A bag contains only black red and gold balls. Any ball is equally like

Question

A bag contains only black red and gold balls. Any ball is equally likely to be picked. If you pick one ball at random, is it more likely to be red than black?

(1) If you remove half of the red balls, the probability of picking the black ball would be 50%
(2) If you remove the third of black balls, the probability of picking a red ball would be 60%

Eeni · Answer

The question says we have 3 kinds of balls- Red (R), Black (B) and Golden (G) and each ball is equally likely to be drawn.

We are asked if the probability of red ball being drawn is more than that of a black ball.

Statement 1:
 
It says if we remove half the number of red balls, so we are left with only half of the red balls, R/2.
If we had R/2, B, G, the prob of black ball=0.5
Black ball/total Balls= 1/2
B/(B+R/2+G)= 1/2
Cross multiply, 2B=B+R/2+G
B=R/2+G,
If R=2, G=1, B=2 => Black or red equally likely.
If R=2, G=3, B=4 => Black is more likely.
We cannot conclude if Statement 1 is sufficient.

Statement 2: 
It says if we remove a third of the number of black balls, so we are left with only two-thirds of the black balls, 2B/3.
If we had R, 2B/3, G, the probability of red ball=0.6
Red ball/total Balls= 0.6
R/(2B/3+R+G)= 3/5
Cross multiply, 5R=3R+2B+3G
2R=2B+3G 
R=B+1.5G
Clearly Red is more in number.
We can conclude that Statement 2 is sufficient.

JerryAtDreamScore · Answer

Breaking Down the Info:

Note that we have three possible colors. We can always replace any red/black ball for gold when our information is incomplete.

Statement 1 Alone:

We may start with the maximum red ball case. That would be 50% black and 50% red after removing. Then before removing we would have had  Black : Red = 50\%:100\% , hence more red than black. We could also have 1 red and let Black > Red. Then this statement is insufficient.

Statement 2 Alone:

We can start with the maximum black ball case.  Red : Black = 60\%:40\%  after removing.

We removed  \frac{1}{3}  of the black balls, which is the same as multiplying the amount of black balls by  \frac{2}{3} .

Then we need to multiply the number of black balls by  \frac{3}{2}  to return to the "before remove" state. Then  Red : Black = 60\%:60\%  before removal.

Recall that this was the maximum black ball case, so Red > Black for the rest of the cases. The current  Red : Black = 60\%:60\%  case is also a hypothetical case since we're supposed to have at least 1 gold ball. Then for all real cases, we have Red > Black, and this statement is sufficient.

Answer: B

Bunuel · Answer

This question is a part of  Are You Up For the Challenge: 700 Level Questions  collection.

Akashmitra97 · Answer

Okay so I have a query. I solved this question based on logic and was able to narrow down on B as my choice which was right in this case. I want to know whether it was the right approach. So the stem says the probability of picking any ball is the same. So, to answer the question of whether we would get a red ball at random will only hold true when the number of red balls increase or the number of black and gold balls decrease. And in option B it clearly states that by removing a third of the black balls, the chances of getting a red ball increases. Whereas the probability of getting a gold ball still remains the same. So we can righty say the only statement 2 is sufficient. Am I right here or is there a problem anywhere? Any insights would be helpful. Thanks!

Fdambro294 · Answer

We can look at the maximum and minimum of each statement and determine if we have enough information to answer the question:

Are there more RED balls > than BLACK Balls?

S1: take away (1/2) of the Red balls and 50% of all the balls will be Black

Case 1:
After the 1/2 of the Red Balls are taken away we can have

Black = 50
Red = 40
Gold = 10 ———> after the (1/2) red are taken away, P(Black) = 50%

(As we are told that there are balls of each type in the box: “each are equally likely to be picked” implied that there must be balls of each color in the box)

This means originally, the balls in the box would be:

Black = 50
Red = (2) (40) = 80
Gold = 10

Red > Black ——- Yes

However, we could just as easily make more of the balls Gold:

Case 2:
After the (1/2) Red balls are taken away we can have:

Black = 50
Red = 10
Gold = 40 ——-> after the (1/2) are taken away, we would have P(Black) = 50%

Which means originally we would have:

Black = 50
Red = (2) (10) = 20
Gold = 40

In this case 2 ——> R < B

We get a NO answer to the question in addition to a YES answer
S1 not sufficient

Statement 2: if we take away (1/3) of Black balls, the Probability of a Red ball would be = 60%

We can MAXIMIZE the number of Balls that could be Black, by Minimizing the number of Balls that could be Gold.

While there must be at least 1 gold ball in the box, we can assume for now that there are NO Gold balls:

MAX Case for Black: let’s say there are 100 balls - after removing (1/3) of the Black Balls, there must be 60% Red balls or:

Red = 60
And
Black = 40
And
Let Gold = 0

In this maximum possible case (which is not possible), how many black balls would we have to start?

Blank ——-> B - (1/3)B = 40

(2/3) B = 40

B = 60

So the original distribution would be:

Black = 60
Red = 60
Gold = 0

Since this case is not possible as there must be at least one gold (“equal chance to pick any of the balls”)

If lowering the Black Balls by (1/3) leads to there being 60% of the Red balls in the box ——-> the ORIGINAL Number of Black Balls MUST BE Less Than < the Number of Red Balls

Answers the Question Yes:

In the original box, there must be More Red Balls > than Black Balls

Or

P(Red) > P(Black)

B statement 2 sufficient alone

Posted from my mobile device

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A bag contains only black red and gold balls. Any ball is equally like

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