In 4 throws of a pair of dice, what is the probability of throwing doublets at least twice? A. 7/36

Question

In 4 throws of a pair of dice, what is the probability of throwing doublets at least twice?

A. \(\frac{7}{36}\)

B. \(\frac{17}{144}\)

C. \(\frac{19}{144}\)

D. None of these

Answer 1

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 we know that the favourable outcomes of getting at least doublets twice are, successes will be, getting a doublet, i.e.,

p = \(\frac{1}{6},\) q = \(\frac{5}{6}\)

The probability of success is \(\frac{1}{6}\) and of failure is also \(\frac{5}{6}\)

\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)

The probability of getting at least 2 successes = P(2)+P(3)+P(4)