# joint probability

#### teacher2be

##### New Member
Three marbles are chosen without replacement from a box containing 5 white and 7 blue marbles. Let Xi = 1 if the ith marble is white and xi = 0 otherwise.
a.) Give the joint probability function of X1 and X2.
b.) Give the marginal probability functions for X1 and X2.
c.) Find the probability that the first marble is white given that the second one is blue.

To find the joint probability for part (a), should I be trying to determine how often a white marble will be drawn first? If so would the probability of getting a white marble be 5/35 or 1/7?

I am not sure I am approaching this correctly, so any guidance would be appreciated.

#### BioStatMatt

##### TS Contributor
The joint probability states how likely both events are to happen simultaneously. In this case, I would try to list all of the outcomes that are possible:

You can have
(x1 = 0, x2 = 0)
(x1 = 1, x2 = 0)
(x1 = 0, x2 = 1)
(x1 = 1, x2 = 1)

The next step would be to find out how likely these events are to occur. When I see things being done in order, I think "Bayes's Rule" For example:

P(x1 = 0 and x2 = 0) = P(x2 = 0 given x1 = 0) * P(x1 = 0)

The second part, P(x1 = 0) is simply the probability of drawing a blue marble on the first draw. This is 7/12 since there are 7 blue marbles and 12 marbles total.

If the first draw yields a blue marble, then the probability of drawing another blue is 6/11 since there are only 6 blue marbles remaining and only 11 total. So,

P(x2 = 0 given x1 = 0) = 6/11

Put this together, and:

P(x1 = 0 and x2 = 0) = (7/12) * (6/11) = 0.318

You can follow this exact process for the other three possibilities and write the joint probability function as:

f(x1, x2) = {0.318 when x1 = 0, x2 = 0
????? when x1 = 1, x2 = 0
... and so on}

You may also solve this problem using combination notation by enumerating the number of ways that each of the events could happen, and divide that by the total number of ways that the 4 seperate events could happen.

You can also solve this problem using combinations. If this is desired, please specifiy in another question.

Hope this helps,

~Matt