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In a set of 24 cards, each card is numbered with a different positive

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Bunuel
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In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   25 Jun 2018, 05:51
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In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24



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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   25 Jun 2018, 06:00
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Solution



Given:
    • In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24
    • One card will be drawn at random from the set

To find:
    • The probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7

Approach and Working:
    • From 1 to 24, inclusive, the number of cases where we can get a number which is divisible by both 2 and 3 = numbers divisible by 6 = 4 numbers (6, 12, 18, and 24)
    • From 1 to 24, inclusive, the number of cases where we can get a number which is divisible by 7 = 3 numbers (7, 14, and 21)
    • Total number of favourable cases = 4 + 3 = 7
    • Therefore, the required probability = \(\frac{7}{24}\)

Hence, the correct answer is option C.

Answer: C
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BrentGMATPrepNow
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   25 Jun 2018, 07:22
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Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24


P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes)

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = number of integers that are divisible by both 2 and 3 OR are divisible by 7/24

number of integers that are divisible by both 2 and 3 OR are divisible by 7
The integers that meet this condition are: 6, 7, 12, 14, 18, 21, 24
There are 7 such numbers

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = 7/24

Answer: C

Cheers,
Brent
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   25 Jun 2018, 07:31
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Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24



NEW question from GMAT® Official Guide 2019


(PS02127)


Total number 24
Number div by 2 and 3 = multiple of LCM of 2 and 3 till 24
LCM of 2 and 3 =2*3=6
Multiple of 6 till 24=6,12,18,24...4 number
Number div by 7 =7,14,21...3 numbers
Total 3+4=7

Probability to pick these 7 out of 24 is 7/24
C
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   26 Jun 2018, 18:13
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Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24


The numbers that are divisible by both 2 and 3 are 6, 12, 18, and 24.

The numbers that are divisible by 7 are 7, 14, and 21.

So the probability is 7/24.

Answer: C
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   03 Jan 2019, 20:11
how come 1 is not included? is one not divisible by 2,3, and 7?
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   02 Jul 2019, 16:22
miketide wrote:
how come 1 is not included? is one not divisible by 2,3, and 7?



I'm trying to figure out the same thing. I included 1 and got 8/24. Here I am thinking "oh, you're not gonna get me" and I was wrong -_-

Can anyone explain why 1 is not included? 1 is divisible by all three numbers.
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In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   01 Jan 2020, 18:05
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justgottamakeit1592 wrote:
miketide wrote:
how come 1 is not included? is one not divisible by 2,3, and 7?



I'm trying to figure out the same thing. I included 1 and got 8/24. Here I am thinking "oh, you're not gonna get me" and I was wrong -_-

Can anyone explain why 1 is not included? 1 is divisible by all three numbers.


1 is not divisible by 2,3,7. But 2,3 and 7 are divisible by 1.

e.g 1/7 = non integer ; 7/1 = 7
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   17 Jan 2020, 21:04
from 1 to 24: div by 2 and 3 means divisible by 6 = 6,12,18,24

Div by 7 = 7, 14, 21

So total number div by 2,3 or 7 = 4+3 =7

And total number of cards = 24

So probably = number of success/ total number = 7/24
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   24 Dec 2020, 06:27
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Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24


The numbers that are divisible by both 2 and 3; will be divisible by \( 2*3=6\)

Numbers are divisible by \(6=\frac{24-6}{6}+1=4\\
\)
Numbers are divisible by \(7=\frac{21-7}{7}+1=3\)

The number divisible by \(2 \ and 3 \ or \ 7 =4+3=7\)

The probability \(= \frac{7}{24}\)

The answer is \(C\)
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   02 May 2021, 22:09
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Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24



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Answer: Option C

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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   18 Sep 2021, 03:04
Correct option : 7/24 - c
Numbers divisible by 2,and 3 or by 7 in range 1 to 24 are {6,12,18,24,7,14,21} = 7 nos
P = 7 / 24
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   29 May 2023, 17:45
BrentGMATPrepNow wrote:
Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24


P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes)

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = number of integers that are divisible by both 2 and 3 OR are divisible by 7/24

number of integers that are divisible by both 2 and 3 OR are divisible by 7
The integers that meet this condition are: 6, 7, 12, 14, 18, 21, 24
There are 7 such numbers

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = 7/24

Answer: C

Cheers,
Brent



BrentGMATPrepNow
Thank you for this! I have a hypothetical question. What if the question asked for cards from 1 through 42. Because 42 is divisible by both 2, 3, and 7, would you count it twice within the odds? In other words, would the answer be 13/42? I am just confused with how to deal with the "or" if there is overlap with the specifications should there be a similar problem in this nature.

Divisible by 2 & 3: 6, 12, 18, 24, 30, 36, 42
Divisible by 7: 7, 14, 21, 28, 35, 42

7/42 + 6/42 = 13/42

Thanks so much!
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] New post   30 May 2023, 08:01
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gonna_get_there72 wrote:
BrentGMATPrepNow wrote:
Bunuel wrote:
In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ?

A. 3/24

B. 4/24

C. 7/24

D. 8/24

E. 17/24


P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes)

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = number of integers that are divisible by both 2 and 3 OR are divisible by 7/24

number of integers that are divisible by both 2 and 3 OR are divisible by 7
The integers that meet this condition are: 6, 7, 12, 14, 18, 21, 24
There are 7 such numbers

So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = 7/24

Answer: C

Cheers,
Brent


BrentGMATPrepNow
Thank you for this! I have a hypothetical question. What if the question asked for cards from 1 through 42. Because 42 is divisible by both 2, 3, and 7, would you count it twice within the odds? In other words, would the answer be 13/42? I am just confused with how to deal with the "or" if there is overlap with the specifications should there be a similar problem in this nature.

Divisible by 2 & 3: 6, 12, 18, 24, 30, 36, 42
Divisible by 7: 7, 14, 21, 28, 35, 42

7/42 + 6/42 = 13/42

Thanks so much!


First of all, if a number is divisible by 2 and 3, then we can just say that the number is divisible by 6.

So we want to find: P(number is divisible by 6 OR divisible by 7)

P(A or B) = P(A) + P(B) - P(A and B)
So, P(number is divisible by 6 OR divisible by 7) = P(number is divisible by 6) + P(number is divisible by 7) - P(number is divisible by 6 AND 7)
= 7/42 + 6/42 - 1/42
= 12/42
= 2/7
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Re: In a set of 24 cards, each card is numbered with a different positive [#permalink]
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