an insurance company reckons the probability of an accident occuring during the year is 0.1 for a low risk person and .35 for a high risk person. 25% of the customers are low risk and the rest are high risk. Mr. Z is randomly selected from the customers, what is the probability he has an accident during the year?
2006-11-08 22:27:06 · 12 answers · asked by kate 1 in Science & Mathematics ➔ Mathematics
the answer is 0.2875
EXPLANATION:
Low risk(LR) = 0.1,
percentage Number of low risk Customers(PNLR) = 25% ( 0.25)
High risk(HRLN) = 0.35,
percentage Number of high risk customer = 75% ( 0.75)
Mr. Z can either be low risk or High risk because he was selected at random.
Prob. of Mr Z having accident =( LR x PNLR) or (HR x PNHR)
=(0.1 x 0.25) + ( 0.35 x 0.75)
= 0.2875
2006-11-08 23:04:52 · answer #1 · answered by demraf 2 · 0⤊ 0⤋
Probability
= 25% x 0.1 + 75% x 0.35
=1/4 *10/100 +3/4* 35*100
=10/400 + 105/400
=115/400
=0.2875
2006-11-09 01:47:02 · answer #2 · answered by toasterpastry 1 · 0⤊ 0⤋
He has a 3 in 4 (75%) chance of being a high risk person - and then he has a 0.35% chance of having an accident.
Which gives 0.2625 ((3/4) * 0.35)
He also has a 1 in 4 chance of being low risk and then a 0.1% chance of having an accident
Which gives 0.025 ((1/4) * 0.1)
Add the two together and he has a 0.2875% chance of having an accident in the year.
2006-11-08 22:33:14 · answer #3 · answered by mark 7 · 2⤊ 0⤋
You have to MULTIPLY the probabilities of things which have to BOTH happen [an AND], and ADD the probabilities of the resulting options [the ORs].
So either:
He's a high risk person (0.75) AND has an accident (0.35)
OR
He's a low risk person (0.25) AND has an accident (0.1)
So total probability = (0.75 x 0.35) + (0.25 x 0.1)
2006-11-08 22:35:18 · answer #4 · answered by gvih2g2 5 · 0⤊ 0⤋
It is either 1 or 0. We do not know which until he has gone through the year, but we will be able to say for definite once we have observed the year.
I don't like probability as a mathematical construct, as I'm sure you've guessed. Mr Z will or will not crash based on whether he drives well on the roads, not because of a set statistical liklihood that he will.
2006-11-09 02:27:46 · answer #5 · answered by Anonymous · 0⤊ 0⤋
The probability of him being low risk and having an accident is 0.25*0.1=0.025
and the probability of him being high risk and having an accident is 0.75*0.35=0.2625
so the probability of him being one or the other is 0.025+0.2625=0.2875 or 28.75%
2006-11-08 22:45:29 · answer #6 · answered by Little Miss Proper 1 · 0⤊ 0⤋
Mr. Z is my grandad. i can tell you he is very accident prone at the moment. so the likleyhood of him having an accident is very very high. put that as your answer...
***** five gold stars for you clever clever
2006-11-08 22:42:17 · answer #7 · answered by queenie 1 · 0⤊ 0⤋
Depends on what category he's in.
2006-11-08 22:29:05 · answer #8 · answered by Anonymous · 0⤊ 0⤋
depends upon what category he's in!
2006-11-08 22:54:22 · answer #9 · answered by Keerthana S 2 · 0⤊ 0⤋
I'm gonna guess... 28.75%?
2006-11-08 22:34:04 · answer #10 · answered by tracethelostboy 2 · 0⤊ 0⤋