Answer:
a) 0.778
b) 0.9222
c) 0.6826
d) 0.3174
e) 2 drivers
Step-by-step explanation:
Given:
Sample size, n = 5
P = 40% = 0.4
a) Probability that none of the drivers shows evidence of intoxication.
[tex] P(x=0) = ^nC_x P^x (1-P)^n^-^x[/tex]
[tex]P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0[/tex]
[tex] P(x=0) = ^5C_0 (0.4)^0 (0.60)^5[/tex]
[tex] P(x=0) = 0.778 [/tex]
b) Probability that at least one of the drivers shows evidence of intoxication would be:
P(X ≥ 1) = 1 - P(X < 1)
[tex] = 1 - P(X = 0) [/tex]
[tex] = 1 - ^5C_0 (0.4)^0 * (0.6)^5[/tex]
[tex] = 1 - 0.0778 [/tex]
[tex] = 0.9222 [/tex]
c) The probability that at most two of the drivers show evidence of intoxication.
P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)
[tex] ^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3 [/tex]
[tex] = 0.6826 [/tex]
d) Probability that more than two of the drivers show evidence of intoxication.
P(x>2) = 1 - P(X ≤ 2)
[tex] = 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3] [/tex]
[tex] = 1 - 0.6826 [/tex]
[tex] = 0.3174 [/tex]
e) Expected number of intoxicated drivers.
To find this, use:
Sample size multiplied by sample proportion
n * p
= 5 * 0.40
= 2
Expected number of intoxicated drivers would be 2
