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Each of the 25 balls in a certain box is either red, blue or

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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 26 Nov 2017, 19:33
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - \(P(W)\);
Probability ball: even - \(P(E)\);
Probability ball: white and even - \(P(W&E)\).

Probability ball picked being white or even: \(P(WorE)=P(W)+P(E)-P(W&E)\).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> \(P(W&E)=0\) (no white ball with even number) --> \(P(WorE)=P(W)+P(E)-0\). Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \(P(W)-P(E)=0.2\), multiple values are possible for \(P(W)\) and \(P(E)\) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine \(P(WorE)\).
)
(1)+(2) \(P(W&E)=0\) and \(P(W)-P(E)=0.2\) --> \(P(WorE)=2P(E)+0.2\) --> multiple answers are possible, for instance: if \(P(E)=0.4\) (10 even balls) then \(P(WorE)=1\) BUT if \(P(E)=0.2\) (5 even balls) then \(P(WorE)=0.6\). Not sufficient.

Answer: E.

Hope it's clear.


Hi Bunuel..

How did u get Probability ball picked being white or even: P(W or E)=P(W)+P(E)-P(W&E)?
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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 26 Nov 2017, 21:06
1
zanaik89 wrote:
Bunuel wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

Probability ball: white - \(P(W)\);
Probability ball: even - \(P(E)\);
Probability ball: white and even - \(P(W&E)\).

Probability ball picked being white or even: \(P(WorE)=P(W)+P(E)-P(W&E)\).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> \(P(W&E)=0\) (no white ball with even number) --> \(P(WorE)=P(W)+P(E)-0\). Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \(P(W)-P(E)=0.2\), multiple values are possible for \(P(W)\) and \(P(E)\) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine \(P(WorE)\).
)
(1)+(2) \(P(W&E)=0\) and \(P(W)-P(E)=0.2\) --> \(P(WorE)=2P(E)+0.2\) --> multiple answers are possible, for instance: if \(P(E)=0.4\) (10 even balls) then \(P(WorE)=1\) BUT if \(P(E)=0.2\) (5 even balls) then \(P(WorE)=0.6\). Not sufficient.

Answer: E.

Hope it's clear.


Hi Bunuel..

How did u get Probability ball picked being white or even: P(W or E)=P(W)+P(E)-P(W&E)?


OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: \(P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\).

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then \(P(A \ and \ B)=0\) and the formula simplifies to: \(P(A \ or \ B) = P(A) + P(B)\).

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: \(P(A \ and \ B) = P(A)*P(B)\).

This is basically the same as Principle of Multiplication: if one event can occur in \(m\) ways and a second can occur independently of the first in \(n\) ways, then the two events can occur in \(mn\) ways.

22. Probability



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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 04 Feb 2018, 13:33
Can someone tell me where my logic went wrong? I selected A, with the thought that since P(W&E)=0, therefore there must be 5 white balls, all of which have an odd number on them (so 5 white balls total). Therefore, P(W)=5/25, P(E)=10/25, and P(W&E)=0. Is it wrong to assume that the numbers don't repeat on each ball color (i.e. the blue balls with all of 1 number cannot be true?).
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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 04 Feb 2018, 22:57
ulanky wrote:
Can someone tell me where my logic went wrong? I selected A, with the thought that since P(W&E)=0, therefore there must be 5 white balls, all of which have an odd number on them (so 5 white balls total). Therefore, P(W)=5/25, P(E)=10/25, and P(W&E)=0. Is it wrong to assume that the numbers don't repeat on each ball color (i.e. the blue balls with all of 1 number cannot be true?).


Hi

P(W&E) = 0, means that there is NO ball which is both white and has an even number on it. But there could be various white balls with odd numbers on them, and there could be various red/blue balls with even numbers on them. We need to take both these kinds of balls into account.

