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Math Expert V
Joined: 02 Sep 2009
Posts: 58336
Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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13 00:00

Difficulty:   65% (hard)

Question Stats: 54% (01:43) correct 46% (01:42) wrong based on 231 sessions

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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

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

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7
1
We are given that, out of the three balls selected 2 are odd and 1 is even.
We can have the following arrangements: OOE, OEO, EOO
P(1st ball is odd numbered) = 2/3

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

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The question seeks, what is the prob that the first ball picked is odd...?

When we start, total no of odd balls = 50
When we start, total no of ball = 100.

Required probability is = Ways of selecting 1 odd ball / Ways of selecting one ball = 50 C 1 / 100 C 1 = 50/100 = 1/2.

The answer is C.
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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Hi,

The probability of any number to be picked is independent of whether it is an odd or an even.

So P(O) = 1
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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Probability that the first ball picked is odd numbered = total number of odd balls/ total number of balls
= 50/100
= 1/2

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

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Balls are numbered 1 to 100. that means 50 odd numbers and 50 even numbers.
Then probability of selecting odd or even at first instance is 1/2. Hence answer is 1/2.
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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Probability of picking 1 even an 2 odd balls = 1/2 * 1/2 * 1/2
Now we have 2 repetitions of odd balls so we have to multiply the probability with 3!/2!
Total probability = 1/2*1/2*1/2*3
=3/8

Probability of picking the 1st ball as odd = 1/2 * 1/2 * 1/2
Now the 2nd and 3rd draw could be counted in 2! ways. Also, we need to divide the entire equation by 2! since we have 2 repetitions (odd nos). So, total probability is 1/8.

This is a conditional probability. Given that we have 2 odd and 1 even ball, what is the probability of picking the 1st ball as odd

1/8 / 3/8
1/8 * 8/3
=1/3

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

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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

IMO B is correct
As its stated out of 100 we have to chose 2 odd numbered and 1 even numbered ball
so prob of this =(3!/2! )* 1/2 * 1/2 * 1/2 = 3/8
Now probability of choosing an odd ball
= 1/2 * 1/2 * 1/2
=1/8

So probability of choosing odd ball at first place = (1/8)/( 3/8)
= 1/3
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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probability of getting odd or even number is 1/2
we have 2 odds and 1 even. the probability of the first to be odd will thus be 1/2 * 2/3 or 1/3
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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IMO : C

As it's a case of replacement, each time one will have the same sample space and thus whether first or second or third pick is an odd numbered ball can simply calculated as

$$\frac{{count(1,3,5.....99)}}{{count(1,2,3,4....100)}}$$ --> $$\frac{50}{100}$$ --> $$\frac{1}{2}$$.

Please correct if wrong.
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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1
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

Selecting the balls either Even or Odd is having probability 50/100=1/2
We have already selected 3 balls with 2 odd numbers and 1 even number.
So we have 3 combinations OOE, OEO, EOO. We have 3 outcomes and 2 are favourable as in 2 cases 1st number is odd. So 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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If three ball are selected order can be anything

here order is important. so 3 balls two odd and one even can be arranged in 3 ways.

first ball can be even number in 1 way...probability is 1/3

so first ball odd number is 1-1/3=2/3

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

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1
1
The question essentially asks the probability of getting the first ball as odd out of three possible outcomes: OOE, EOO, OEO.
Here, we can see that the first ball is odd in two instances out of three.
Therefore, the probability of getting the first ball as odd is: 2/3
Ans. D
Manager  B
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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Is it 2/3 ? D

Total numbers are 3, out of which 2 are odd and 1 is even.

so first number to be odd, it should be 2/3
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GMAT 1: 650 Q50 V27 Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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P(odd)= 1/2
Probability of odd out of 3 (2 odd and 1 even) = 2/3

P(odd)*P(odd out of 3)= 1/2*2/3=1/3
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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If the 3 numbers on the balls selected contain two odd and one even. The occurances can be
O,O,E
O,E,O
E,O,O

So the probability that the first ball picked up is odd numbered 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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2
Correct answer is D i.e. 2/3 because two odd and one even can be arranged in 3!/2!=3 ways OOE,OEO and EOO out of these only two cases are favorable. So probability will be 2/3
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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1
As per the given question, there are two "Odd" numbers and one "even" number picked.(with replacement-----the most important part)

There for the scenarios can be

Odd, Odd, Even....(a)
Odd, Even, Odd....(b)
Even, Odd, Odd....(c)

Therefor for the first number to be odd, required probability

= 2/3
Option (D)
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Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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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

Soln: All the middle part is irrelevent, since the probability of first ball being odd numbered is not affected by those. So the ans is 1/2 -> C
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GMAT 1: 720 Q48 V41 Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered  [#permalink]

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1/2
Probability of 1st ball being odd= total no of odd balls divided by total no of balls in box
= 50/100 = 1/2 Re: Jamboree and GMAT Club Contest: A box contains 100 balls, numbered   [#permalink] 12 Nov 2015, 23:11

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