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A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from [#permalink]
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Updated on: 18 Apr 2018, 12:14
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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.



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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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09 Apr 2018, 08:20
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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09 Apr 2018, 14:48
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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10 Apr 2018, 17:21
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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Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from [#permalink]
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11 Apr 2018, 05:56
=> The total number of ways the two balls may be selected is 4C 2 =\(\frac{( 4 * 3 )}{( 1 * 2 )} = 6.\) 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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Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from [#permalink]
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11 Apr 2018, 09:33
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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18 Apr 2018, 00:32
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



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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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18 Apr 2018, 03:23
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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Re: A jar contains 4 balls, labeled 1,2,3, and 4. A ball is selected from [#permalink]
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18 Apr 2018, 04:21
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
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