I think we cannot assume that there must be 5 white balls, its possible that there is only one white ball in the entire box. And yes, I think we also cannot assume that the numbers cannot repeat on same coloured balls (because its nowhere given). Its possible that all blue coloured balls have only number 1 painted on them.
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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 24 Apr 2018, 11:14
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lexis wrote:
Each of the 25 balls in a certain box is either red, blue or white and has a number from 1 to 10 painted on it. If one ball is to be selected at random from the box, what is the probability that the ball selected will either be white or have an even number painted on it?

(1) The probability that the ball will both be white and have an even number painted on it is 0

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2


Target question: What is the value of P(white or even)?

To solve this, we'll use the fact that P(A or B) = P(A) + P(B) - P(A & B)
So, P(white or even) = P(white) + P(even) - P(white & even)

Statement 1: P(white & even) = 0
We can add this to our probability equation to get: P(white or even) = P(white) + P(even) - 0
Since we don't know the value of P(white) and P(even), we cannot determine the value of P(white or even)
NOT SUFFICIENT

Statement 2: P(white) - P(even)= 0.2
We have no idea about the sum of P(white) and P(even), and we don't know the value of P(white & even)
NOT SUFFICIENT

Statements 1 and 2 combined:
Given P(white) - P(even)= 0.2 does not tell us the individual values of P(white) and P(even), and it doesn't tell us the value of P(white) + P(even).

So, since we can't determine the value of P(white) + P(even) - P(white & even), the statements combined are NOT SUFFICIENT.

Answer: E

Cheers,
Brent
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Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 17 Jun 2018, 07:56
Bunuel wrote:
jananijayakumar wrote:
But how can this be solved in less than 2 mins???


You can solve this problem in another way. Transform probability into actual numbers and draw the table.

Given:
Attachment:
1.JPG


So we are asked to calculate \(\frac{a+b-c}{25}\) (we are subtracting \(c\) not to count twice even balls which are white).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> \(c=0\) --> \(\frac{a+b}{25}=?\). Not sufficient.
Attachment:
4.JPG


(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \(\frac{white}{25}-\frac{even}{25}=0.2\) --> \(white-even=25*0.2=5\) --> \(a-b=5\) --> \(b=a-5\) --> \(\frac{a+a-5-c}{25}=?\). Not sufficient.
Attachment:
2.JPG


(1)+(2) \(c=0\) and \(b=a-5\) --> \(\frac{a+a-5+0}{25}=\frac{2a-5}{25}\). Not sufficient.
Attachment:
3.JPG

Answer: E.



Hi Bunuel,

This is an interesting approach to this problem.

Please let me know when can we use this Transformation of probability into actual numbers.

And also if you have any similar questions in which I can apply this technique for practice.

Thanks in advance!!
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Re: Each of the 25 balls in a certain box is either red, blue or  [#permalink]

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New post 17 Jun 2018, 08:51
SudhanshuSingh wrote:
Bunuel wrote:
jananijayakumar wrote:
But how can this be solved in less than 2 mins???


You can solve this problem in another way. Transform probability into actual numbers and draw the table.

Given:
Attachment:
1.JPG


So we are asked to calculate \(\frac{a+b-c}{25}\) (we are subtracting \(c\) not to count twice even balls which are white).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> \(c=0\) --> \(\frac{a+b}{25}=?\). Not sufficient.
Attachment:
4.JPG


(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> \(\frac{white}{25}-\frac{even}{25}=0.2\) --> \(white-even=25*0.2=5\) --> \(a-b=5\) --> \(b=a-5\) --> \(\frac{a+a-5-c}{25}=?\). Not sufficient.
Attachment:
2.JPG


(1)+(2) \(c=0\) and \(b=a-5\) --> \(\frac{a+a-5+0}{25}=\frac{2a-5}{25}\). Not sufficient.
Attachment:
3.JPG

Answer: E.



Hi Bunuel,

This is an interesting approach to this problem.

Please let me know when can we use this Transformation of probability into actual numbers.

And also if you have any similar questions in which I can apply this technique for practice.

Thanks in advance!!


PS Overlapping + Probability questions

DS Overlapping + Probability questions

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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