Question

A box contains 5 red marbles and 2 yellow marbles. Consider the two–stage experiment of randomly selecting a marble from a box, NOT replacing it, and then selecting a second marble. Determine the probabilities of the following events:

a. Selecting 2 red marbles.

P (red) =

b. Selecting a red marble on the first draw and a yellow marble on the second.

P (red then yellow) =

c. Selecting at least 1 yellow marble.

P (at least 1 yellow) =

Answer 1

a) 10/21 b) 5/21 c)1/21

Explanation:

a) your have 7 choice and 5 or them or red, 5/7. You do not replace the red marble, so for your second draw, you now only have 6 choices and of those 6 choices only 4 of them are now red or 4/6 or 2/3. We multiple these two numbers together:

5/7(2/3) = 10/21

b) For the first draw, you have 5 red marbles available out of a total 7 marbles or 5/7. The second draw now only has 6 marbles and 2 of them or yellow to make 2/6 or 1/3 (simplified). We multiply these together.

5/7(1/3) =5/21

c) I am assuming that you are still selecting 2 marbles. The first draw would be 2/7. This time I am selecting yellow and there are only 2 choices for yellow. My next draw will only have 6 marbles and only 1 of them will be yellow, so 1/6. We multiply these together

2/7(1/6) = 2/42 or 1/21