Questions: Suppose that we have a bag full of marbles, identical except for color. The bag contains 7 red marbles, 20 green marbles, and 11 blue marbles. We draw two marbles from the bag, one after the other, without placing the drawn marbles back into the bag. Let A1 be the event that the first marble drawn is red, A2 the event that the first marble drawn is green, and A3 the event that the first marble drawn is blue. Let B1 be the event that the second marble drawn is red, B2 the event that the second marble drawn is green, and B3 the event that the second marble drawn is blue. What is P(B2 A3) ? (In other words, what is the probability that the second marble drawn is green, given that the first marble drawn was blue?) 0.289474 0.526316 0.184211 0.297297

Suppose that we have a bag full of marbles, identical except for color. The bag contains 7 red marbles, 20 green marbles, and 11 blue marbles.

We draw two marbles from the bag, one after the other, without placing the drawn marbles back into the bag.

Let A1 be the event that the first marble drawn is red, A2 the event that the first marble drawn is green, and A3 the event that the first marble drawn is blue.

Let B1 be the event that the second marble drawn is red, B2 the event that the second marble drawn is green, and B3 the event that the second marble drawn is blue.

What is P(B2 A3) ? (In other words, what is the probability that the second marble drawn is green, given that the first marble drawn was blue?)
0.289474
0.526316
0.184211
0.297297

Transcript text: Suppose that we have a bag full of marbles, identical except for color. The bag contains 7 red marbles, 20 green marbles, and 11 blue marbles. We draw two marbles from the bag, one after the other, without placing the drawn marbles back into the bag. Let $A_{1}$ be the event that the first marble drawn is red, $A_{2}$ the event that the first marble drawn is green, and $A_{3}$ the event that the first marble drawn is blue. Let $B_{1}$ be the event that the second marble drawn is red, $B_{2}$ the event that the second marble drawn is green, and $B_{3}$ the event that the second marble drawn is blue. What is $P\left(B_{2} \mid A_{3}\right)$ ? (In other words, what is the probability that the second marble drawn is green, given that the first marble drawn was blue?) 0.289474 0.526316 0.184211 0.297297

Solution

Solution Steps

Step 1: Define the Problem

We need to find the conditional probability $ P(B_2 \mid A_3) $, which represents the probability that the second marble drawn is green given that the first marble drawn was blue.

Step 2: Identify the Total Number of Marbles

Initially, the bag contains:

Red marbles: $ R = 7 $
Green marbles: $ G = 20 $
Blue marbles: $ B = 11 $

The total number of marbles is: \[ N = R + G + B = 7 + 20 + 11 = 38 \]

Step 3: Determine Remaining Marbles After the First Draw

After drawing one blue marble, the remaining marbles are:

Red marbles: $ R = 7 $
Green marbles: $ G = 20 $
Blue marbles: $ B = 10 $ (since one blue marble has been drawn)

Thus, the total number of remaining marbles is: \[ N' = R + G + B = 7 + 20 + 10 = 37 \]

Step 4: Calculate the Probability Using Hypergeometric Distribution

We want to find the probability of drawing 1 green marble from the remaining 37 marbles, where there are still 20 green marbles left. The hypergeometric probability formula is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] where:

$ N' = 37 $ (total remaining marbles)
$ K = 20 $ (remaining green marbles)
$ n = 1 $ (number of marbles drawn in the second draw)
$ k = 1 $ (number of green marbles we want to draw)

Substituting the values into the formula: \[ P(B_2 \mid A_3) = \frac{\binom{20}{1} \binom{17}{0}}{\binom{37}{1}} = \frac{20 \cdot 1}{37} = \frac{20}{37} \approx 0.540541 \]