Question

Monica has a bag of marbles. There are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble. What is the probability of Monica picking a red and then blue marble?

Answer 1

Answer:

$\frac{7}{64}\approx 0.11$

Step-by-step explanation:

We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.

Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.          

Since the probability of two independent compound events is always the product of probabilities of both events.

$P(\text{A and B})=P(A)*P(B)$

Now let us find probability of picking a red marble out of 16 (4+7+5) marbles.

$P(Red)=\frac{\text{Total red marbles}}{\text{Total marbles}}$

$P(Red)=\frac{4}{16}$

Probability of picking blue ball out of 16 (4+7+5) marbles:

$P(Blue)=\frac{\text{Total blue marbles}}{\text{Total marbles}}$

$P(Blue)=\frac{7}{16}$

Now let us find probability of Monica picking a red and then a blue marble.

$P(\text{Red and Blue})=\frac{4}{16}\times \frac{7}{16}$

$P(\text{Red and Blue})=\frac{1}{4}\times \frac{7}{16}$

$P(\text{Red and Blue})=\frac{7}{4*16}$

$P(\text{Red and Blue})=\frac{7}{64}$

$P(\text{Red and Blue})=0.109375\approx 0.11$

Therefore, the probability of picking a red and then blue marble is $\frac{7}{64}\approx 0.11$.