Edward has four pearls – two white and two black – which he can distri

Question

Edward has four pearls – two white and two black – which he can distribute as he chooses between two identical bags. He must then choose a bag at random and pick one pearl at random from the bag he chose. If Edward distributes the pearls so as to maximize his chances of pulling a black pearl, what is the probability that Edward will pick a black pearl?
 
A. 1/2
B. 3/5
C. 2/3
D. 3/4
E. 1

BrentGMATPrepNow · Accepted Answer

NOTE: Since the bags are considered identical, we can ignore equivalent distributions like Bag X: BWW | Bag Y: B, which is the same as the 2nd distribution in the list.

Some students may be able to automatically eliminate some answer choices upon examination. 
Let's examine each case and determine P(Edward selects a black ball)

1) Bag X: 0 marbles | Bag Y: BBWW  
P(black ball) = P(select bag Y  AND  select one of the black balls in that bag)
= P(select bag Y)  x  P(select one of the black balls in that bag)
= 1/2  x  2/4
= 1/4

2) Bag X: B | Bag Y: BWW  
P(black ball) = P(select bag X  AND  select black ball in that bag   OR   select bag Y  AND  select black ball in that bag]
=  x  P(select black ball in that bag)]  +   x  P(select black ball in that bag)]
=  x  1]  +   x  1/3]
= 1/2  +  1/6
= 4/6
=   2/3

3) Bag X: W | Bag Y: BBW  
P(black ball) = P(select bag Y  AND  select one of the black balls in that bag)
= P(select bag Y)  x  P(select one of the black balls in that bag)
= 1/2  x  2/3
= 1/3

4) Bag X: BB | Bag Y: WW 
P(black ball) = P(select bag X  AND  select one of the black balls in that bag)
= P(select bag X)  x  P(select one of the black balls in that bag)
= 1/2  x  1
= 1/2

5) Bag X: BW | Bag Y: BW  
P(black ball) = P(select bag X  AND  select black ball in that bag   OR   select bag Y  AND  select black ball in that bag)]
=  x  P(select black ball in that bag)]  +   x  P(select black ball in that bag]
=  x  1/2]  +   x  1/2]
= 1/4  +  1/4
= 1/2

Of all possible distributions, the highest probability is   2/3  
Answer: C

Cheers,
Brent

EgmatQuantExpert · Answer

Solution 
A beautiful question to apply your understanding of the topic. 
One needs to visualize different situations and then should be able to manipulate how one should be distributing the pearls.

Given

• 2 black pearls and 2 white pearls to be distributed in two identical bags.
• Once done, one needs to pick one bag and pick a pearl from that.

To Find

• The probability that the pearl picked up is black.

Approach and Working Out

Let’s assume a case where we distribute them as – 
  o Bag 1 – (1B, 1W)
o Bag 2 – (1B, 1W)

• The probability of picking the first bag =  ½  
• Probability of picking a black pearl is also  ½   so 
  o For the probability that a black pearl is picked from the first bag is  ½  x  ½  =  ¼ .

• Similarly, the probability that a black pearl is picked from the second bag is also  ¼  .

So total probability =  ¼  +  ¼  =  ½  .

Now, we can see that net probability is coming as,
 (Probability of picking first bag x  Probability of picking a black pearl from it)  + (Probability of picking second bag x  Probability of picking a black pearl from it )

• Now only the underlined part can be manipulated as the remaining part is always  ½ . The best we can do is to make it 1 by putting only black pearls. 
  o So we put only black pearls in the first bag.

• Now the question is should we put 1 black pearl or 2 black pearls.
 o If we put 2 black pearls then in the first bag then we can put no black balls in the second.
o So we should put only 1 black pearl in the first one and in the second bag 1 black pearl and 2 white pearls.

Optimized distribution –

• Bag 1 – (1B)
• Bag 2 – (1B and 2W)

Probability 
= (Probability of picking first bag x Probability of picking a black pearl from it) + (Probability of picking second bag x Probability of picking a black pearl from it)
= ½ x 1 + ½ x  \frac{1}{3} 
=  \frac{2}{3}

Correct Answer: Option C

mohshu · Answer

GMATPrepNow

please tell how to eliminate the choices one examination...i cud eliminate E,,tats it

BrentGMATPrepNow · Answer

1) Bag X: 0 marbles | Bag Y: BBWW 
In order to select a black pearl, we must select Bag Y (50% chance of doing so) AND we must select one of the two black pearls (which further decreases the likelihood of selecting a black pearl). So, the probability is LESS than 1/2. 
The smallest answer choice is 1/2, so ELIMINATE A

2) Bag X: B | Bag Y: BWW 
Looks promising - KEEP

3) Bag X: W | Bag Y: BBW 
In order to select a black pearl, we must select Bag Y (50% chance of doing so) AND we must select one of the two black pearls (which further decreases the likelihood of selecting a black pearl). So, the probability is LESS than 1/2. 
The smallest answer choice is 1/2, so ELIMINATE C

4) Bag X: BB | Bag Y: WW 
In order to select a black pearl, all we need to do is select Bag Y (50% chance of doing so)
So, probability = 1/2
This is one of the answer choices, so KEEP

5) Bag X: BW | Bag Y: BW 
The bags are identical, so it doesn't really matter which one you select. 
Once you select ANY bag, the probability of selecting a black pearl = 1/2
This is one of the answer choices, so KEEP

Does that help?

Cheers,
Brent

AnthonyRitz · Answer

I disagree with this. Just because the greatest probability could (hypothetically) be achieved in two ways, that fact would not eliminate it from consideration. It can, of course, be eliminated because another case is better, but you do still have to establish that fact if you want to pick something other than 1/2.

