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If there are 20 cards that have either a red number or a black number

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Intern
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Joined: 25 Jun 2018
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If there are 20 cards that have either a red number or a black number  [#permalink]

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New post 27 Jun 2018, 06:50
1
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

44% (02:13) correct 56% (01:54) wrong based on 73 sessions

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If there are 20 cards that have either a red number or a black number that is between 1 and 15 inclusive, what is the probability that a card will either be red or odd?

(1) The probability that the card is red and odd is 10%
(2) The probability that the card will be red minus the probability that it will be odd is 20%


OA is E.

A bit confused as from 1. we can deduce P(Red)*P(Odd) = 10% (insufficient standalone)
From 2 we can deduce P(Red)-P(Odd) = 20% (insufficient standalone).

Answer is either C or E.

If we combine, we get: P(Red)-0.10/P(Red) = 0.20%. P(Red) is circa 43.2%

Then we can imply P(odd)

From Here, Probability of Red OR Probability of Odd = P(Red)+P(Odd)-P(red and Odd).

Hence I chose C. This is a source which is routinely wrong on answers - can I confirm other people also get C.

Thanks!
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Re: If there are 20 cards that have either a red number or a black number  [#permalink]

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New post 27 Jun 2018, 07:48
From statement 1
P(R&O)= P(R) + P(O)- P(R OR O)= 1/10
not sufficient

from statement 2
P(R)- P(O) = 1/5
Not sufficient

Statement 1 + statement 2 ....not sufficient

E is answer.
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Re: If there are 20 cards that have either a red number or a black number  [#permalink]

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New post 27 Jun 2018, 12:19
1
1
cr7notmessi wrote:
If there are 20 cards that have either a red number or a black number that is between 1 and 15 inclusive, what is the probability that a card will either be red or odd?

(1) The probability that the card is red and odd is 10%
(2) The probability that the card will be red minus the probability that it will be odd is 20%


Since you're concerned about C versus E specifically, I'll just focus on what happens when we put the two statements together.

Both statements together:

The first statement tells us that there are 2 odd, red cards.

The second statement tells us: "there are four more red cards than there are odd cards."

We want to know the probability that a card will "either be red or odd." There are actually two ways to read this, and they're ambiguous. We could be wondering what the probability is that it's red, odd, or both. Or, we could be wondering what the probability is that it's just red or just odd, but not both. (The math term here is "inclusive or" versus "exclusive or".)

Let's suppose we're reading it the first way. Then here are two possibilities for the set of cards:

1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 : probability of either red or odd (inclusive) is 6/20 = 30%

1 1 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 : probability of either red or odd (inclusive) is 10/20 = 50%

Different answers, so it's insufficient and the answer is E.

Or maybe we're reading it the second way. We want to know how many cards are just red or just odd, but not both. Then, we can use the same two cases:

1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 : probability of either red or odd (exclusive) is 4/20 = 20%

1 1 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 : probability of either red or odd (exclusive) is 8/20 = 40%

Different answers, so it's insufficient and the answer is E.

The answer is E either way you read it.
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Re: If there are 20 cards that have either a red number or a black number &nbs [#permalink] 27 Jun 2018, 12:19
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