Each of the 25 balls in a bag is red, blue or white and is numbered 1 to 10. If a ball is picked what is the probablity that it will be either white or will have an even number printed on it?
1) Probability that the ball will be white and have an even number on it is 0
2) Prob of ball being white less prob of ball having an even number on it is 0.2
OA is E
Here is how I approached it:
Let prob of ball being white = P(W)
Prob ball having even number is = P(E)
1) P(E) * p(W) = 0 therefore either P(E) = 0 or P(W) = 0
Not enough
2) P(W) - P(E) = 0.2
Not enough
Combining both, P(E) = 0 since prob cant be negative therefore P(W) = 0.2 and P(E) + P(W) = 0.2
What is my mistake?
Thanks
Prob Doubt
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Target question: What is the value of P(white or even)?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.
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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Nijo wrote:Each of the 25 balls in a bag is red, blue or white and is numbered 1 to 10. If a ball is picked what is the probablity that it will be either white or will have an even number printed on it?
1) Probability that the ball will be white and have an even number on it is 0
2) Prob of ball being white less prob of ball having an even number on it is 0.2
OA is E
Here is how I approached it:
Let prob of ball being white = P(W)
Prob ball having even number is = P(E)
1) P(E) * p(W) = 0 therefore either P(E) = 0 or P(W) = 0
Not enough
2) P(W) - P(E) = 0.2
Not enough
Combining both, P(E) = 0 since prob cant be negative therefore P(W) = 0.2 and P(E) + P(W) = 0.2
What is my mistake?
Thanks
Your formula is incomplete.
It should be: P(white or even) = P(white) + P(even) - P(white & even)
Cheers,
Brent
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An alternate approach is to TEST CASES.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.
Let:
W = the total number of white marbles.
E = the total number of even marbles.
WE = the total number of white even marbles.
Statement 1:
Thus, WE = 0.
No way to determine the value of W or E.
INSUFFICIENT.
Statement 2:
Implication:
W-E = (0.2)(25) = 5.
Case 1: W=10, E=5, WE=0
Here, P(W or E) = (10+5)/25 = 15/25 = 3/5.
Case 2: W=11, E=6, WE=0
Here, P(W or E) = (11+6)/25 = 17/25.
Since P(W or E) can be different values, INSUFFICIENT.
Statements combined:
Cases 1 and 2 satisfy both statements.
Since P(W or E) can be different values, INSUFFICIENT.
The correct answer is E.
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Answer: Option ENijo wrote:Each of the 25 balls in a bag is red, blue or white and is numbered 1 to 10. If a ball is picked what is the probablity that it will be either white or will have an even number printed on it?
1) Probability that the ball will be white and have an even number on it is 0
2) Prob of ball being white less prob of ball having an even number on it is 0.2
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