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
