A die is rolled 12 times. Find the probability of rolling no more than 4 fives.
+118609 P(no more thatn 4 fours) = P(no fours)+P(1four)+P(2 fours)+P(3 fours)+P(4 fours)
$$\\\left(\frac{5}{6}\right)^{12}
\;\;+\;\;^{12}C_1\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{11}
\;\;+\;\;^{12}C_2 \left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{10}\\\\
\;\;+\;\;^{12}C_3\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{9}
\;\;+\;\;^{12}C_4\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{8}$$
+128475 This means we either rolll no fives, one five, two fives, three fives or 4 fives
The probability of rolling no fives is C(12,0)(1/6)^0*(5/6)^12 = 0.1121566547846151
The probability of rolling one five is C(12,1)(1/6)^1*(5/6)^11 = 0.0000000044516784
The probability of rolling two fives is C(12,2)(1/6)^2*(5/6)^10 = 0.2960935686313838
The probability of rolling three fives is C(12,3)(1/6)^3*(5/6)^9 = 0.1973957124209225
The probability of rolling four fives is C(12,4)(1/6)^4*(5/6)^8 = 0.0888280705894151
And summing these we have......about 69.4% chance of rolling no more than 4 fives
+118609 P(no more thatn 4 fours) = P(no fours)+P(1four)+P(2 fours)+P(3 fours)+P(4 fours)
$$\\\left(\frac{5}{6}\right)^{12}
\;\;+\;\;^{12}C_1\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{11}
\;\;+\;\;^{12}C_2 \left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{10}\\\\
\;\;+\;\;^{12}C_3\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{9}
\;\;+\;\;^{12}C_4\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{8}$$
