0, 1, 2, …n

= ^{6}C_{x} (1/2)^{6-x}(1/2)^{x}

= ^{6}C_{x} (1/2)^{6}

(i) Probability of getting 5 successes = P(X = 5)

= ^{6}C_{5}(1/2)^{6}

(ii) Probability of getting at least 5 successes = P(X ≥ 5)

= P(X = 5) + P(X = 6)

= ^{6}C_{5}(1/2)^{6} + ^{6}C_{5} (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 – ^{6}C_{6} (1/2)^{6}

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