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555-605 Level|   Probability|      
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If 5 integers are to be selected at random and without replacement 14 Feb 2024, 07:41
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­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

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If 5 integers are to be selected at random and without replacement 14 Feb 2024, 07:49
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Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

 
Show more
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:
  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
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If 5 integers are to be selected at random and without replacement 23 Feb 2024, 08:10
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Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60
Show more
If 5 distinct integers are selected from the first 20 positive integers, what is the probability that 10 and 20 are among the 5 integers selected?

A pool of 20 integers, 5 slots to fill.
The probability that 10 occupies one of these 5 slots = 5/20
19 integers remain, 4 slots left to fill.
The probability that 20 occupies one of these 4 slots = 4/19
Thus:
P(both 10 and 20 are selected) = 5/20 * 4/19 = 1/19

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If 5 integers are to be selected at random and without replacement 16 Feb 2024, 10:58
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Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60


 
Show more
\(­P = \frac{good-outcomes}{all-possible-outcomes}\)

All possible outcomes = all possible 5-integer combinations:
From 20 integers, the number of ways to choose 5 = 20C5 = \(\frac{20*19*18*17*16}{5*4*3*2*1}\)­

Good outcomes = 5-integer combinations that include 10 and 20:
We must select 3 integers from 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, and 19 to combine with 10 and 20.
From the 18 integers in blue, the number of ways to choose 3 to combine with 10 and 20 = 18C3 = \(\frac{18*17*16}{3*2*1}\)

To divide the second fraction by the first, multiply the second fraction by the reciprocal of the first:
\(\frac{18*17*16}{3*2*1} * \frac{5*4*3*2*1}{20*19*18*17*16} = \frac{5*4}{20*19} = \frac{1}{19}\)­­­­­­­

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If 5 integers are to be selected at random and without replacement Updated on: 15 Jul 2024, 13:38
Originally posted by QuantMadeEasy on 16 Feb 2024, 02:32.
Last edited by QuantMadeEasy on 15 Jul 2024, 13:38, edited 1 time in total.
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5 integers are to be selected at random, as the order of selection does not matter, the question is on combination
Total number of ways in which 5 integers can be selected out of 20 integers is 20C5

Integers 10 and 20 to be there means remaining 3 integers to be selected out of 18 integers.
Total number of ways in which 3 integers can be selected out of 18 is 18C3

Required probability = 20C5/ 18C3
Simplifying to get 1/19

IMO D

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If 5 integers are to be selected at random and without replacement 22 Feb 2024, 21:44
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Hi Bunuel

Could you explain to me why we can’t do 2/20 * 1/19 * 1 * 1 * 1? Isn’t it 2/20 because we can either choose 10 or 20 and then it is 1/19 because after we choose 10 or 20 it’s whichever one is left?

I do understand how to do 18C3
/ 20C5 but I also get confused sometimes on problems like this and solve it wrong.

Posted from my mobile device
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If 5 integers are to be selected at random and without replacement 23 Feb 2024, 01:21
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kevinhirose
Hi Bunuel

Could you explain to me why we can’t do 2/20 * 1/19 * 1 * 1 * 1? Isn’t it 2/20 because we can either choose 10 or 20 and then it is 1/19 because after we choose 10 or 20 it’s whichever one is left?

I do understand how to do 18C3
/ 20C5 but I also get confused sometimes on problems like this and solve it wrong.

Posted from my mobile device
Show more
­The reason why using 2/20 for selecting either 10 or 20 and then 1/19 for selecting the remaining one is incorrect is because it results in double-counting certain outcomes.

For instance, selecting 10 first and then 20 is one outcome, and selecting 20 first and then 10 is another outcome. However, both outcomes essentially represent the same event: choosing both 10 and 20. By using 2/20, we're inadvertently counting this event twice.
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If 5 integers are to be selected at random and without replacement 14 Mar 2024, 02:04
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Hi Bunuel, what is wrong with 1 - 18C5/20C5?

