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Each of the cards in a deck of 50 cards has a number from 1 to 20
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25 Jun 2016, 05:23
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Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink? (1) The probability that the card will both have an off number and be written in red ink is 0. (2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 P.S. Give a kudo or bookmark the question if you are unable to solve it and I will send you the solution in your inbox
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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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26 Jun 2016, 05:56
For those of you who were not able to answer this, here is the answer. We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x. So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red)  (prob of selecting a card with an odd number written in red) or we can write the above as: x = P(odd) + P(red)  P(odd and red).Statement 1: P(odd and red) = 0Insufficient as we don't know P(odd) + P(red). Statement 2: P(odd)  P(red) = 0.4This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc. Hence, statement 2 is also insufficient! Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)P(red) and P(odd and red). Therefore, the answer will be E.  P.S. Don't forget to give kudos
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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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26 Jun 2016, 08:07
14101992 wrote: For those of you who were not able to answer this, here is the answer. We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x. So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red)  (prob of selecting a card with an odd number written in red) or we can write the above as: x = P(odd) + P(red)  P(odd and red).Statement 1: P(odd and red) = 0Insufficient as we don't know P(odd) + P(red). Statement 2: P(odd)  P(red) = 0.4This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc. Hence, statement 2 is also insufficient! Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)P(red) and P(odd and red). Therefore, the answer will be E.  P.S. Don't forget to give kudos Quite not understood the logic x = P(odd) + P(red)  P(odd and red). 1) says P(odd and red)=0 so x = P(odd) + P(red)  0= x = P(odd) + P(red)2) says P(odd) P(Red)=0.4 P(odd)= 0.5 so P(red)=0.1 combine 1 and 2 x= 0.5+0.1=0.6 are we not getting a value here?



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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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26 Jun 2016, 10:03
RatneshS wrote: 14101992 wrote: For those of you who were not able to answer this, here is the answer. We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x. So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red)  (prob of selecting a card with an odd number written in red) or we can write the above as: x = P(odd) + P(red)  P(odd and red).Statement 1: P(odd and red) = 0Insufficient as we don't know P(odd) + P(red). Statement 2: P(odd)  P(red) = 0.4This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc. Hence, statement 2 is also insufficient! Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)P(red) and P(odd and red). Therefore, the answer will be E.  P.S. Don't forget to give kudos Quite not understood the logic x = P(odd) + P(red)  P(odd and red). 1) says P(odd and red)=0 so x = P(odd) + P(red)  0= x = P(odd) + P(red)2) says P(odd) P(Red)=0.4 P(odd)= 0.5 so P(red)=0.1 combine 1 and 2 x= 0.5+0.1=0.6 are we not getting a value here? Hi! RatneshS, Statement 2 can have many combinations: P(odd) P(Red)=0.4 P(odd)= 0.5 so P(red)=0.1 or P(odd)= 0.2 so P(red)=0.2 and so on. We are not getting a unique value by combining both statement.
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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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22 May 2017, 11:14
Divyadisha wrote: RatneshS wrote: 14101992 wrote: For those of you who were not able to answer this, here is the answer. We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x. So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red)  (prob of selecting a card with an odd number written in red) or we can write the above as: x = P(odd) + P(red)  P(odd and red).Statement 1: P(odd and red) = 0Insufficient as we don't know P(odd) + P(red). Statement 2: P(odd)  P(red) = 0.4This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc. Hence, statement 2 is also insufficient! Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)P(red) and P(odd and red). Therefore, the answer will be E.  P.S. Don't forget to give kudos Quite not understood the logic x = P(odd) + P(red)  P(odd and red). 1) says P(odd and red)=0 so x = P(odd) + P(red)  0= x = P(odd) + P(red)2) says P(odd) P(Red)=0.4 P(odd)= 0.5 so P(red)=0.1 combine 1 and 2 x= 0.5+0.1=0.6 are we not getting a value here? Hi! RatneshS, Statement 2 can have many combinations: P(odd) P(Red)=0.4 P(odd)= 0.5 so P(red)=0.1 or P(odd)= 0.2 so P(red)=0.2 and so on. We are not getting a unique value by combining both statement. Odd numbers between 1 and 20 is 10. So P(odd) = 10/50= 0.2 Am I doing something wrong?



