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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
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(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
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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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 that the ball is white: \(P(W)\)
Probability that the ball has an even number: \(P(E)\)
Probability that the ball is white and has an even number: \(P(W\&E)\)
Probability that the ball selected is white or has an even number:
\(P(W \ or \ E)=P(W)+P(E)-P(W\&E)\)
(1) The probability that the ball will both be white and have an even number painted on it is 0 -->
This implies \(P(W\&E)=0\), meaning there is no white ball with an even number.
\(P(W \ or \ E)=P(W)+P(E)-0\)
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 -->
This implies \(P(W)-P(E)=0.2\). Multiple values are possible for \(P(W)\) and \(P(E)\), such as 0.6 and 0.4 OR 0.4 and 0.2.
Cannot determine \(P(W \ or \ E)\).
Not sufficient.
(1)+(2) \(P(W\&E)=0\) and \(P(W)-P(E)=0.2\), so:
\(P(W \ or \ E)=P(W)+P(E)-P(W\&E)=(P(E)+0.2)+P(E)-0=2P(E)+0.2\)
Still, multiple answers are possible. For instance, if \(P(E)=0.4\) (10 even balls), then \(P(W \ or \ E)=1\). But if \(P(E)=0.2\) (5 even balls), then \(P(W \ or \ E)=0.6\).
Not sufficient.
Answer: E.
Hope it helps.
_________________
Probability that the ball is white: \(P(W)\)
Probability that the ball has an even number: \(P(E)\)
Probability that the ball is white and has an even number: \(P(W\&E)\)
Probability that the ball selected is white or has an even number:
\(P(W \ or \ E)=P(W)+P(E)-P(W\&E)\)
(1) The probability that the ball will both be white and have an even number painted on it is 0 -->
This implies \(P(W\&E)=0\), meaning there is no white ball with an even number.
\(P(W \ or \ E)=P(W)+P(E)-0\)
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 -->
This implies \(P(W)-P(E)=0.2\). Multiple values are possible for \(P(W)\) and \(P(E)\), such as 0.6 and 0.4 OR 0.4 and 0.2.
Cannot determine \(P(W \ or \ E)\).
Not sufficient.
(1)+(2) \(P(W\&E)=0\) and \(P(W)-P(E)=0.2\), so:
\(P(W \ or \ E)=P(W)+P(E)-P(W\&E)=(P(E)+0.2)+P(E)-0=2P(E)+0.2\)
Still, multiple answers are possible. For instance, if \(P(E)=0.4\) (10 even balls), then \(P(W \ or \ E)=1\). But if \(P(E)=0.2\) (5 even balls), then \(P(W \ or \ E)=0.6\).
Not sufficient.
Answer: E.
Hope it helps.
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lexis
I got E.
The question is asking for:
P(white) + P(even) - P(white&even) = ?
(1) is saying: P(white&even) = 0
Still cannot find the answer
INSUFFICIENT
(2) is saying: P(white) - P(even) = 0.2
We don't know P(white&even), INSUFFICIENT
Together, you have
P(white) - P(even) = 0.2
and want to find: P(white) + P(even)=?
cannot complete the calculation with information given.
INSUFFICIENT
jananijayakumar
You can solve this problem in another way. Transform probability into actual numbers and draw the table.
Given:
Attachment:
1.JPG [ 8.64 KiB | Viewed 131906 times ]
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 [ 8.7 KiB | Viewed 131382 times ]
(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 [ 8.68 KiB | Viewed 131075 times ]
(1)+(2) \(c=0\) and \(b=a-5\) --> \(\frac{a+a-5+0}{25}=\frac{2a-5}{25}\). Not sufficient.
Attachment:
3.JPG [ 8.82 KiB | Viewed 130853 times ]
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lexis
We are given that there are 25 balls in a box and each one is either red, blue, or white and has a number from 1 to 10 painted on it. We need to determine the probability of selecting a white ball or an even-numbered ball.
Since each ball is colored and has a number painted on it, selecting a ball with a number or color is not mutually exclusive. Thus, to determine the probability of selecting a white or even ball we use the following formula:
P(white or even ball) = P(white ball) + P(even ball) – P(white and even ball)
Statement One Alone:
The probability that the ball will both be white and have an even number painted on it is 0.
Using the information in statement one, we know that P(white and even ball) = 0. However, we still cannot determine the probability of selecting a white or even ball. We can eliminate answer choices A and D.
Statement Two Alone:
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.
Statement two does not provide enough information to determine the probability of selecting a white or even ball. We can eliminate answer choice B.
