Question

2. A bag contains 5 red marbles, 3 blue marbles, 6 white marbles, and 2 green marbles.

A. You choose one marble at random. Find the probability that you will choose a
white marble.
B. You choose one marble at random. Find the probability that you do not choose a red
marble.
I
C. You choose one marble at random, put the marble aside, then choose a second
marble at random. What is the probability that both marbles are blue?
D. You choose one marble at random, put the marble back in the bag, then choose a
second marble at random. What is the probability that you choose a red marble
and then a green marble?

Answer 1

Explanation:

5 red

3 blue

6 white

2 green

total 16 marbles.

Note: P(E) = probability of event E happening

P(~E) = probability of event E NOT happening.

We usually leave probabilities in fractions whenever possible because they will not be subject to round-off errors.

A. choose one marble at random.

P(W) = 6 white / 16 total = 3/8

B. choose one marble at random.

P(~R) = (16-5)/16 = 11/16

C. choose one marble at random, then a second one without replacement,

there will be 3 blue out of 16 at the beginning, then if successful, there will be 2 blue out of 15.

P(BB) = (3/16)*(2/15) = 1 / 40

D. choose two with replacement

5 red out of 16 for first draw, still 5 red out of 16 for second draw.

P(RR) = (5/16)^2 = 25/256