[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. 1/3
B. 3/4
C. 1/2
D. 5/8
E. 3/8
A jar contains 4 balls, labeled 1,2,3, and 4. A ball is sel
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- Max@Math Revolution
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P = (ways to draw 2 balls with a difference of 1)/(total ways to draw 2 balls).Max@Math Revolution 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. 1/3
B. 3/4
C. 1/2
D. 5/8
E. 3/8
Total ways to draw 2 balls:
Number of options for the first ball = 4. (Any of the 4 numbers.)
Number of options for the second ball = 4. (Any of the 4 numbers.)
To combine these options, we multiply:
4*4 = 16.
Ways to draw 2 balls with a difference of 1:
1,2
2,1
2,3
3,2
3,4
4,3
6 options.
Thus:
P = 6/16 = 3/8.
The correct answer is E.
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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 . . . . . 2 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
2 . . . . . only 4
So, 1 * 2 = 2 ways.
Total 2*2 + 2*1 = 6
Probability = 6/16 = 3/8. Option E.
Regards!
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 . . . . . 2 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
2 . . . . . only 4
So, 1 * 2 = 2 ways.
Total 2*2 + 2*1 = 6
Probability = 6/16 = 3/8. Option E.
Regards!
- Max@Math Revolution
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=>
The total number of ways the two balls may be selected is 4C2 = ( 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 3/6 or 1/2
Therefore, C is the answer.
Answer: C
The total number of ways the two balls may be selected is 4C2 = ( 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 3/6 or 1/2
Therefore, C is the answer.
Answer: C
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The following are favorable outcomes for the two draws:
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. 1/3
B. 3/4
C. 1/2
D. 5/8
E. 3/8
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: E
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