Answer:
(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.
(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.
(c) The probability that exactly 1 of the next three vehicles passes is 0.189.
(d) The probability that at most 1 of the next three vehicles passes is 0.216.
(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.
Step-by-step explanation:
Let <em>X</em> = number of vehicles that pass the inspection.
The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.
(a)
Compute the probability that all the next three vehicles inspected pass the inspection as follows:
P (All 3 vehicles pass) = [P (X)]³
Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.
(b)
Compute the probability that at least 1 of the next three vehicles inspected fail as follows:
P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)
Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.
(c)
Compute the probability that exactly 1 of the next three vehicles passes as follows:
P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)
= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)
+ P (Only 3rd vehicle passes)
Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.
(d)
Compute the probability that at most 1 of the next three vehicles passes as follows:
P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)
+ P (0 vehicles passes)
Thus, the probability that at most 1 of the next three vehicles passes is 0.216.
(e)
Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.
Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:
![P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525](https://tex.z-dn.net/?f=P%28X%7CY%29%3D%5Cfrac%7BP%28X%5Ccap%20Y%29%7D%7BP%28Y%29%7D%20%3D%5Cfrac%7BP%28X%29%7D%7BP%28Y%29%7D%20%3D%5Cfrac%7B%280.70%29%5E%7B3%7D%7D%7B%5B1-%280.30%29%5E%7B3%7D%5D%7D%20%3D0.3525)
Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.
