Question
"In vase A, there are 3 red balls and 5 white balls. In vase B, there are 6 red balls and 4 white balls. Two balls are to be drawn at random, without replacement, from each vase. What is the probability of selecting 4 red balls?"
(Answer: 1/28)
So I thought about doing 4 cases and then adding them up (don't pick red ball in A, don't pick red ball in B...pick red ball in A, pick red ball in B...etc...).
I did this:
(3/8)(3/7)(6/10)(6/9) +
(3/8)(2/7)(6/10)(5/9) +
.....
But it doesn't work. What am I doing wrong?
(Answer: 1/28)
So I thought about doing 4 cases and then adding them up (don't pick red ball in A, don't pick red ball in B...pick red ball in A, pick red ball in B...etc...).
I did this:
(3/8)(3/7)(6/10)(6/9) +
(3/8)(2/7)(6/10)(5/9) +
.....
But it doesn't work. What am I doing wrong?
Answers
Since you pick 2 from each vase, and you want 4 reds, there is only one way, that is, you pick 2 from A and 2 from B.
which is C(3,2)/(C(8,2) * C(6,2)/C(10,2)
= 3/28 * 15/45 = 1/28
which is C(3,2)/(C(8,2) * C(6,2)/C(10,2)
= 3/28 * 15/45 = 1/28
Oh yeah! Thank you!
Related Questions
An urn contains white balls and red balls. If Juan chooses balls at random from the urn, what is...
Use the table to answer the question. Bowl A: 5 white balls and 15 red balls. Bowl B: 16 white balls...
Bowls a and b contain a number of white and red balls. Clark repeatedly selected 100 balls from both...
A floral vase is shown. The body of the vase is white with flowers in a vase painting on the side...