Statements One and Two Together:
Using the information from statements one and two we still only know the following:
P(white or even ball) = P(white ball) + P(even ball) – P(white and even ball)
P(white or even ball) = P(white ball) + P(even ball) – 0
Without knowing the sum of P(white ball) and P(even ball), we cannot answer the question.
Answer: E
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We can also translate each statement into plain English:
1) There are no white balls with even numbers on them. Therefore, if we can find the chance of white and the chance of even, we can add them without having to worry about overlap. However, we don't have much to go on from this statement alone. If they asked for the chance that the ball was white AND even, clearly this would be sufficient.
2) The chance of white exceeds the chance of even by 20%. This isn't the same as saying 'The chance is 20% greater.' For instance, the chances could be white 50% and even 30%, but not white 12% and even 10%. In any case, we have no idea what the actual numbers are. Insufficient.
1 & 2) We know that the total chance = w + e. Since w = e + .2, we can say that the total chance = 2e + .2. However, since we have no idea what e is, this is insufficient. More simply, we can just say that we don't have an actual number for either probability, so there is no way to add them up.
I hope this helps!
_________________
1) There are no white balls with even numbers on them. Therefore, if we can find the chance of white and the chance of even, we can add them without having to worry about overlap. However, we don't have much to go on from this statement alone. If they asked for the chance that the ball was white AND even, clearly this would be sufficient.
2) The chance of white exceeds the chance of even by 20%. This isn't the same as saying 'The chance is 20% greater.' For instance, the chances could be white 50% and even 30%, but not white 12% and even 10%. In any case, we have no idea what the actual numbers are. Insufficient.
1 & 2) We know that the total chance = w + e. Since w = e + .2, we can say that the total chance = 2e + .2. However, since we have no idea what e is, this is insufficient. More simply, we can just say that we don't have an actual number for either probability, so there is no way to add them up.
I hope this helps!
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Posts: 4,849
There are 25 balls in a box, each of which could be red, blue or white in color. Each ball also has a number from 1 to 10 on it.
Let’s understand this in more detail by taking a few examples.
Can we have a red ball with a 7 on it? Yes we can. Can 7 be on a green ball as well? That’s possible as well.
Now that the situation is clearer, let’s proceed to understand the question stem. We need to find the probability that the one ball picked at random from the box is white or has an even number painted on it.
Using Venn diagrams is a great way to solve this question from this stage onwards.
16th June 2021 - Reply 7.JPG [ 72.75 KiB | Viewed 45594 times ]
From the Venn diagram, it’s clear that to find the required probability, we need the respective probabilities on the RHS of the equation.
From statement I alone, P(White and Even) = 0. No information about P(White) and P(Even).
Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.
From statement II alone, P(White) – P(Even) = 0.2 This is insufficient to find out the value of P(White) + P(Even). Additionally, no information about P(White & Even).
Statement II alone is insufficient. Answer option B can be eliminated. Possible answer options are C or E.
Combining statements I and II, we have the following:
From statement I alone, P(White & Even) = 0
From statement II alone, P(White) – P(Even) = 0.2.
We do not have the value of P(White) + P(Even).
The combination of statements is insufficient. Answer option C can be eliminated.
The correct answer option is E.
Hope that helps!
Aravind B T
Let’s understand this in more detail by taking a few examples.
Can we have a red ball with a 7 on it? Yes we can. Can 7 be on a green ball as well? That’s possible as well.
Now that the situation is clearer, let’s proceed to understand the question stem. We need to find the probability that the one ball picked at random from the box is white or has an even number painted on it.
Using Venn diagrams is a great way to solve this question from this stage onwards.
Attachment:
16th June 2021 - Reply 7.JPG [ 72.75 KiB | Viewed 45594 times ]
From the Venn diagram, it’s clear that to find the required probability, we need the respective probabilities on the RHS of the equation.
From statement I alone, P(White and Even) = 0. No information about P(White) and P(Even).
Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.
From statement II alone, P(White) – P(Even) = 0.2 This is insufficient to find out the value of P(White) + P(Even). Additionally, no information about P(White & Even).
Statement II alone is insufficient. Answer option B can be eliminated. Possible answer options are C or E.
Combining statements I and II, we have the following:
From statement I alone, P(White & Even) = 0
From statement II alone, P(White) – P(Even) = 0.2.
We do not have the value of P(White) + P(Even).
The combination of statements is insufficient. Answer option C can be eliminated.
The correct answer option is E.
Hope that helps!
