Answer: (a) 0.684
(b) 0.884
(c) 0.001
(d) 0.99
Explanation:
Recall from the properties of probability
(i) p(x)≤1
(ii) p(x) ≥0
(iii) ∑p(x) = 1
That is , if the probability of success is denoted with p and probability of failure is denoted with q , then p + q = 1
Given from the question;
In one year period
Probability of a randomly chosen car to be repaired once 0.2
That means the probability that it won’t be repaired once = 1 – 0.2 = 0.8
Probability of a randomly chosen car to be repaired twice = 0.1
That means the probability that it won’t be repaired twice = 1 – 0.1 = 0.9
Probability of a randomly chosen car to be repaired three or more = 0.05
That means the probability that it won’t be repaired three or more times = 1 – 0.05 = 0.95
(a) Probability of a randomly chosen car not to be repaired mean; it won't be repaired once, it won't be repaired twice and it won't be repaired three or more times, and you should note that "and" in probability means multiplication so,
p( no repair) =0.8×0.9×0.95
=0.684
(b) Probability of not more than one repair means the probability of no repair or the probability of one repair , which means
P( no more than one repair) = p( no repair) + p ( 1 repair)
= 0.6884 + 0.2
= 0.884
(c) probability of some repair means , the probability of 1 , the probability of 2 and the probability of more than 3 repair , which implies
P( some repairs ) = p( 1 repair) x p ( 2 repairs ) x p ( s or more repair )
= 0.2 x 0.1 x 0.05
= 0.01
(d) p ( neither will need repair) = 1 – p ( some repair )
= 1 – 0.01
= 0.99