a bowl contains 8 blue marbles and 5 green marbles. monique randomly draws 3 marbles from the bowl she does not replace the marbles after each draw.

Question

what is the probability that sge draws 1 blue marble and 2 green marbles?

Answer 1

Given 8 blue marbles and 5 green marbles. There are 13 marbles altogether.

 

The probability of drawing a blue marble is 8/13.

 

The probability of drawing the first green marble is 5/12, since only 12 marbles remain.

 

The probability of drawing the second green marble is 4/11, since only 11 marbles remain.

 

Therefore the probability of choosing 1 blue marble and 2 green marbles is (8/13) * (5/12) * (4/11) = 40/429

 

Let (pick1, pick2, pick3) represent the order that the marbles were chosen.

With this sample 1 blue (b) and 2 green (g) can be chosen in the following order (b, g, g), (g, b, g), (g, g, b)

 

Since there are 3 distinct ways to choose 1 blue and 2 green marbles, the probability of choosing them is three times choosing them in a particular order, or 40/143

Answer 2

Probability of drawing 1 blue and 2 green marbles = Probability of 1 blue * Probability 1 green * Probability 1 green

 

 

Probability of drawing 1 blue marble = 8/13 (because 8 blue marbles, 13 marbles total)

 

Probability of drawing the 1st green marble = 5/12 (5 green marbles; only 12 marbles left after you take out the 1st blue marble)

 

Probability of drawing the 2nd green marble = 4/11 (4 green marbles left since you took out 1 green marble; 11 total marbles in the bowl left)

 

So: 8/12 * 5/12 * 4/11 

= 0.6153824 * 0.4166666 * 0.36363636 

=0.0932396

 

or 9.323%