Answer:
P(X1=1, X2=1) = 1/15
P(X1=1, X2=0) = 7/30
P(X1=0, X2=1) = 7/30
P(X1=0, X2=0) = 7/15
Step-by-step explanation:
Let Xi = 1 if the i-th white ball is selected. In this question the 3 white balls are marked 1,2 and 3.
We need to know the possible combination between X1 and X2 i.e. for the white ball 1 and 2 being chosen in the event.
We also need to note that the event is dependent which that after a ball is being chosen, it will not be put back hence affecting the probability of picking the next ball.
Consider all the possible combination between X1 and X2
a) both being chosen P(X1=1, X2=1)
= (3/10) x (2/9) = 1/15
Note that the first probability is the probability before any ball is being picked. The chances for ball white to be pick is 3/10 (3 white ball from the total 10 balls).
After 1 white ball being selected, that ball is not again out back into the urn making white ball 2 and total ball 9. Hence probability of picking another white ball is 2/9
b) only X1 chosen P(X1=1, X2=0)
= (3/10) x (7/9) = 7/30
After the white ball was picked, the probability of white not being pick again is the same as red being picked. Since there is still 7red balls and a total of 9 balls, the probability is 7/9
c) only X2 chosen P(X1=0, X2=1)
= (7/10) x (3/9) = 7/30
The white is not being picked first, making the probability of picking red is 7/10. Then the probability of white being picked is 3/9
d) both not chosen
P(X1=0, X2=0)
= (7/10) x (6/9) = 7/15
In other word only red being chosen. So the first probability is 7 red out of 10 balls (7/10), and the next red ball being picked next is 6/9
, but 1.618 should suffice.