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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7349
GMAT 1: 760 Q51 V42 GPA: 3.82
A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 60% (01:44) correct 40% (01:36) wrong based on 69 sessions

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[GMAT math practice question]

A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from the jar, and its number is recorded before it is returned to the jar. If a second ball is then selected from the jar, what is the probability that the difference between the numbers on the two balls selected is 1?

$$A. \frac{1}{3}$$
$$B. \frac{3}{8}$$
$$C. \frac{1}{2}$$
$$D. \frac{5}{8}$$
$$E. \frac{3}{4}$$

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Originally posted by MathRevolution on 09 Apr 2018, 02:28.
Last edited by Bunuel on 18 Apr 2018, 12:14, edited 3 times in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 7672
Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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1
MathRevolution wrote:
[GMAT math practice question]

A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from the jar, and its number is recorded before it is returned to the jar. If a second ball is then selected from the jar, what is the probability that the difference between the numbers on the two balls selected is 1?

$$A. \frac{1}{3}$$
$$B. \frac{3}{4}$$
$$C. \frac{1}{2}$$
$$D. \frac{5}{8}$$
$$E. \frac{3}{8}$$

The choices are not in any order and OA is wrong..

way you can choose first ball - 4 ways
Next ball - 4 ways again as ball has been returned..
ways -$$4*4=16$$

ways in which difference is 1..
A) if the first ball is 2 or 3, next ball can be picked in TWO ways
2.... 1 or 3
3..... 2 or 4
so 2*2=4 ways
B) but if the first ball is 1 or 4, next ball can be picked in ONE way
1..... only 2
4..... only 3
so 1*2=2 ways

total = 2*2+2*1 = 6

Probability = $$\frac{6}{16} = \frac{3}{8}$$....
B
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Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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Total possibilities = 4*4 = 16

Favorable outcomes:
If the ball with #1 is chosen = 1 favorable
If the ball with #2 is chosen = 2 favorable
If the ball with #3 is chosen = 2 favorable
If the ball with #4 is chosen = 1 favorable

6/16 = 3/8

answer B
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Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from the jar, and its number is recorded before it is returned to the jar. If a second ball is then selected from the jar, what is the probability that the difference between the numbers on the two balls selected is 1?

$$A. \frac{1}{3}$$
$$B. \frac{3}{8}$$
$$C. \frac{1}{2}$$
$$D. \frac{5}{8}$$
$$E. \frac{3}{4}$$

The following are favorable outcomes for the two draws:

1, 2

2, 1

3, 2

2, 3

3, 4

4, 3

We see that we have 6 favorable outcomes. The total number of outcomes is 4 x 4 = 16, since there are always 4 balls in the jar for each of the two picks.

P(that the difference between the numbers on the two balls selected is 1) = 6/16 = 3/8.

Answer: B
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7349
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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=>

The total number of ways the two balls may be selected is 4C2 =$$\frac{( 4 * 3 )}{( 1 * 2 )} = 6.$$
There are three ways in which the numbers on the two balls selected can have a difference of $$1: ( 1, 2 ), ( 2, 3 )$$, and $$( 3, 4 )$$.

Thus, the probability is $$\frac{3}{6}$$ or $$\frac{1}{2}$$

Therefore, C is the answer.
Answer: C
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Posts: 7672
Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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MathRevolution wrote:
=>

The total number of ways the two balls may be selected is 4C2 =$$\frac{( 4 * 3 )}{( 1 * 2 )} = 6.$$
There are three ways in which the numbers on the two balls selected can have a difference of $$1: ( 1, 2 ), ( 2, 3 )$$, and $$( 3, 4 )$$.

Thus, the probability is $$\frac{3}{6}$$ or $$\frac{1}{2}$$

Therefore, C is the answer.
Answer: C

Hi..

you are wrong on two counts..

1) since ball is being returned, ways of picking two balls will be 4*4 and NOT 4C2
2) ways difference is one = (1,2); (2,1) ; (2,3) ; (3,2) ; (3,4); (4,3) so 6 ways
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Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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MathRevolution wrote:
=>

The total number of ways the two balls may be selected is 4C2 =$$\frac{( 4 * 3 )}{( 1 * 2 )} = 6.$$
There are three ways in which the numbers on the two balls selected can have a difference of $$1: ( 1, 2 ), ( 2, 3 )$$, and $$( 3, 4 )$$.

Thus, the probability is $$\frac{3}{6}$$ or $$\frac{1}{2}$$

Therefore, C is the answer.
Answer: C

You are wrong.

Total Outcomes : 16
two balls with repetition : 4 * 4 = 16

Favourable Outcomes : 6
4,3 / 3,2 / 2,1 or its inverse that is 3,4 / 2,3 / 1,2

Probability : 6/16 or 3/8
Manager  S
Joined: 28 Nov 2017
Posts: 144
Location: Uzbekistan
Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from the jar, and its number is recorded before it is returned to the jar. If a second ball is then selected from the jar, what is the probability that the difference between the numbers on the two balls selected is 1?

$$A. \frac{1}{3}$$
$$B. \frac{3}{8}$$
$$C. \frac{1}{2}$$
$$D. \frac{5}{8}$$
$$E. \frac{3}{4}$$

Answer is B. Totally agree with the reasoning given by chetan2u
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Tulkin.
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Joined: 08 Sep 2016
Posts: 113
Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from  [#permalink]

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The answer is definitely B because the balls are being replaced.

Total possibilities are 4*4 = 16

The favorable outcomes that fit the question are:
1&2 - 2 ways
2&3 - 2 ways
3&4 - 2 ways

total of 6 favorable outcomes.

6/16 = 3/8

answer B Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from   [#permalink] 18 Apr 2018, 04:21
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# A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from

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