I take 18C5 to be choose 5 numbers when 10 and 20 are excluded. So in the equation 1 - 18C5/20C5 --> the solution gets us the probability when 10, 20 are included.

Thank you in advance for your kind response.
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Show Spoiler
­
­
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:

  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 19/20 * 18/19 * 17/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
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If 5 integers are to be selected at random and without replacement 14 Mar 2024, 02:14
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unicornilove
Hi Bunuel, what is wrong with 1 - 18C5/20C5?

I take 18C5 to be choose 5 numbers when 10 and 20 are excluded. So in the equation 1 - 18C5/20C5 --> the solution gets us the probability when 10, 20 are included.

Thank you in advance for your kind response.
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Show Spoiler
­
­
Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:


  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 19/20 * 18/19 * 17/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
­
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­The opposite of choosing both 10 and 20 is not neither 10 nor 20. It's neither 10 nor 20, plus 10 but not 20, and plus 20 but not 10. So, in this case, it's better to stick with the direct approach.
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If 5 integers are to be selected at random and without replacement 19 Sep 2024, 09:43
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Just use the combination formula to count your total possible selections, and then the desired selections whereby you do select 10 and 20:

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If 5 integers are to be selected at random and without replacement 06 Dec 2024, 02:45
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Out of of 20 integers, 5 to be chosen

The probability that 10 occupies one of these 5 choices =1/20 + 1/20 + 1/20 + 1/20 + 1/20 = 5/20

19 integers remain, 4 slots left to fill.

The probability that 20 occupies one of these 4 slots = 1/19 + 1/19 + 1/19 + 1/19 = 4/19

Thus: P(both 10 and 20 are selected) = 5/20 * 4/19 = 20/20*19 = 1/19
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If 5 integers are to be selected at random and without replacement 12 Dec 2024, 23:52
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Hello ,
5 integers are to be selected at random
Total number of ways in which 5 integers can be selected out of 20 integers is 20C5

Integers 10 and 20 to be there means remaining 3 integers to be selected out of 18 integers.
Total number of ways in which 3 integers can be selected out of 18 is 18C3*2C2

Required probability = 18C3*2C2/20C5
when you solve we get
1/19
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If 5 integers are to be selected at random and without replacement 25 Mar 2025, 08:45
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Sorry I couldn't get my head around the logic and I am confused on why the 2/20*1/19 would not work in this case, could you please help me understand the logic
Perhaps if you could quote a question where this works and 1/20*1/19 doesn't, it could really help.

Many thanks
Bunuel
kevinhirose
Hi Bunuel

Could you explain to me why we can’t do 2/20 * 1/19 * 1 * 1 * 1? Isn’t it 2/20 because we can either choose 10 or 20 and then it is 1/19 because after we choose 10 or 20 it’s whichever one is left?

I do understand how to do 18C3
/ 20C5 but I also get confused sometimes on problems like this and solve it wrong.

Posted from my mobile device
­The reason why using 2/20 for selecting either 10 or 20 and then 1/19 for selecting the remaining one is incorrect is because it results in double-counting certain outcomes.

For instance, selecting 10 first and then 20 is one outcome, and selecting 20 first and then 10 is another outcome. However, both outcomes essentially represent the same event: choosing both 10 and 20. By using 2/20, we're inadvertently counting this event twice.
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If 5 integers are to be selected at random and without replacement 25 Mar 2025, 11:11
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jacob46
I am confused on why the 2/20*1/19 would not work in this case, could you please help me understand the logic.

Many thanks
kevinhiro
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This approach is viable if only two numbers are being selected.
If two distinct numbers are selected from the first 20 positive integers, what is the probability that both 10 and 20 are selected?

