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A die is thrown 6 times. If ‘getting an even number’ is a success, find the probability of getting

(i) exactly 5 successes

(ii) at least 5 successes

(iii) at most 5 successes

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(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)

As the die is thrown 6 times the total number of outcomes will be 66.

And we know that the favourable outcomes of getting exactly 5 successes will be, either getting 2, 4 or 6 i.e., 1/6 probability of each, total, \(\frac{3}{6}\) probability, p = \(\frac{1}{2},\) q = \(\frac{1}{2}\)

The probability of success is \(\frac{3}{6}\) and of failure is also \(\frac{3}{6}.\)

Thus, the probability of getting exactly 5 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)

As the die is thrown 6 times the total number of outcomes will be \(6^6\)

And we know that the favourable outcomes of getting at least 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, \(\frac{3}{6}\) probability, p = \(\frac{3}{6}\), q = \(\frac{3}{6}\)

The probability of success is \(\frac{3}{6}\) and of failure is also \(\frac{3}{6}.\)

Thus, the probability of getting at least 5 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)

As the die is thrown 6 times the total number of outcomes will be \(6^6.\)

And we know that the favourable outcomes of getting at most 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, \(\frac{3}{6}\) probability of success.

The probability of success is \(\frac{3}{6}\) and of failure is also \(\frac{3}{6}.\)

Thus, the probability of getting at most 5 successes will be

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