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# Jamboree and GMAT Club Contest: A box contains 100 balls, numbered

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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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13 Nov 2015, 06:03
1
A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability that the first ball picked up is odd numbered?

(A) 0
(B) 1/3
(C) 1/2
(D) 2/3
(E) 1
Explanation:-
Total no of possibilities is 3C1(selecting 1 form 2 odd & 1 even)
No of favourable outcomes is -2C1(selecting 1 from 2 odd balls)
P=2C1/3C1=2/3 .

Ans is( D).
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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13 Nov 2015, 07:29
1
The question asks us to determine the probability that 1st ball picked up will be odd numbered, if the 3 numbers on the balls selected contain two odd and one even.
3 balls, with 2 odd numbered and 1 even numbered can be represented as Odd Odd Even, Odd Even Odd and Even Odd Odd, i.e. 3 ways.

Out of these 3, our favorable ways (1st ball is Odd numbered) are 2 i.e. Odd Odd Even and Odd Even Odd

Hence the probability is 2/3. Answer D.
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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13 Nov 2015, 09:16
1
This is a case of conditional probability.
P(B) = Probability of getting 2 odd balls and 1 even ball . There are 3 scenarios here
Odd, Odd,Even
Odd, Even, Odd
Even, Odd, Odd

1/2*1/2*1/2 + 1/2*1/2*1/2 + 1/2*1/2*1/2 = 3/8

P(A) = Probability of getting first odd ball which is
Odd, Odd,Even
Odd, Even, Odd
1/2*1/2*1/2 + 1/2*1/2*1/2 = 2/8

For conditional probability that event B has occurred and for event A to occur
P (A∩B)/P(B)
P (A∩B) = 2/8 (common scenario between events A and B) P(B) =3/8
Probability = 2/8 /3/8 = 2/3 which is choice D.

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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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13 Nov 2015, 10:11
C: 1/2. The result does not matter here, and when you started to pick from the box the number of odd and even is equal. So the probability that you can pick an odd, or an even number is 1/2.
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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13 Nov 2015, 23:33
Since we are picking balls with replacement probability or chance of picking an odd numbered ball is same as chance of picking and even numbered ball as there are equal number of odd and even numbered balls. There probability that the first ball will be odd is 1/2. Ans. C
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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14 Nov 2015, 05:10
As we are picking the balls with replacement, all the 3 picks are independent of each other. Therefore, in effect, what number occurs in the first pick has got nothing to do with what occurs on the second or the third pick.
Therefore this question can be restated as-
"What is the probability of picking an odd number from a lot of 100 numbers from 1 to 100"

$$P(getting an odd number) = \frac{Number of odd numbers in the range 1 to 100}{Total number of numbers in the range 1 to 100}$$

$$P = \frac{50}{100}$$
$$P = \frac{1}{2}$$

Option C
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Posts: 50621
Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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14 Nov 2015, 08:41
Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #5:

A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability that the first ball picked up is odd numbered?

(A) 0
(B) 1/3
(C) 1/2
(D) 2/3
(E) 1

Check conditions below:

For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

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Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!

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JAMBOBREE OFFICIAL SOLUTION:

If 3 balls are selected such that 2 of them have odd numbers on them and the other has even number, then there are 3 cases:
Odd Odd Even - OOE,
Odd Even Odd - OEO,
Even Odd Odd - EOO

Clearly, of these total 3 cases, there are 2 favorable cases in which the first ball picked is odd. Thus, required probability is 2/3.

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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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14 Nov 2015, 08:52
1
My answre is D

Stated: Three number selected: Odd, Odd, Even.

Looking for: Prob 1st ball is odd.

Total balls = 3.

How many ways can we have the first ball selected be Even? Only One - even, odd, odd. Therefore, probability of even first is even# / total # or 1/3.

Since we have two odds, either one can be selected to be first, therefore the probability of this happening is = 1 - [unfavorable] or:

1 - 1/3 = 2/3
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Posts: 7035
Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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14 Nov 2015, 21:02
Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #5:

A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability that the first ball picked up is odd numbered?

(A) 0
(B) 1/3
(C) 1/2
(D) 2/3
(E) 1

Check conditions below:

For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)

Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!

There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

Hi,
we all need to know the concept of conditional probability..
Most of us have answered 1/2... yes probability of picking one number(even/odd) in equal number of even and odd numbers will be 1/2..

But here we have moved further from this logic by having an event of already picking up 2 odd and one even number..
The Q asks us probability, which is now limited to these 3 numbers and not entire 100 numbers..
and that is why we get 2/3 as the answer..
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Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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12 Jul 2017, 00:46
chetan2u wrote:
Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #5:

A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability that the first ball picked up is odd numbered?

(A) 0
(B) 1/3
(C) 1/2
(D) 2/3
(E) 1

Check conditions below:

For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)

Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!

There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

Hi,
we all need to know the concept of conditional probability..
Most of us have answered 1/2... yes probability of picking one number(even/odd) in equal number of even and odd numbers will be 1/2..

But here we have moved further from this logic by having an event of already picking up 2 odd and one even number..
The Q asks us probability, which is now limited to these 3 numbers and not entire 100 numbers..
and that is why we get 2/3 as the answer..

So the answer is gonna be 1/2 only in the case we don't know the outcome of other selections isn't it?

One more question. Is the language to be interpreted as "any odds are and any even number is selected" or "some specific odds are and a specific even number is selected" ?

Although the answer to both will be same as I calculated both, but still what must be the interpretation here?

Also, I'm not sure that the question is properly worded. I first fell for 1/2, as every drawing is independent, but then calculated it. The question is kind of badly worded to straightaway indicate that conditional probability is into play.

Thanks
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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22 Mar 2018, 03:47
Q) Given that three balls (02 Odd and 01 Even) are picked, what is the probability of the first being odd?

Rest all info is unnecessary. What we require is this -

Odd - Even - Odd
Odd - Odd - Even
Even - Odd - Odd

So, total number of outcomes - 3
Required number of outcomes - 2

Ans) 2/3

D

Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #5:

A box contains 100 balls, numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box. If the 3 numbers on the balls selected contain two odd and one even. What is the probability that the first ball picked up is odd numbered?

(A) 0
(B) 1/3
(C) 1/2
(D) 2/3
(E) 1

Check conditions below:

For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)

Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!

There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered &nbs [#permalink] 22 Mar 2018, 03:47

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