Conditional Probability: picking three big blue marbles

I have been trying to understand this one problem for weeks. No matter what route I take to solve it, I can't come up with the right answer. It states p(x) p(y/x) = p(y) p(x/y).

A bag contains 40 blue marbles, 40 red marbles and 20 green marbles. 60 of the marbles are big. Three marbles are selected at random without replacement. What is the probability of selecting three big blue marbles?

The most logical answer I can come up with, which is wrong, is:

P(Blue) P(Big) + P(Blue) P(Big) + P(Blue) P(Big)

(40/100)(60/100) + (39/99)(59/99) + (38/98)(58/98)
2400/10000 + 2301/9801 + 2204/9604
.2400 + .2345 + .2295

.704 My given answers are .0125, .0240, .2323, and .2400. I have also used the Bayes Theorem without success in resoving this problem. Please help. Thanks.

cjc72671 said:

A bag contains 40 blue marbles, 40 red marbles and 20 green marbles. 60 of the marbles are big. Three marbles are selected at random without replacement. What is the probability of selecting three big blue marbles?

Click to expand...

Are you quite sure that you have written the problem correctly?

As written, nothing is said about numbers with respect to being ‘BIG’.
Therefore, as written the number of “BIG BLUE” marbles could be any number from 0 to 40. In that case, the probability of selecting three big blue marbles having chosen three without replacement is
\(\displaystyle \frac{{\sum\limits_{k = 3}^{40} {\left( \begin{array}{c} {\rm{k}} \\ {\rm{3}} \\ \end{array}} \right)} }}{{\left( {\begin{array}{c} {100} \\ 3 \\\end{array}} \right)}} =0.6262834\).

(I can not get the LaTex right!)

Thanks. I am sure. I rechecked the problem. It was the last variable of the problem. It said that 60 of the marbles were big.

I am working on the same problem these are my results similar to yours but instead of adding my three results I multiplied.

Chance of first pick being blue is 40/100 and chance that it is big is 60/100.
So chance that it is both big and blue is 2/5 * 3/5 = 6/25 = 0.24
Second marble has chances 39/99 * 59/99 = 0.2348
Third marble has chances 38/98 * 58/98 = 0.2295
Multiplying all three together gets 0.0129 which is very close to the .0125

cjc72671 said:

A bag contains 40 blue marbles, 40 red marbles and 20 green marbles. 60 of the marbles are big. Three marbles are selected at random without replacement. What is the probability of selecting three big blue marbles?

Click to expand...

Since that's the way the problem should read (so you say) and no info is given as to the "big" proportions,
then just assume that the 60 "bigs" are the reds and the greens; so you have NO big blues; so probability = 0.
Such an answer is as silly as the problem :shock: