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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 10 Oct 2017, 08:55
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



Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

...........| White | Not White
------------------------------
.....Even.| a | b
---------------------------------
Not Even.| c | d

=> Q: ( a + b + c ) / 25 ?

a + b + c + d = 25

Since we have 4 variables and 1 equation, E could be the answer most likely.

1) & 2)
From the condition 1), we have a = 0.
From the condition 2), ( a + c ) / 25 - ( a + b ) / 25 = ( c - b ) / 25 = 0.2
c - b = 5

Since we have only 2 equations a = 0 and c - b = 5, we can't figure out b + c and solve this question.

Therefore, the answer is E as expected.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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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, 18:33
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, 20:06
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] 26 Nov 2017, 20:06

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