Question

Amy has a box containing 6 white, 4 red, and 8 black marbles. She picks a marble randomly. It is red. The second time, she picks a white marble. In the third attempt, Amy picks a white marble again. Amy has not replaced the marbles she picked.
The probability that the fourth marble Amy picks is black is . If Amy picks a black marble in the fourth attempt, the probability that in the fifth attempt she will pick a red or a black marble is .

Answer 1

Answer: The probability that the fourth marble Amy picks is black $=\frac{8}{15}$

The probability that in the fifth attempt she will pick a red or a black marble P(R∩B) =$\frac{5}{7}$

Step-by-step explanation:

To find  the probability that in the fifth attempt she will pick a red or a black marble P(R∩B) .

The total marble in beginning=6+4+8=18

After 3 attempts, the total  marbles left=18-3=15

The number of black marbles =8

The probability that the fourth marble Amy picks is black $=\frac{8}{15}$

As four marble has already picked up, then the total marbles left in the box

=18-4=14

Now, as a red marble is already picked in 1st attempt , the number of red marbles in the box now= 4-1=3

And a black marble is already picked in fourth attempt , the number of black marbles in the box now= 8-1=7

Now P(R∩B)=P(R)+P(B)=$\frac{3}{14}+\frac{7}{14}=\frac{10}{14}=\frac{5}{7}$