Thread: Probability - Drivers and accidents

1. Probability - Drivers and accidents

Another probability question, I've just started doing stats and probability, I've already done 23 questions now haha, this is number 42 in a longgggg list of examples im trying to master to form a masterbook so i can solve most problems . I've got as far as defining what I know etc, i.e. P(XA)=0.02 (probability that X has accident) any help would be cool

An insurance company classifies drivers as class X, Y or Z. Experience indicates that the probability a class X driver has at least one accident in any one year is 0.02, while the corresponding probabilities for Y and Z are 0.04 and 0.1 respectively. The class of a new driver who applies to them for cover is not known, but from experience they have found that
30% of applicants are class X, 60% are class Y , and 10% are class Z.

(a) What the the probability of a new client having an accident in their first year?

(b) If a new client has no accidents in their first year, what is the probability that they are a class X driver?
(c) What is the probability of a new client having no accidents in their first 5 years?

2. Originally Posted by mitch_nufc
Another probability question, I've just started doing stats and probability, I've already done 23 questions now haha, this is number 42 in a longgggg list of examples im trying to master to form a masterbook so i can solve most problems . I've got as far as defining what I know etc, i.e. P(XA)=0.02 (probability that X has accident) any help would be cool

An insurance company classifies drivers as class X, Y or Z. Experience indicates that the probability a class X driver has at least one accident in any one year is 0.02, while the corresponding probabilities for Y and Z are 0.04 and 0.1 respectively. The class of a new driver who applies to them for cover is not known, but from experience they have found that
30% of applicants are class X, 60% are class Y , and 10% are class Z.

(a) What the the probability of a new client having an accident in their first year?

(b) If a new client has no accidents in their first year, what is the probability that they are a class X driver?

(c) What is the probability of a new client having no accidents in their first 5 years?
Pr(at least 1 accident | X) = 0.02

Pr(at least 1 accident | Y) = 0.04

Pr(at least 1 accident | Z) = 0.1

Pr(X) = 0.3

Pr(Y) = 0.6

Pr(Z) = 0.1

(a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

(b) Calculate Pr(X | no accident).

Note: Pr(no accident) = 1 - answer to (a) = ....
Pr(X and no accident) = Pr(no accident | X) Pr(X) = ....

(c) (1 - answer to a)^5, assuming each year is independent.

3. I got 3.2 for part b), this obviously isnt right at all lol

4. Originally Posted by mitch_nufc
I got 3.2 for part b), this obviously isnt right at all lol

5. P(no accident|X)=P(X & no accident)/P(X)

=0.96/0.3

as P(X & no accident)=P(X)P(no accident|X)

Should I have got 0.288?

6. Originally Posted by mitch_nufc
Another probability question, I've just started doing stats and probability, I've already done 23 questions now haha, this is number 42 in a longgggg list of examples im trying to master to form a masterbook so i can solve most problems . I've got as far as defining what I know etc, i.e. P(XA)=0.02 (probability that X has accident) any help would be cool

An insurance company classifies drivers as class X, Y or Z. Experience indicates that the probability a class X driver has at least one accident in any one year is 0.02, while the corresponding probabilities for Y and Z are 0.04 and 0.1 respectively. The class of a new driver who applies to them for cover is not known, but from experience they have found that
30% of applicants are class X, 60% are class Y , and 10% are class Z.

(a) What the the probability of a new client having an accident in their first year?

(b) If a new client has no accidents in their first year, what is the probability that they are a class X driver?
(c) What is the probability of a new client having no accidents in their first 5 years?
Mitch,

On (b), use
$P(\text{X} \; | \; \text{no accident}) =\frac{P(\text{X and no accident})}{P(\text{no accident})}$

On (c), use
$P(\text{no accident})^5$.

(I think Mr. F. slipped a little on that last one, if I understand the problem correctly.)

7. Originally Posted by awkward
[snip]On (c), use
$P(\text{no accident})^5$.

(I think Mr. F. slipped a little on that last one, if I understand the problem correctly.)
Yes, a careless slip-up. I've fixed it up and edited in what I meant.

8. Could anyone show me how b) and c) are done with numbers put it I'm really tired and have a banging headache, all ive done all day is stats and bayesian stats and my head is mush thanks

9. Originally Posted by mr fantastic
Pr(at least 1 accident | X) = 0.02

Pr(at least 1 accident | Y) = 0.04

Pr(at least 1 accident | Z) = 0.1

Pr(X) = 0.3

Pr(Y) = 0.6

Pr(Z) = 0.1

(a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

(b) Calculate Pr(X | no accident).

Note: Pr(no accident) = 1 - answer to (a) = ....
Pr(X and no accident) = Pr(no accident | X) Pr(X) = Mr F adds: (0.98) (0.3).

(c) (1 - answer to a)^5, assuming each year is independent.
Originally Posted by mitch_nufc
Could anyone show me how b) and c) are done with numbers put it I'm really tired and have a banging headache, all ive done all day is stats and bayesian stats and my head is mush thanks
Advice: Take a break. Then either re-do the calculation using what's in this post or .....

Post your calculation with all the details. The calculations you posted did not have detail to make it easy for me to see what you've done:

Originally Posted by mitch_nufc
P(no accident|X)=P(X & no accident)/P(X)

=0.96/0.3

as P(X & no accident)=P(X)P(no accident|X)
Then, if you've made a mistake, it will be easy to find.

Note that I have given you all the numbers that go into the formula that awkward provided and which it appears you're using. The rest is just arithmetic.

10. (a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

(b) Calculate Pr(X | no accident).

Note: Pr(no accident) = 1 - answer to (a) = ....
Pr(X and no accident) = Pr(no accident | X) Pr(X) =

(c) (1 - answer to a)^5, assuming each year is independent.

so from part a) we get 0.04.
therefore 1- a) = 1-0.04 = 0.96 not 0.98 yeah?
so part b) = 0.96x0.3
& part c) = (0.96)^5

11. Originally Posted by mitch_nufc
(a) (0.02)(0.3) + (0.04)(0.6) + (0.1)(0.1) = ....

(b) Calculate Pr(X | no accident).

Note: Pr(no accident) = 1 - answer to (a) = ....
Pr(X and no accident) = Pr(no accident | X) Pr(X) = Mr F adds: (0.98) (0.3).

(c) (1 - answer to a)^5, assuming each year is independent.
so from part a) we get 0.04.
therefore 1- a) = 1-0.04 = 0.96 not 0.98 yeah? Mr F says: No.

so part b) = 0.96x0.3
& part c) = (0.96)^5
Pr(no accident) is not the same as Pr(no accident | X).

Pr(no accident | X) = 1 - Pr(at least 1 accident | X) = 1 - 0.02 = 0.98 like I said.

12. FINALLY done this Q!

Thanks to Mr. F and awkward for combined help.