The Question
Suppose you randomly draw two marbles, without replacement, from a bag containing six green, four red, and three black marbles.
a) Determine the probability that both marbles are red.
b) Determine the probability that you pick at least one green marble.
My Approach
a) I did 4/13 * 3/12 = 1/13.
b) I have no idea how to go about solving this one
Can someone check if my approach for a) is correct and also show me how to do b)?
Best Answer
There are 13 marbles, six green so 13- 6= 7 'not green'. So the probability the first marble is "not green" is 7/13. Then there are 12 marbles 6 "not green" so the probability the second is "not green" is 6/12= 1/2. The probability neither marble is green is (7/13)(1/2)= 7/26. The probability "at least one marble is green" is 1- 7/26= 26/26- 7/26= 19/26.
You could also do this though it is harder: The probability the first marble is green is 6/13. There are then 12 marbles, 5 green, so 7 "not green". The probability the second marble is not green is 7/12. The probability of one green and one not green, in that order, is (6/13)(7/12)= (1/13)(7/2)= 7/26.
The probability the first marble is not green is 7/13. There are then 12 marbles, 6 green so the probability the second marble is not green is 6/12= 1/2. The probability of one not green and one green, in that order, is (7/13)(1/2)= 7/26.
Since "at least one green" includes "both green", the probability the first marble is green is 6/13. There are then 12 marbles, 5 green so the probability both marbles are green is 5/12 is (6/13)(5/12)= (1/13)(5/2)= 5/26.
The probability of "at least one green marble" is 7/26+ 7/26+ 5/26= 19/26.