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 5 times the total number of outcomes will be 65.
And we know that the favourable outcomes of getting at least 4 successes will be, either getting 1, 3 or 5 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 \(\frac{3}{6}\)
\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)
The probability of getting at least 4 successes = P(4)+P(5)
