Answer:
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)
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