BrentGMATPrepNow · Answer

Good point. I was confusing the 5 possible arrangements with the 5 answer choices. 
I have deleted that portion of my explanation.

Thanks!

Cheers,
Brent

VerbalKnight · Answer

If Edward decides to split the two bags into one bag containing a single pearl black and the other bag containing a black pearl and 2 white pearls, then Edward will have a 50% chance of picking the one-pearl bag and immediately getting a black pearl, plus a 50% chance of picking the three-pearl bag and having a 1/3 chance at a black pearl

The total probability of a black pearl is  \frac{1}{2}∗1+\frac{1}{2}∗\frac{1}{3}=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}

This question made me remember a similar interesting problem called "Monty Hall Problem". If you are interested then you can find it in the movie "21" or just a google search "Monty hall problem" would suffice.

rvgmat12 · Answer

OE: Right off the bat, we should note that the answer 1/2 is far too obvious to actually be correct.

It’s easy to get 1/2, of course; put two pearls in each bag, and no matter which colors are where, Edward will end up with a 50% chance at a black pearl. Putting all four pearls in one bag is even worse (there’s a 50% chance of getting nothing at all).

The only remaining possibility is putting three pearls in one bag and a single pearl in the other. In that case, the solo pearl can be either black or white.

If Edward chooses for the solo pearl to be white, then there’s a 50% chance that Edward will pick the bag with the white pearl and immediately be stuck with that, and a 50% chance that Edward will pick the three-pearl bag and have a 2/3 chance at a black pearl. The total probability of a black pearl is 1/2∗0+1/2∗2/3=1/3
.
If Edward decides to make the solo pearl black, then Edward will have a 50% chance of picking the one-pearl bag and immediately getting a black pearl, plus a 50% chance of picking the three-pearl bag and having a 1/3 chance at a black pearl (the three-pearl bag now contains two white pearls and one black pearl). The total probability of a black pearl is 1/2∗1+1/2∗1/3=1/2+1/6=2/3. This is the maximum possible probability, and it’s answer C.

Fdambro294 · Answer

List out the possibilities in which we can distribute the 4 balls within two identical groupings. There are 4 scenarios:

Scenario 1:

(None) ————- (W-W-B-B)

Scenario 2:

(W) ——————— (W-B-B)

Scenario 3:

(W-W) ——————-(B-B)

Scenario 4:

(W-W-B) ——————(B)

Scenario 5:

(B-W) ————— (B-W)

It is equally likely that he will pick up either bag ——- so, probability of picking up any 1 bag out of the two = (1/2)

We can then find the expected probability of pulling a black (B) ball in each scenario

Scenario 1:

Half the time he will grab the bag with no balls: P(black) = 0

Half the time he will grab the other bag and the prob. of picking a black ball will be: P(black) = 2/4 = 1/2

Scenario 1 expected probability =

(1/2) (0) + (1/2) (1/2) = 1/4

Scenario 2:

Half the time he picks bag with the one white ball: P(black) = 0

Half the time he picks the other back with 2 black balls: P(black) = 2/3

(1/2)(0) + (1/2)(2/3) =

1/3 probability for scenario 2

Scenario 3:

Half the time he grabs bag with two white balls: P(black) = 0

Half the time he grabs bag with two black balls: P(black) = 1

(1/2)(0) + (1/2)(1) =

1/2 probability for scenario 3

Scenario 4:

Half the time he picks the bag with just one black ball: P(black) = 1

Half the time he picks the bag with one black and two white balls: P(black) = 1/3

Expected probability of picking a black ball:

(1/2) (1) + (1/2) (1/3) =

(1/2) + (1/6) = (6 + 2) / (12) = 8/12 = 2/3

Scenario 5:

Half the time he will pick one bag: P(black) = 1/2

Half the time he will pick the other bag = P(black) = 1/2

(1/2)(1/2) + (1/2)(1/2) = 1/2

Scenario 4 would provide the greatest chance of picking a black ball.

2/3 is the answer

Posted from my mobile device

JeffTargetTestPrep · Answer

Since we can consider the two bags and the pearls of the same color identical, there are only five cases of “pearls in one bag & pearls in another bag” to consider:
 
BBWW & None
BBW & W
BWW & B
BB & WW
BW & BW
 
For such 2-step selection problems, we can use the total probability formula:
 
P(A) = ΣP(A|Bi)P(Bi)
 
It looks complicated, but it’s relatively easy to use.
 
 In the first case: 
 
P(black) = P(black|one bag)P(one bag) + P(black|another bag)P(another bag)
 
P(black|one bag) = 2/4 = 1/2, since the number of favorable outcomes is 2, and the total number of outcomes is 4.
 
P(one bag) = 1/2, since we have to choose between two bags at random.
 
P(black|another bag) = 0, since it is an impossible event to choose a black pearl from an empty bag.
 
P(another bag) = 1/2, since we have to choose between two bags at random.
 
P(black) = 1/2 × 1/2 + 0 × 1/2 = 1/4
 
 In the second case: 
 
P(black) = 2/3 × 1/2 + 0 × 1/2 = 1/3
 
 In the third case: 
 
P(black) = 1/3 × 1/2 + 1 × 1/2 = 2/3
 
 In the fourth case: 
 
P(black) = 1 × 1/2 + 0 × 1/2 = 1/2
 
 In the fifth case: 
 
P(black) = 1/2 × 1/2 + 1/2 × 1/2 = 1/2
 
We see that the greatest probability, which is 2/3, belongs to the third case.
 
 Answer: C

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Edward has four pearls – two white and two black – which he can distri

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