Question

The probability of picking two black marbles from a box at random without replacement is 1091 . The probability of drawing a black marble first is 514 . What is the probability of picking a second black marble, given that the first marble picked was black?

Answer 1

The correct answer to the question above is 4/13. Below it shows how it's achieved...

EQUATION: 
P(A|B) = P(A ∩ B)
                 P(B)

Step 1: Identify 
P(B) = 5/14 
P(A ∩ B) = 10/91

Step 2: Insert into the equation and Solve
10/91 ÷ 5/14
Use reciprocal to divide
10/91 × 14/5 
Simplify the answer 
140/455
Simplified (Final Answer)
4/13 

Answer 2

We want to get the probability of picking a second black marble given that the first one was also black marble. It is Q = 24/91.

So, we know that:

Probability of picking a black marble first is P = 5/14

Probability of picking a black marble after we picked one black marble = Q

Joint probability (probability of both of these events happening) is just the product between the individual probabilities:

J = P*Q = (5/14)*Q

And we know that it is equal to 10/91, so:

10/91 = (5/14)*Q

(10/91)*(14/5) = Q = 140/455 = 28/91

So the probability of picking a second black marble, given that the first marble picked was black is 28/91

If you want to learn more about probability, you can read:

https://brainly.com/question/1349408