http://www.ck12.org Chapter 1. Independent and Dependent Events
On the third draw, the probability of drawing a red marble is:
P(red) =
5
8
Therefore, the probability of drawing 2 purple marbles and 1 red marble is:
P(1 purple and 1 purple and 1 red) =P( 1 P∩ 1 P∩ 1 R)
=P 1 (purple)×P 2 (purple)×P(red)
=
5
10
×
4
9
×
5
8
=
100
720
=
5
36
Example C
In Example B, what is the probability of first drawing all 5 red marbles in succession and then drawing all 5 purple
marbles in successionwithout replacement?
The probability of first drawing all 5 red marbles in succession can be calculated as follows:
P(1 red and 1 red and 1 red and 1 red and 1 red) =P( 1 R∩ 1 R∩ 1 R∩ 1 R∩ 1 R)
=P 1 (red)×P 2 (red)×P 3 (red)×P 4 (red)×P 5 (red)
=
5
10
×
4
9
×
3
8
×
2
7
×
1
6
=
120
30240
=
1
252
At this point, there are 5 marbles left, and all of them are purple. This means that with each remaining draw, the
probability of drawing a purple marble is 1. Therefore, the probability of first drawing all 5 red marbles in succession
and then drawing all 5 purple marbles in succession is 2521 × 1 × 1 × 1 × 1 × 1 = 2521.