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

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

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Updated on: 05 Feb 2019, 06:00
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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

Originally posted by lexis on 01 May 2008, 11:01.
Last edited by Bunuel on 05 Feb 2019, 06:00, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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13 Aug 2010, 03:42
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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 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.

Hope it's clear.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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01 May 2008, 13:37
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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 have an eve number painted on it is 0.2

I got E.

P(white) + P(even) - P(white&even) = ?

(1) is saying: P(white&even) = 0
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
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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13 Aug 2010, 09:27
16
4
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 [ 8.64 KiB | Viewed 66328 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 66311 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 66314 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 66280 times ]

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

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17 Jul 2011, 21:51
4
linda577ford wrote:
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.

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

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15 May 2011, 10:50
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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.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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11 Oct 2016, 16:55
3
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

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.

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

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08 Oct 2010, 16:35
2
2
BlitzHN 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.

Hope it's clear.

how did you get this?
$$P(WorE)=2P(E)+0.2$$

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

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16 May 2011, 13:03
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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!
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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04 May 2015, 11:14
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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
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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23 May 2016, 19:34
2
rjivani 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.

Hope it's clear.

---

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)

What says that numbers are allotted in sequence? There could be 5 balls with number 1 on them, 10 balls with number 2 on them, 3 balls with number 3 on them and then rest of the 7 balls with numbers 4 to 10 on them. Or any other combination.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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01 May 2008, 13:43
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I agree that the answer is E. Here is my logic:

The question is asking for what is P(W) or P(E).

Statement 1 tells you that P(W) and P(E) is mutually exclusively. Thus P(W+E) = 0 So not enough info on its own.

Statement 2 tells you that P(W)-P(E) is 0.2. That is not sufficient either. In order to find P(W) or P(E), we need P(W) + P(E). However there is no information given concerning P(W) and P(E).

Please let me know if my analysis is not correct. Probability is not my strong suit.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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05 Aug 2013, 10:54
1
Stiv 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.

Hope it's clear.

You say in the second statement that multiple values are possible (0.6 and 0.4 or 0.4 and 0.2)
Are these values only possible? If they are please explain why.
Why they cannot be 0.3 and 0.1 or 0.5 and 0.3?

solution:

White balls = w
Red = R
Blue = b
Total ball = 25
Sum total of even numbered ball = E

We have to evaluate = w/25 + (E)/25

From st(1) , we only know there is no white ball which contains even number. Even we still don’t know about red and blue balls have how many even numbered balls in them. So all are in mystery and doubt.

From st(2), w/25 – E/25 = 0.2 , again all unknown.

From both statement, we can assume several different things, I mean double case.
So both are insufficient. Answer is (E)
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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05 May 2015, 04:38
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swaggerer wrote:
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.

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

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10 Oct 2016, 09:36
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ariiet wrote:
This is my approach. Please correct me where I went wrong.
I chose A for this one.....

- Label from 1 to 10 >> maximum number of each balls = 10

My curious is, from statement 1, can't we conclude that white balls are only marked with odd numbers?

Then, the total number of balls is 25 thus there are 5 white balls, 10 blue balls, and 10 red balls.
Blue balls and red balls have 5 even balls each. >>> Even balls = 10
White balls = 5

P(white)+P(Even) = 10/25+5/25 = 3/5

Yes, we can conclude that white balls are only marked with odd numbers. But we have numbers on the balls from 1 to 10 only and we have total 25 balls. SO, it is obvious that numbers are repeated. Now, We don't have any information whether we do have only 5 ODD numbered white balls or more than 5 ODD numbered white balls. So, we cannot conclude the number of white balls. Hence, Statement 1 is also insufficient.

I hope it is clear now. Let me know if you have more doubts.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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04 Feb 2017, 05:55
1
ajay2121988 wrote:
How do you get to this formula.:

$$P(WorE)=P(W)+P(E)-P(W&E)$$

Any underlying concept ?

Yes, it is a general formula to determine the probability of Events A or B when both A and B are NOT mutually exclusive.

For example, say I have the numbers from 1 to 10.

Event A : All even numbers
Event B : All multiples of 3.

Probability of A or B will be All even + All multiple of 3 - Those that have been repeated(only 6 in this case).

I hope it is clear now.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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07 Feb 2017, 09:49
1
ajay2121988 wrote:
How do you get to this formula.:

$$P(WorE)=P(W)+P(E)-P(W&E)$$

Any underlying concept ?

Exactly same as that of two overlapping sets.

n(A or B) = n(A) + n(B) - n(A and B)

Why do you subtract n(A and B) out of it? Because it is double counted - once in n(A) and another time in n(B).

Similarly, P(A and B) is double counted - once in probability of A and another time in probability of B.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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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.

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

For more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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01 May 2008, 21:24
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.
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Re: Each of the 25 balls in a certain box is either red, blue or white and  [#permalink]

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08 Oct 2010, 16:31
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.

Hope it's clear.

how did you get this?
$$P(WorE)=2P(E)+0.2$$
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Re: Each of the 25 balls in a certain box is either red, blue or white and   [#permalink] 08 Oct 2010, 16:31

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