Box 1 has 2 white marbles and 1 black marble. Box 2 has 1 white marble and 3 black marbles.
A) A marble is randomly selected from box 1 and placed in box 2. A marble is then randomly chosen from box 2. What is the probability that the marble selected from box 2 is black.
B) If a black marble is selected from box 2, what is the probability that a white marble was transferred from box 1 to box 2.
Please show work.
A) A marble is randomly selected from box 1 and placed in box 2. A marble is then randomly chosen from box 2. What is the probability that the marble selected from box 2 is black.
B) If a black marble is selected from box 2, what is the probability that a white marble was transferred from box 1 to box 2.
Please show work.
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a) From box 1, there is a 1/3 chance of selecting a black marble and a 2/3 chance of selecting a white marble.
So draw a tree diagram where the first branches are 1/3 and 2/3.
If a black is chosen and added to box 2, then box 2 will have 1 white and 4 black marbles, meaning that there will be a 4/5 chance of choosing a black marble.
So make the branch that stems from the 1/3 become a 4/5 and 1/5
If a white marble is chosen and added to box 2, then box 2 will have 2 white and 3 black marbles, meaning that there will be a 3/5 chance of choosing a black marble.
So make the branch that stems from the 2/3 become a 3/5 and 1/5.
The probability of choosing a black marble in the end is:
(1/3 * 4/5) + (2/3 * 3/5)
= 4/15 + 2/5
= 2/3
b) The odds of having a black marble selected is a 2/3 chance.
Pr(A given B) = Pr(A intersecting B) / Pr(B)
Pr(white marble transferred given that a black marble was selected) = Pr(A black marble has been selected and a white marble has been transferred simultaneously) / Pr(a black marble was selected)
We worked out in part A that Pr(black marble selected) = 2/3
We also worked out that Pr(white marble transferred) = 2/3
We also worked out the probability that both happened simultaneously is (2/3 * 3/5) = 2/5.
So using the formula:
Pr(white transferred given black was selected) = (2/5) / (2/3)
Pr(white transferred given black was selected) = 2/5 * 3/2
Pr(white transferred given black was selected) = 3/5
So draw a tree diagram where the first branches are 1/3 and 2/3.
If a black is chosen and added to box 2, then box 2 will have 1 white and 4 black marbles, meaning that there will be a 4/5 chance of choosing a black marble.
So make the branch that stems from the 1/3 become a 4/5 and 1/5
If a white marble is chosen and added to box 2, then box 2 will have 2 white and 3 black marbles, meaning that there will be a 3/5 chance of choosing a black marble.
So make the branch that stems from the 2/3 become a 3/5 and 1/5.
The probability of choosing a black marble in the end is:
(1/3 * 4/5) + (2/3 * 3/5)
= 4/15 + 2/5
= 2/3
b) The odds of having a black marble selected is a 2/3 chance.
Pr(A given B) = Pr(A intersecting B) / Pr(B)
Pr(white marble transferred given that a black marble was selected) = Pr(A black marble has been selected and a white marble has been transferred simultaneously) / Pr(a black marble was selected)
We worked out in part A that Pr(black marble selected) = 2/3
We also worked out that Pr(white marble transferred) = 2/3
We also worked out the probability that both happened simultaneously is (2/3 * 3/5) = 2/5.
So using the formula:
Pr(white transferred given black was selected) = (2/5) / (2/3)
Pr(white transferred given black was selected) = 2/5 * 3/2
Pr(white transferred given black was selected) = 3/5
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qA
P[white from 1 & black from 2] = 2/3 *3/5 = 6/15
P[black from 1 & black from 2] = 1/3 *4/5 = 4/15
P[black from 2] = 10/15 = 2/3 <-------
qB
using bayes rule,
P[white from 1 | black from 2] = (6/15)/(10/15) = 3/5 <-------
P[white from 1 & black from 2] = 2/3 *3/5 = 6/15
P[black from 1 & black from 2] = 1/3 *4/5 = 4/15
P[black from 2] = 10/15 = 2/3 <-------
qB
using bayes rule,
P[white from 1 | black from 2] = (6/15)/(10/15) = 3/5 <-------
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A) (1 black / 3 total) x (3 black / 4 total) = .25 or 25%
B) 2 white / 3 total = .67 or 67%
Not quite sure, but yeah.
B) 2 white / 3 total = .67 or 67%
Not quite sure, but yeah.