Question

A bag contains 28 blue marbles and 12 red marbles. If Milton removes two marbles without replacing them, what is the probability that the first marble will be blue and the second one will be red?

Answer 1

Probability that the first marble will be blue and the second one will be red is 0.2154.

Step-by-step explanation:

We are given that a bag contains 28 blue marbles and 12 red marbles.

Probability of an event = $\frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}$

Also, Probability is never negative and does not exceed 1 which means the value of probability of any event lies between 0 and 1.

Now, we are given that Milton removes two marbles without replacing them means that when he picks one marble from the bag, he does not put it back and picks another marble from the bag.

Total number of marbles in the bag = Number of blue marbles + Number of red marbles = 28 + 12 = 40

Probability of drawing first marble to be blue = $\frac{\text{Number of blue marbles in the bag}}{\text{Total number of marbles in the bag}}$

                                                                              = $\frac{28}{40}$ = 0.7

Now, after drawing first marble from the bag, the total number of marbles left are = 40 - 1 = 39

Probability of drawing second marble be red = $\frac{\text{Number of red marbles in the bag}}{\text{Total number of marbles in the bag}}$

                                                                              = $\frac{12}{39}$ = 0.31

So, probability that the first marble will be blue and the second one will be red = Probability of drawing first marble to be blue $\times$ Probability of drawing second marble to be red

            = $\frac{28}{40}\times \frac{12}{39}$

            = $\frac{14}{65}$  = 0.2154

Hence, if Milton removes two marbles without replacing them, the probability that the first marble will be blue and the second one will be red is 0.2154.