(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