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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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24 Apr 2018, 05:21
Hi, Can someone explain this question in a simpler way? I'm sorry, I'm not able to understand the logic or correct me if I'm making some wrong assumptions. Statement 1 says: The probability that the card will both have an odd number and be written in red ink is 0. Which means that the red cards are all even. Had there been one red card, which is odd then the probability wouldn't be 0. So my understanding is that, all the red cards are even. So 10 cards. then in the remaining 40 cards it'd be 20 blue cards of even and odd. 20 black cards of even and odd. The probability of odd cards is 30/50 and probability of red card is 10/50. So is the final answer 4/5? and is statement 1 not sufficient to answer?



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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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24 Apr 2018, 05:50
Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will e ither have an odd number or be written in red ink? (1) The probability that the card will both have an off number and be written in red ink is 0. P(O) . P(R) = 0(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 P(O)  P(R) = 0.4Clearly 1 and 2 individually not sufficient. combining both P(O)  P(R) = 0.4 : Squaring both sides we get \([P(O)]^2\) + \([P(R)]^2\)  2\(P(O)\).\(P(R)\) = \(0.16\) from statement 2 we know P(O).P(R) = 0
hence \([P(O)]^2\) + \([P(R)]^2\) = \(0.16\) = rewriting the LHS
\([ P(O) + P(R) ]^2\) 2\(P(O)\).\(P(R)\) = \(0.16\)
\(P(O).P(R) = 0\) from statement 2
\([ P(O) + P(R) ]^2\) = \(0.16\)
\([ P(O) + P(R) ]\) = \(0.4\)  which is a definite answerWhat is wrong with my approach?
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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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07 May 2018, 23:30
14101992 wrote: For those of you who were not able to answer this, here is the answer. We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x. So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red)  (prob of selecting a card with an odd number written in red) or we can write the above as: x = P(odd) + P(red)  P(odd and red).Statement 1: P(odd and red) = 0Insufficient as we don't know P(odd) + P(red). Statement 2: P(odd)  P(red) = 0.4This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc. Hence, statement 2 is also insufficient! Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)P(red) and P(odd and red). Therefore, the answer will be E.  P.S. Don't forget to give kudos Correct me if I'm wrong: P(odd)  P(red) = 0.4 squaring both sides: [P(odd)  P(red)]^2 = 0.4^2 P(odd)^2 + P(red)^2 2P(odd)xP(red) = 0.4^2 Since P(odd and red) = 0 (from statement 1): 2P(odd)xP(red) = 0 P(odd)^2 + P(red)^2 = 0.4^2 Now we have to calculate P(odd) + P(red) squaring this expression we get: P(odd)^2 + P(red)^2 + 2P(odd)xP(red) Since P(odd and red) = 0 (from statement 1): 2P(odd)xP(red) = 0; we are left with P(odd)^2 + P(red)^2 which we have proved above to be 0.4^2



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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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25 Jun 2018, 18:42
14101992 wrote: For those of you who were not able to answer this, here is the answer. We know there are cards written with numbers in different probability colours. We are asked for the probability of randomly selecting a card that has either an odd or is written in red. Let us assume this prob to be x. So, x= (prob of selecting a card with an odd number) + (prob of selecting a card written in red)  (prob of selecting a card with an odd number written in red) or we can write the above as: x = P(odd) + P(red)  P(odd and red).Statement 1: P(odd and red) = 0Insufficient as we don't know P(odd) + P(red). Statement 2: P(odd)  P(red) = 0.4This can be true for various combination of P(odd) and P(red), for ex. P(odd)=0.5 and P(red)=0.1 or P(odd)=0.6 and P(red)=0.2 etc. Hence, statement 2 is also insufficient! Combining 1 and 2 also we do not have exact value of P(odd) and P(red). We just have values for P(odd)P(red) and P(odd and red). Therefore, the answer will be E.  P.S. Don't forget to give kudos I agree with the rest of the comments. What you want is P(odd n red) = P(odd) + P(red)  P(odd u red) [note: n stands for intersect, u stands for union] Statement 1 clearly tells us that P(odd u red) is 0. So we need to know what P(odd) + P(red) is. In statement 2, I think you meant to to exploit that some people wouldn't know the actual formula. There are several other questions like this, so it's not a bad question. But if you randomly select a card that must be either even or odd, and we are given a finite number of evens and odds, I can tell you that P(odd)=(.5) From there: P(odd) P(red) = .4 .5  P(red) = .4 P(red) = .1 P(red) = .1 Thus P(red n odd) = .5 + .1  0 = .6 There is only one answer, so C is correct.