Aravind B T
lexis
Answer: Option E
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lexis
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
General Discussion
Posts: 1
I am new to this but this is a probablity Math question. There are 25 balls. 8 of each color with the probability of
3 even numbers per color. 3 divided by 8 is 0.375 divided by 25 is 0.015 which is .02.
But where did that other ball go.
3 even numbers per color. 3 divided by 8 is 0.375 divided by 25 is 0.015 which is .02.
But where did that other ball go. 
Bunuel
how did you get this?
\(P(WorE)=2P(E)+0.2\)
BlitzHN
From (1) \(P(WorE)=P(W)+P(E)-0\) --> \(P(WorE)=P(W)+P(E)\);
From (2) \(P(W)-P(E)=0.2\) --> \(P(W)=P(E)+0.2\);
Substitute \(P(W)\) --> \(P(WorE)=P(W)+P(E)=P(E)+0.2+P(E)=2P(E)+0.2\).
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probability of the ball to be white = A
probability of the ball to be an even numbered ball = B
A U B = A + B - (A intersection B)
a. A intersection B = 0. Not sufficient as A and B not given.
b. A-B = 0.2 not sufficient for solving the equation.
a+b
A U B = B+0.2 + B - 0
gives value in terms of B. Hence not sufficient.
E.
probability of the ball to be an even numbered ball = B
A U B = A + B - (A intersection B)
a. A intersection B = 0. Not sufficient as A and B not given.
b. A-B = 0.2 not sufficient for solving the equation.
a+b
A U B = B+0.2 + B - 0
gives value in terms of B. Hence not sufficient.
E.
linda577ford
It is not essential that there will be 8 balls of each color. Each ball is red, blue or white. Overall, we could have 20 red balls, 2 blue balls and 3 white balls or 10 red balls, 10 blue balls and 5 white balls or some other combination. The point is that it is not essential that there are an equal number of balls with the same color. Similarly, the numbers on the balls will also be random. Say 9 balls could have 1-9 written on them and the rest of the balls could have 10 written on them. So you cannot find the probability of selecting a white ball or an even numbered ball until and unless you have some other data.
We know that P(Even OR White) = P(Even) + P(White) - P(Even AND White)
Both statements together don't give us the value of P(Even OR White).
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pacifist85
How did you get P(E) = 5/25?
Did you say that there are 5 even numbers in 1 to 10 and that is why 5/25? What says that the numbers cannot be repeated. In fact, each ball has a number from 1 to 10 on it - i.e. there are 10 numbers but there are 25 balls. So numbers must be repeated. So we don't know exactly how many are even.
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Why is the following logic wrong?
1) P(W and E) = P(W)*P(E) = 0 --> P(W) or P(E) or both are zero. Insufficient.
2) P(W) - P(E) = .2. Insufficient.
1+2) P(W)=.2, P(E)=0. Therefore P(W) + P(E) - P(W and E) = .2
1) P(W and E) = P(W)*P(E) = 0 --> P(W) or P(E) or both are zero. Insufficient.
2) P(W) - P(E) = .2. Insufficient.
1+2) P(W)=.2, P(E)=0. Therefore P(W) + P(E) - P(W and E) = .2
swaggerer
Think about it:
Under what conditions is P(W and E) = P(W)*P(E)? Is it always true?
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swaggerer
Think about it like this. (Represent the given info in a tree structure as shown below). We are given that every ball is either Red or Blue or White. We are also given that ball of each color can have either an even number or an odd number written on them.
Now from statement 1, we can see that the number of white balls that have an even number written on them is 0. But this doesn't mean that the total number of white balls is 0.
Similarly, this doesn't also mean that there are no even numbers written on other colored balls. (Referring to your inference that P(E) is 0).
The correct way to write the above formula would be:
P(choosing a white ball with an even number) = P(choosing a white ball out of all balls) * P(choosing an even numbered ball out of all white balls).
Notice that the second term is not P(E). Since you chose the short notation, you incorrectly inferred that the P(E) is 0. In fact, the correct inference would be P(choosing Even numbered ball from white balls) = 0)
Hope this helps.
Regards,
Krishna
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bkk145
I have a query related this explanation.
Doesn't statement 1 tell us that either P(white) or P(even) is ZERO?
If it does, then from statement 2, we can conclude that P(white) + P(even) is also 0.2, and hence the overall answer becomes 0.2; both statements combined become sufficient (C).
Explain please!
Bunuel
---
Isnt P(E) = 12 / 25 (each ball has numbers 1- 10, assuming 5 from first 10 balls, another 5 from second 10 balls, and 2 from the remaining 5 balls) which will let us calculate P(W), and then calculate P(W) + P(E) - (PWnE = 0)
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