Hi again! another probability problems!:) thanks!?

Question

1. Urn I contains three red chips and one white chip. Urn II contains two red chips and two white chips. One chip is drawn from each urn and transferred to the other urn. What is the probability that the chip ultimately drawn from Urn I is red?

2. John takes a 20-question multiple choice exam where each question has five answers. Some of the answers he knows, while others he gets right just by making lucky guesses. Suppose that the conditional probability of his knowing the answer given that he got it right is 0.92. How many of the 20 questions was he prepared of?

3. An insurance company has three clients - one in Manila, one in Davao and one in Cebu - whose estimated chances of living to the year 2010 is, 0.7, 0.9 and 0.3, respectively. What is the probability that by the end of 2009, the company will have had to pay death benefits to exactly one of the three?

Thanks! :)

Answer 1

1. I assume this means chips are simultaneously selected from each urn and swapped, and that we then select one chip from urn I. If you take one chip from urn I, put it into urn II, and then select one chip from urn II to put into urn I it becomes more complicated because of the possibility that you select the chip from urn II that was originally from urn I (and, of course, a similar situation applies if you do it the other way around).

We have a 3/4 probability of withdrawing a red chip from urn I and a 1/4 probability of withdrawing a white one. Similarly, we have a 1/2 probability of adding a red chip and a 1/2 probability of adding a white one. So we get:
take R, add R: (3/4)(1/2) = 3/8
take R, add W: (3/4)(1/2) = 3/8
take W, add R: (1/4)(1/2) = 1/8
take W, add W: (1/4)(1/2) = 1/8
So we have a 3/8 probability of having RRWW, a 1/2 probability of RRRW and a 1/8 probability of RRRR.

So the probability of then selecting a red chip is
(3/8) (1/2) + (1/2) (3/4) + (1/8) (1)
= 11/16.

2. We assume P(right | knows answer) = 1 and P(right | doesn't know) = 1/5 - that is, he either knows exactly or has no clue whatsoever. Suppose he knows N answers. Then, for a random question,
P(knows) = N/20
P(doesn't know) = 1 - N/20
P(knows and right) = N/20
P(doesn't know and right) = (1-N/20) (1/5) = 1/5 - N/100
P(right) = P(knows and right) + P(doesn't know and right)
= N/20 + 1/5 - N/100
= 1/5 + N/25
so P(knows | right) = P(knows and right) / P(right)
= N/20 / (1/5 + N/25)
= 5N / (20 + 4N)
= 0.92
so 5N = 0.92 (20 + 4N)
= 18.4 + 3.68N
<=> N = 13.9 (1 d.p.), so he was prepared for 14 questions.


3. P(exactly one dies)
= P(first dies, other two live) + P(second dies, other two live) + P(third dies, other two live)
= (1-0.7)(0.9)(0.3) + 0.7(1-0.9)(0.3) + 0.7(0.9)(1-0.3)
= 0.543.

Answer 2

This year about hmm 3 months-5 months ago we got a cat too that had tapeworms, fleas, and pretty horrible un-smooth fur. We have him all of his shots and medicine and he got better too. Nothing like that has happened to him, but when we had our other cat before she passed away, she was 21 years old and she would get hair kind of like that. Except, her hair was in clumps so we would cut it so that she wouldn't fight to smooth it. I think maybe you should see another vet as well and seek their opinion so that the vet you are seeing now is just not prescribing anything because they want you to come in until the last minute so they can get paid more. With the news you receive with the two different vets, you could combine them and if they say the same thing, maybe just see if they do anything. I hope your cat gets better, best wishes! Sincerely, Taylor

Answer 3

1. simultanous swapping or rotational swapping?
2. no good with that kind of question
3. 0.3*0.3*0.9+0.7*0.7*0.9+0.7*0.3*0.1
=0.9(0.09+0.49)+0.021
=0.543