Question

A die is thrown 4 times. ‘Getting a 1 or a 6’ is considered a success, Find the probability of getting

(i) exactly 3 successes

(ii) at least 2 successes

(iii) at most 2 successes

Answer 1

(i) Using Bernoulli’s Trial P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)

x = 0, 1, 2, ………n and q = (1-p)

We know that the favourable outcomes of getting exactly 3 successes will be, either getting 1 or a 6 i.e, total, \(\frac{2}{6}\) probability

The probability of success is \(\frac{2}{6}\) and of failure is \(\frac{4}{6}.\)

Thus, the probability of getting exactly 3 successes will be

(ii) Using Bernoulli’s Trial P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)

x = 0, 1, 2, ………n and q = (1-p)

We know that the favourable outcomes of getting at least 2 successes will be, either getting 1 or a 6 i.e, total, \(\frac{2}{6}\) probability

The probability of success is \(\frac{2}{6}\) and of failure is \(\frac{4}{6}.\)

Thus, the probability of getting at least 2 successes will be

(iii) Using Bernoulli’s Trial P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)

x = 0, 1, 2, ………n and q = (1-p)

We know that the favourable outcomes of getting at most 2 successes will be, either getting 1 or a 6 i.e, total, \(\frac{2}{6}\) probability

The probability of success is \(\frac{2}{6}\) and of failure is \(\frac{4}{6}.\)

Thus, the probability of getting at most 2 successes will be