Drunken drivers

40% of drivers driving between 11 pm and 5 am are drunken drivers. In a random sample of 20 drivers driving between 11 pm and 5 am, find the probability that:
A) Exactly 12 will be drunken drivers
B) At least 7 will be drunken drivers
C) At most 5 will be drunken drivers

Correct answer:

p₁₂ = 0.0355
b = 0.75
c = 0.1256

Step-by-step explanation:

p = 40 % = \frac{4 0}{1 0 0} = \frac{2}{5} = 0.4 n = 20 p_{12} = (1 2 n) \cdot p^{12} \cdot (1 - p)^{n - 12} = (1 2 2 0) \cdot 0 . 4^{12} \cdot (1 - 0.4)^{20 - 12} = 125970 \cdot 0 . 4^{12} \cdot 0 . 6^{8} = 0.0355

p_{0} = (0 n) \cdot p^{0} \cdot (1 - p)^{n - 0} = (0 2 0) \cdot 0 . 4^{0} \cdot (1 - 0.4)^{20 - 0} = 1 \cdot 0 . 4^{0} \cdot 0 . 6^{20} ≐ 3.6562 \cdot 1 0^{- 5} p_{1} = (1 n) \cdot p^{1} \cdot (1 - p)^{n - 1} = (1 2 0) \cdot 0 . 4^{1} \cdot (1 - 0.4)^{20 - 1} = 20 \cdot 0 . 4^{1} \cdot 0 . 6^{19} ≐ 0.0005 p_{2} = (2 n) \cdot p^{2} \cdot (1 - p)^{n - 2} = (2 2 0) \cdot 0 . 4^{2} \cdot (1 - 0.4)^{20 - 2} = 190 \cdot 0 . 4^{2} \cdot 0 . 6^{18} ≐ 0.0031 p_{3} = (3 n) \cdot p^{3} \cdot (1 - p)^{n - 3} = (3 2 0) \cdot 0 . 4^{3} \cdot (1 - 0.4)^{20 - 3} = 1140 \cdot 0 . 4^{3} \cdot 0 . 6^{17} ≐ 0.0123 p_{4} = (4 n) \cdot p^{4} \cdot (1 - p)^{n - 4} = (4 2 0) \cdot 0 . 4^{4} \cdot (1 - 0.4)^{20 - 4} = 4845 \cdot 0 . 4^{4} \cdot 0 . 6^{16} ≐ 0.035 p_{5} = (5 n) \cdot p^{5} \cdot (1 - p)^{n - 5} = (5 2 0) \cdot 0 . 4^{5} \cdot (1 - 0.4)^{20 - 5} = 15504 \cdot 0 . 4^{5} \cdot 0 . 6^{15} ≐ 0.0746 p_{6} = (6 n) \cdot p^{6} \cdot (1 - p)^{n - 6} = (6 2 0) \cdot 0 . 4^{6} \cdot (1 - 0.4)^{20 - 6} = 38760 \cdot 0 . 4^{6} \cdot 0 . 6^{14} ≐ 0.1244 p_{7} = (7 n) \cdot p^{7} \cdot (1 - p)^{n - 7} = (7 2 0) \cdot 0 . 4^{7} \cdot (1 - 0.4)^{20 - 7} = 77520 \cdot 0 . 4^{7} \cdot 0 . 6^{13} ≐ 0.1659 c = p_{0} + p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = 3.6562 \cdot 1 0^{- 5} + 0.0005 + 0.0031 + 0.0123 + 0.035 + 0.0746 ≐ 0.1256 b = 1 - (c + p_{6}) = 1 - (0.1256 + 0.1244) = 0.75

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