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A bag contains a total of four billiard balls. The billiard [#permalink]
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Updated on: 09 Jul 2013, 10:43
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A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7. Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd? Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd?
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Originally posted by slingfox on 26 Oct 2009, 00:39.
Last edited by Bunuel on 09 Jul 2013, 10:43, edited 1 time in total.
Renamed the topic and edited the question.



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A bag contains a total of four billiard balls. The billiard [#permalink]
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26 Oct 2009, 00:54
slingfox wrote: Can you guys help me figure out the most efficient way to do this type of problem?:
A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7.
Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?
Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd? Scenario 1:P(sum odd) = (Favorable outcomes)/(Total # of outcomes) Total # of outcomes = 4C2 = 6, # of selection of 2 balls out of 4 balls. Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3, 5, 7. So two ball must be (2,3), (2,5), (2,7). So we have 3 favorable outcomes. P(sum odd) = (Favorable outcomes)/(Total # of outcomes) = 3/6 = 1/2. Scenario 2: Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3, 5, 7. This can happen in TWO ways: even (2), odd (3,5,7); OR odd(3,5,7), even(2). P(sum odd) = 2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls) = 2*1/4*3/4 = 3/8
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Re: Probability Question [#permalink]
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26 Oct 2009, 05:29
Thanks Bunuel. You are my hero!
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Re: Probability Question [#permalink]
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26 Oct 2009, 07:23
Bunuel wrote: slingfox wrote: Can you guys help me figure out the most efficient way to do this type of problem?:
A bag contains a total of four billiard balls. The billiard balls are numbered as follows: 2, 3, 5 and 7.
Scenario 1: Two billiard balls are pulled out of the bag. What is the probability that the sum of the numbers appearing on the balls is odd?
Scenario 2: A billiard ball is pulled out of the bag then placed back in. Another billiard ball is pulled out of the bag (which may or may not be identical to the first) then placed back in. What is the probability that the sum of the two balls is odd? Scenario 1: P(sum odd)=Favorable outcomes/Total # of outcomes Total # of outcomes=4C2=6, # of selection of 2 balls out of 4 balls. Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3,5,7. So two ball must be (2,3),(2,5),(2,7). So we have 3 favorable outcomes.P(sum odd)=Favorable outcomes/Total # of outcomes=3/6=1/2 Scenario 2: Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3,5,7. This can happen in TWO ways: even (2), odd (3,5,7); OR odd(3,5,7), even(2). P(sum odd)=2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls)=2*1/4*3/4=3/8 In the first case we consider (2,3),(2,5),(2,7) but why do we not consider (3,2), (5,2), (7,2). But then total no. of cases will be 6 and Probability cannot be 6/6 = 1. I m too confused Help!!!



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Re: Probability Question [#permalink]
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26 Oct 2009, 09:03
rvthryet wrote: In the first case we consider (2,3),(2,5),(2,7) but why do we not consider (3,2), (5,2), (7,2). But then total no. of cases will be 6 and Probability cannot be 6/6 = 1. I m too confused Help!!! Yes, that is confusing. But consider this: Scenario 1: P(sum odd)=Favorable outcomes/Total # of outcomes Total # of outcomes=4C2=6, # of selection of 2 balls out of 4 balls. Favorable outcome: two numbers can total odd if one number is even 2 and another odd 3,5,7. So two ball must be (2,3),(2,5),(2,7). So we have 3 favorable outcomes. P(sum odd)=Favorable outcomes/Total # of outcomes=3/6=1/2 Favorable outcome is the PAIR of balls, therefore it doesn't matter it'll be (2,3) or (3,2), it's still ONE pair. On the other hand for the total # of outcomes we are also counting number of drawing of TWO balls out of 4. BUT we can count this in another way (the way we did for scenario 2): First ball 2, seond any odd =1/4*3/3(as after drawing the even ball there are 3 balls left and all are odd)=1/4 First ball any odd, second ball 2 =3/4*1/3(as after drawing the odd ball there are 3 balls left and only one is even)=1/4. Final P=1/4+1/4=1/2 OR: 2*1/4*3/3=1/2 Scenario 2: Different situation. But again the odd sum can occur if either of drawing gives even 2 and another odd 3,5,7. This can happen in TWO ways: even (2), odd (3,5,7); OR odd(3,5,7), even(2). P(sum odd)=2(as we have 2 ways favorable sum can occur)*1/4(probability of picking even 2, 1 ball out of 4 balls)*3/4(probability of picking odd 3,5 or 7, 3 out of 4 balls)=2*1/4*3/4=3/8 We also can count the probability here with combination: 2*1C1*3C1/4C1*4C1=2*(1*3)/(4*4)=3/8 1C1=# of selections of even out of one 3C1=# of selections of odd out of three 4C1=TOTAL # of selections of first ball out of 4 4C1=TOTAL # of selections of second ball out of 4 (as after drawing the first one we put it back, so there are still 4 balls in the bag) Multiplying this by 2 as there are two ways: ood/even or even/odd. Hope it's clear.
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Re: Probability Question [#permalink]
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26 Oct 2009, 16:40
Bunuel wrote: BUT we can count this in another way (the way we did for scenario 2): First ball 2, seond any odd =1/4*3/3(as after drawing the even ball there are 3 balls left and all are odd)=1/4 First ball any odd, second ball 2 =3/4*1/3(as after drawing the odd ball there are 3 balls left and only one is even)=1/4. Final P=1/4+1/4=1/2
I like this method more. And i know u get this a lot these days but this is awesome stuff.. Thnks a ton..



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Re: Probability Question [#permalink]
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09 May 2011, 00:33
1 3/6 = 1/2 2 let 1st ball be 2 then 2nd ball has to be 3,4,5 thus probability = 1/4 * 3/4 let 1st ball be either 3,4 or 5 and 2nd ball be 2 then = 3/4 * 1/4 thus 2* 1/4 * 3/4
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Re: Probability Question [#permalink]
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10 May 2011, 07:43
Scenario 1: 1 ball has to be 2 (1 * 3)/(4C2) = 3/6 = 1/2 Scenario 2: P(2)*P(any other number) + P(Any other number) * P(2) = 2 * 1/4 * 3/4 = 6/16 = 3/8
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Re: A bag contains a total of four billiard balls. The billiard [#permalink]
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Re: A bag contains a total of four billiard balls. The billiard
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