Probability question for actuarial exam P: From 27 pieces of luggage...

Steven G

Steven G

Elite Member
Joined
Dec 30, 2014
Messages
14,382
I am not seeing this one
From 27 pieces of luggage, an airline luggage handler damages a random
sample of four. The probability that exactly one of the damaged pieces of
luggage is insured is twice the probability that none of the damaged pieces

are insured. Calculate the probability that exactly two of the four damaged
pieces are insured.

I think that the 27 is not needed.

P (exactly 0 Insured) + P ( exactly 1 insured) + P( exactly 2 insured) + ... + P( 4 insured) = 1
so x + 2x + P( exactly 2 insured) + ... + P( 4 insured) =1

I'm not seeing exactly what The probability that exactly one of the damaged pieces ofluggage is insured is twice the probability that none of the damaged pieces
are insured is really saying
 
Jomo said:
I am not seeing this one
From 27 pieces of luggage, an airline luggage handler damages a random
sample of four. The probability that exactly one of the damaged pieces of
luggage is insured is twice the probability that none of the damaged pieces

are insured. Calculate the probability that exactly two of the four damaged
pieces are insured.

I think that the 27 is not needed.

P (exactly 0 Insured) + P ( exactly 1 insured) + P( exactly 2 insured) + ... + P( 4 insured) = 1
so x + 2x + P( exactly 2 insured) + ... + P( 4 insured) =1

I'm not seeing exactly what The probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces
are insured is really saying
Click to expand...
I think that the wording is bad. Using the 27 I found the probability that exactly two of the four damaged pieces are insured is 0.2728.
 
Last edited:
You are correct that the information that these four damaged pieces of luggage are from 27 pieces is irrelevant. We are only asked about the four damaged pieces.

Let "p" be the probability that any one is insured. Then the probability that one is not insured is 1- p. The probability that exactly one of four is damaged is \(\displaystyle 4p(1- p)^3\) and the probability none is damaged is \(\displaystyle (1- p)^4\). We are told that "the probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured" so \(\displaystyle 4p(1- p)^3= 2(1- p)^4\). Divide both sides by \(\displaystyle (1- p)^3\) to get \(\displaystyle 4p= 2(1- p)= 2- 2p\) so \(\displaystyle 6p= 2\). That is, the probability any one damaged piece is insured is \(\displaystyle p= \frac{2}{6}= \frac{1}{3}\).

The rest of the problem should be easy.
 
HallsofIvy said:
You are correct that the information that these four damaged pieces of luggage are from 27 pieces is irrelevant. We are only asked about the four damaged pieces.

Let "p" be the probability that any one is insured. Then the probability that one is not insured is 1- p. The probability that exactly one of four is damaged is \(\displaystyle 4p(1- p)^3\) and the probability none is damaged is \(\displaystyle (1- p)^4\). We are told that "the probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured" so \(\displaystyle 4p(1- p)^3= 2(1- p)^4\). Divide both sides by \(\displaystyle (1- p)^3\) to get \(\displaystyle 4p= 2(1- p)= 2- 2p\) so \(\displaystyle 6p= 2\). That is, the probability any one damaged piece is insured is \(\displaystyle p= \frac{2}{6}= \frac{1}{3}\).

The rest of the problem should be easy.
Click to expand...
Here is what I did and I did use the 27.

I let s = the number of pieces of luggage that are insured.

P( exactly 1 damaged piece is insured) = (sC1)(27-sC3)/(27C4)

P( exactly 0 damaged pieces are insured) = (sC0)(27-sC4)/(27C4)

Since this ratio is 2 we arrive at s=8.

P( exactly 2 damaged pieces are insured) = (8C2)(19C2)/(27C4) ~ .2728
 
Last edited:
HallsofIvy said:
You are correct that the information that these four dam .2963aged pieces of luggage are from 27 pieces is irrelevant. We are only asked about the four damaged pieces.

Let "p" be the probability that any one is insured. Then the probability that one is not insured is 1- p. The probability that exactly one of four is damaged is \(\displaystyle 4p(1- p)^3\) and the probability none is damaged is \(\displaystyle (1- p)^4\). We are told that "the probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured" so \(\displaystyle 4p(1- p)^3= 2(1- p)^4\). Divide both sides by \(\displaystyle (1- p)^3\) to get \(\displaystyle 4p= 2(1- p)= 2- 2p\) so \(\displaystyle 6p= 2\). That is, the probability any one damaged piece is insured is \(\displaystyle p= \frac{2}{6}= \frac{1}{3}\).

The rest of the problem should be easy.
Click to expand...
Continuing....P(exactly two damaged pieces are insured) = (4C2)(1/3)^2(2/3)^2 = 6(4/81) ~.2963

In the end (if I solved your method correctly) we both got different answers. Where is the error?
 
I don't know what you mean by "we both got different answers" because I did not give an answer!

Again, the "27 pieces of luggage" is irrelevant. You are asked for the probability that, out of the four pieces of luggage that were damaged, exactly two were insured. We only need to consider those four pieces. I pointed out that the probability any one of those four is insured is 1/3, so the probability any one is not insured is 2/3. From that, that 2 out of the four are insured, the other 2 not insured, is \(\displaystyle _4C_2(1/3)^2(2/3)^2= 6(1/9)(4/9)= 24/81= 8/27\). That's the same answer you got.
 
Last edited:
HallsofIvy said:
Again, the "27 pieces of luggage" is irrelevant. You are asked for the probability that, out of the four pieces of luggage that were damaged, exactly two were insured. We only need to consider those four pieces. I pointed out that the probability any one of those four is insured is 1/3, so the probability any one is not insured is 2/3. From that, that 2 out of the four are insured, the other 2 not insured, is \(\displaystyle _4C_2(1/3)^2(2/3)^2= 6(1/9)(4/9)= 24/81= 8/27\).
Click to expand...
8/27 is exactly what I got and stated in an earlier post. I still am not convinced that you (or I) are correct. Did you read my solution?
 
You must log in or register to reply here.
Top