SOLUTION: A cage contains six rats, two of them are white (W) and four of them are brown (B). A second cage contains four rats, two white and two brown; and a third cage contains five r

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Question 1185120: A cage contains six rats, two of them are white (W) and four of them are
brown (B).
A second cage contains four rats, two white and two brown;
and a third cage contains five rats, three white and two brown.
If one rat is randomly selected from each cage,
a) What is the probability that all three rats are white?
b) What is the probability that exactly two of the three will be brown?
c) What is the probability of selecting at least two white rats?

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
A cage contains six rats, two of them are white (W) and four of them are brown (B).
2 times out of 6, 2/6 = 1/3 of the time the rat picked from the 1st cage
will be white.
4 times out of 6, 4/6 = 2/3 of the time the rat picked from the 1st cage
will be brown.

A second cage contains four rats, two white and two brown;
 
2 times out of 4, 2/4 = 1/2 of the time the rat picked from the 2nd cage
will be white.
2 times out of 4, 2/4 = 1/2 of the time the rat picked from the 2nd cage
will be brown.

and a third cage contains five rats, three white and two brown.
 
3 times out of 5, 3/5 of the time the rat picked from the 3rd cage
will be white.
2 times out of 5, 2/5 of the time the rat picked from the 3rd cage
will be brown.

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If one rat is randomly selected from each cage,
a) What is the probability that all three rats are white?
WWW
1/3 of the time the rat from the 1st cage will be white.
1/2 of those times the rat from the 2nd cage will also be white.
3/5 of those times the rat from the 3rd cage will also be white.

That's 3/5ths of 1/2 of 1/3rd of the time, which is (3/5)(1/2)(1/3)=1/10th
of the time, all three rats will be white. ANSWER: 1/10


b) What is the probability that exactly two of the three will be brown?
BBW or BWB or WBB
case 1: BBW.
2/3 of the time the rat from the 1st cage will be brown.
1/2 of those times the rat from the 2nd cage will also be brown.
3/5 of those times the rat from the 3rd cage will be white.

That's 3/5ths of 1/2 of 2/3rd of the time, which is (3/5)(1/2)(2/3)=1/5th
of the time.

case 2: BWB

2/3 of the time the rat from the 1st cage will be brown.
1/2 of those times the rat from the 2nd cage will be white.
2/5 of those times the rat from the 3rd cage will be brown.

That's 2/5ths of 1/2 of 2/3rd of the time, which is (2/5)(1/2)(2/3)=2/15th
of the time.

case 3: WBB

1/3 of the time the rat from the 1st cage will be white.
1/2 of those times the rat from the 2nd cage will be brown.
2/5 of those times the rat from the 3rd cage will also be brown.

That's 2/5ths of 1/2 of 1/3rd of the time, which is (2/5)(1/2)(1/3)=1/15th
of the time.

All three cases amounts to 1/5 + 2/15 + 1/15 = 2/5. ANSWER: 2/5


c) What is the probability of selecting at least two white rats?

We find the complement event and subtract its probability from 1.

The complement event is to have exactly 1 white or no whites (BBB).

We've already found the probability for exactly 1 white, because if you have
exactly 2 browns, you automatically have exactly 1 white. We found problem
b) to have answer 2/5.

The probability of no whites (BBB) is (2/3)(1/2)(2/5)=2/15 

So the probability of the complement event is 2/5 + 2/15 = 8/15

We subtract that from 1 and get 7/15.   ANSWER:  7/15

Edwin