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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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25 Jun 2018, 19:26
14101992 wrote: Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink? (1) The probability that the card will both have an off number and be written in red ink is 0. (2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 P.S. Give a kudo or bookmark the question if you are unable to solve it and I will send you the solution in your inbox Hi... Lot of people going wrong .... Reason : Assuming something that is not given. It is not given that even and odd are equal in number. It may just be possible that all are odd or all are even. Don't get into P(O), P(R) etc, the meaning should get you to the correct answer. (1) The probability that the card will both have an odd number and be written in red ink is 0. It just tells us that there is no overlap and thus no odd number is written in red inkInsufficient (2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 So (ODD cards)/Total  (Red cards)/total =0.4 So O/50  R/50 = 0.4 OR=0.4*20=8 O can be 10 and R2 or O50 & R42 etc Insufficient Combined Nothing new.. Now the values can be O10 , R2 etc Only that O +R cannot be >50 Insufficient
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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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25 Jun 2018, 19:48
chetan2u wrote: 14101992 wrote: Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink? (1) The probability that the card will both have an off number and be written in red ink is 0. (2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 P.S. Give a kudo or bookmark the question if you are unable to solve it and I will send you the solution in your inbox Hi... Lot of people going wrong .... Reason : Assuming something that is not given. It is not given that even and odd are equal in number. It may just be possible that all are odd or all are even. Don't get into P(O), P(R) etc, the meaning should get you to the correct answer. (1) The probability that the card will both have an odd number and be written in red ink is 0. It just tells us that there is no overlap and thus no odd number is written in red inkInsufficient (2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 So (ODD cards)/Total  (Red cards)/total =0.4 So O/50  R/50 = 0.4 OR=0.4*20=8 O can be 10 and R2 or O50 & R42 etc Insufficient Combined Nothing new.. Now the values can be O10 , R2 etc Only that O +R cannot be >50 Insufficient Oh wow, my apologies to the author.



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Re: Each of the cards in a deck of 50 cards has a number from 1 to 20
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26 Jun 2018, 00:50
Leo8 wrote: Each of the cards in a deck of 50 cards has a number from 1 to 20 written on it in either black, red or blue ink. If one card is to be selected at random from the deck, what is the probability that the card selected will either have an odd number or be written in red ink?
(1) The probability that the card will both have an off number and be written in red ink is 0. P(O) . P(R) = 0
(2) The probability that the card will have an odd number minus the probability that the card will be written in red ink is 0.4 P(O)  P(R) = 0.4
Clearly 1 and 2 individually not sufficient. combining both
P(O)  P(R) = 0.4 : Squaring both sides we get \([P(O)]^2\) + \([P(R)]^2\)  2\(P(O)\).\(P(R)\) = \(0.16\) from statement 2 we know P(O).P(R) = 0
hence \([P(O)]^2\) + \([P(R)]^2\) = \(0.16\) = rewriting the LHS
\([ P(O) + P(R) ]^2\) 2\(P(O)\).\(P(R)\) = \(0.16\)
\(P(O).P(R) = 0\) from statement 2
\([ P(O) + P(R) ]^2\) = \(0.16\)
\([ P(O) + P(R) ]\) = \(0.4\)  which is a definite answer
What is wrong with my approach? The probability that the card will both have an odd number and be written in red ink is 0. This statement does not mean P(O).P(R) = 0. This statement means that P(OR) = 0. It means that there is no odd numbered card with red ink. P(OR) = P(O intersection R)/P(R) O interesection R = 0 Let's take an example. Say, there are 10 items  5 blue marbles and 5 red chairs. Let's pick a random item now. Probability of picking a blue object = P(B) = 1/2 Probability of picking a chair = P(C) = 1/2 Now, probability of picking a blue chair is not equal to P(B).P(C). That would give us 1/4, whereas, the probability of picking a blue chair is ZERO. Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app




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