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27 Nov 2018, 00:31
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:19) correct
33%
(01:31)
wrong
based on 291
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Twenty-three cards are numbered consecutively from 1 through 23. If one card is to be randomly selected from the 23 cards, what is the probability that the number on the selected card will be a multiple of 3 or a multiple of 4 or a multiple of both 3 and 4 ?
A. 7/23
B. 9/23
C. 11/23
D. 12/23
E. 13/23
A. 7/23
B. 9/23
C. 11/23
D. 12/23
E. 13/23
27 Nov 2018, 03:32
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Multiple of 3 - 3,6,9,12,15,18,21 - 7 numbers.
Probability is 7/23.
Multiple of 4 - 4,8,12,16,20 - 5 numbers.
Probability is 5/23.
Multiple of both 3&4 - 12 - 1 number.
Probability is 1/23.
P(3 U 4) = P(3) + P(4) - P(3 and 4)
= 11/23
C is the answer
Posted from my mobile device
Probability is 7/23.
Multiple of 4 - 4,8,12,16,20 - 5 numbers.
Probability is 5/23.
Multiple of both 3&4 - 12 - 1 number.
Probability is 1/23.
P(3 U 4) = P(3) + P(4) - P(3 and 4)
= 11/23
C is the answer
Posted from my mobile device
Archit3110
Updated on: 28 Nov 2018, 07:11
Originally posted by Archit3110 on 27 Nov 2018, 10:35.
Last edited by Archit3110 on 28 Nov 2018, 07:11, edited 2 times in total.
Last edited by Archit3110 on 28 Nov 2018, 07:11, edited 2 times in total.
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Bunuel
P of multiple of 3 : ( 3,6,9,12,15,18,21) : 7
P of multiple of 4 : ( 4,8,12,16,20): 5
P multiple of both 4 & 3 : ( 12) : 1
since 12 is common in both sets so reduce by 1 for P(3) and P(4)
Total probablity would be 6/23+4/23+1/23= 11/23 IMO C
