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Twenty-three cards are numbered consecutively from 1 through 23. If on

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Twenty-three cards are numbered consecutively from 1 through 23. If on  [#permalink]

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New post 27 Nov 2018, 00:31
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A
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C
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E

Difficulty:

  45% (medium)

Question Stats:

60% (01:13) correct 40% (01:25) wrong based on 56 sessions

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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

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Re: Twenty-three cards are numbered consecutively from 1 through 23. If on  [#permalink]

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New post 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

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Twenty-three cards are numbered consecutively from 1 through 23. If on  [#permalink]

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New post Updated on: 28 Nov 2018, 07:11
Bunuel wrote:
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


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
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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.
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Re: Twenty-three cards are numbered consecutively from 1 through 23. If on  [#permalink]

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New post 27 Nov 2018, 10:36
Afc0892 wrote:
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

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Afc0892 why have you subtracted P of both 3 & 4?
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Twenty-three cards are numbered consecutively from 1 through 23. If on  [#permalink]

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New post 27 Nov 2018, 12:34
Archit3110 wrote:
Afc0892 wrote:
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

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Afc0892 why have you subtracted P of both 3 & 4?


Because u are right about these statements:

There are 5 numbers who are a multiple of 4, including 12.

There are 7 numbers who are a multiple of 3, including 12

However, the multiple of 3&4 = 12. So when the number is 12 it is a multiple of 3, a multiple of 4 and a combined multiple of 3 & 4.

There is still only one card numbered “12” in the deck of 23 cards.

So the total of numbers is multiples of 3 ONLY (without 12) + multiples of 4 ONLY (without 12) + combined multiple of 3&4 (12).

This is: 6+4+1 = 11 numbers
Probability is 11/23 as there are 23 cards.

This results in C

Hope that makes sense
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Twenty-three cards are numbered consecutively from 1 through 23. If on   [#permalink] 27 Nov 2018, 12:34
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