The average number of 5s in 20 throws is given as 4.8. This implies that the expected value for the number of 5s is 4.8, which can be expressed as:
\(20p = 4.8\)
Solving for \(p\), we get:
\(p = \frac{4.8}{20} = 0.24\)
We need to find the probability that the number of 5s is less than 3 in 20 throws. This is a binomial distribution problem where \(n = 20\) and \(p = 0.24\).
The probability of getting exactly \(k\) successes in a binomial distribution is given by:
\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
We need to calculate \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).
Calculating each term:
\(P(X = 0) = \binom{20}{0} (0.24)^0 (0.76)^{20} = (0.76)^{20}\)
\(P(X = 1) = \binom{20}{1} (0.24)^1 (0.76)^{19} = 20 \times 0.24 \times (0.76)^{19}\)
\(P(X = 2) = \binom{20}{2} (0.24)^2 (0.76)^{18} = \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)
Summing these probabilities:
\(P(X < 3) = (0.76)^{20} + 20 \times 0.24 \times (0.76)^{19} + \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)
Calculating the above gives:
\(P(X < 3) = 0.109\)
