The probability of obtaining a total less than 4 when throwing two dice is given by the outcomes (1,1), (1,2), and (2,1). The probability for each of these outcomes is:
\(\frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{3}{36} = \frac{1}{12}\)
We need to find the probability of obtaining a total less than 4 on at least three throws out of 10. This is a binomial probability problem where:
\(n = 10, \quad p = \frac{1}{12}\)
We need to calculate:
\(1 - P(0, 1, 2)\)
Using the binomial probability formula:
\(P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\)
Calculate for 0, 1, and 2 successes:
\(P(X = 0) = \binom{10}{0} \left(\frac{1}{12}\right)^0 \left(\frac{11}{12}\right)^{10}\)
\(P(X = 1) = \binom{10}{1} \left(\frac{1}{12}\right)^1 \left(\frac{11}{12}\right)^9\)
\(P(X = 2) = \binom{10}{2} \left(\frac{1}{12}\right)^2 \left(\frac{11}{12}\right)^8\)
Calculate each:
\(P(X = 0) = 0.418904\)
\(P(X = 1) = 0.380822\)
\(P(X = 2) = 0.155791\)
Sum these probabilities:
\(P(0, 1, 2) = 0.418904 + 0.380822 + 0.155791 = 0.955517\)
Therefore, the probability of obtaining a total less than 4 on at least three throws is:
\(1 - 0.955517 = 0.0445\)
