Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities

(a) P (all of the next three vehicles inspected pass)

(b) P (at least one of the next three inspected fails

(c) P (exactly one of the next three inspected pass)

(d) P (at most one of the next three vehicles inspected passes)

(e) Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability) ?

Question

Mathematics

Dominique Kim

3 June, 07:56

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities

(a) P (all of the next three vehicles inspected pass)

(b) P (at least one of the next three inspected fails

(c) P (exactly one of the next three inspected pass)

(d) P (at most one of the next three vehicles inspected passes)

(e) Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability) ?

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Answers (1)

Know the Answer?

Lucian Lam · Answer 1

a. 0.343

b. 0.657

c. 0.189

d. 0.216

e. 0.353

Step-by-step explanation:

We use the combination formula of probability distribution to solve the question.

Where P (x=r) = nCr * p^r * q^n-r

Where n = number of trials = 3 vehicles

r = desired outcome of trial which varies.

p = probability of success = 70% = 0.7

q = probability of failure = 1-p = 0.3

a. Probability that all 3 vehicles passed = P (X=3)

= 3C3 * 0.7^3 * 0.3^0 = 1 * 0.343 * 1

= 0.343.

b. Probability that at least one fails = 1 - (probability that none failed)

And probability that none failed = probability that all 3 vehicles passed.

Hence Probability that at least one fails = 1 - (probability that all 3 vehicles passe)

= 1 - 0.343

= 0.657

c.) probability that exactly one pass = P (X=1)

= 3C1 * 0.7¹ * 0.3² = 3 * 0.7 * 0.09

= 0.189

d.) probability that at most One of the vehicles passed = probability that none passed + probability of one passed.

Probability that none of the vehicles passed = P (X=0)

= 3C0 * 0.7^0 * 0.3^3 = 1*1*0.027

=0.027

Probability that one passed as calculated earlier = 0.189

Hence probability that at most one vehicle passed = 0.189 + 0.027 = 0.216

e.) Probability that all three Vehicles pass given that at least one pass = (probability of all three vehicles passes) / (probability that at least one passes)

Probability that at least one pass = 1 - probability that none passed.

= 1 - 0.027

= 0.973

Hence,

Probability that all three Vehicles passed given that at least one passed = 0.343/0.973

= 0.3525 = 0.353 (3. d. p)