P(first number is 10 or 20) = 2/20 (Of the 20 integers, two are 10 or 20)
P(second number is 10 or 20) = 1/19 (Of the 19 remaining integers, only one is 10 or 20)
Since we want both events to happen, we multiply the fractions:
2/20 * 1/19 = 1/190

Using combinatorics:
From 20 integers, the number of ways to choose 2 = 20C2 = \(\frac{20*19}{2*1} =190\)
Since there is only 1 good outcome -- choosing both 10 and 20 -- we get:
P(good outcome) = 1/190

If only two integers are selected, the probability that both 10 and 20 are chosen is quite small (1/190).
But in the problem at hand, five integers are selected.
Since in this case there are five opportunities to select 10 and 20 instead of only two, the probability that both 10 and 20 are among the selected integers is much greater than 1/190.
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If 5 integers are to be selected at random and without replacement 07 May 2025, 01:37
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einstein801
Hi Bunuel, what is wrong with 1 - 18C5/20C5?

I take 18C5 to be choose 5 numbers when 10 and 20 are excluded. So in the equation 1 - 18C5/20C5 --> the solution gets us the probability when 10, 20 are included.

Thank you in advance for your kind response.
Show more
If we use this approach then our answer will be:

Choosing both 10 and 20 = Total probability - choosing neither 10 nor 20 - choosing 10 but not 20 - choosing 20 but not 10

Total probability = 1

Choosing neither 10 nor 20 =\( \frac{18C5 }{ 20C5} = \frac{18*17*16*15*14}{5! }* \frac{ 5!}{20*19*18*17*16} =\frac{ 15*14 }{ 20*19} = \frac{21 }{ 38}\\
\)
Choosing 10 but not 20 (since 10 is chosen and 20 is not allowed, so 18 possible integers remain) = \(\frac{18C4 }{ 20C5 }= \frac{18*17*16*15}{4! } * \frac{ 5!}{20*19*18*17*16} = \frac{15*5}{20*19 }= \frac{15}{76}\)

Choosing 20 but no 10 = \(\frac{18C4 }{ 20C5} = \frac{15}{76}\)

Choosing both 10 and 20\( = 1 - \frac{ 21}{38 } - \frac{ 15}{76 } - \frac{ 15}{76} = \frac{(76 - 42 - 15 - 15)}{76 } = \frac{ 4}{76 } = \frac{ 1}{19}\)

Answer D.
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If 5 integers are to be selected at random and without replacement 07 May 2025, 01:47
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Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60
Show more
Total Outcomes = 20C5

Favorable outcomes = 3 more numbers out of 18 keeping 10 and 20 fixed = 18C3

Required Probability = 18C3 / 20C5 = 1/19

Answer: Option D

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If 5 integers are to be selected at random and without replacement 26 May 2025, 06:49
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Correct answer is option D.

Basically, we need to find out the probability of selecting 2 numbers from the set {10,20} and 3 numbers from the set of all numbers from 1 to 20 except for 10 and 20 (that is 18 numbers). The answer to this is same as the answer to what is the probability that 2 of the 5 selected integers will be 10 and 20.

number of ways in which 2 numbers can be selected from {10,20} = 2C2 = 1 (duh)
number of ways in which 3 numbers can be selected from the remaining 18 numbers = 18C3
number of ways in which you can select 5 numbers from the overall set of 20 numbers = 20C5

required probability = 18C3/20C5
if you know how to expand 18C3 and 20C5 then the calculation becomes very simple and it turns out to be
= [ 18*17*16 ]/[ 3*2*1 ] * [5*4*3*2*1]/[20*19*18*17*16]
everything cancels out and we get = 1/19
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

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HammeredDeepu97
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If 5 integers are to be selected at random and without replacement 26 Jul 2025, 07:00
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Hey Bunuel,

Why is it 5! / 3! for distribution? The 3 numbers that we choose apart from 10 and 20, do we consider the (10&20) as a box and the rest of the 3 numbers as movable? Also why is the probability of choosing these numbers as 1?
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:
  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
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If 5 integers are to be selected at random and without replacement 27 Jul 2025, 04:48
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HammeredDeepu97
Hey Bunuel,

Why is it 5! / 3! for distribution? The 3 numbers that we choose apart from 10 and 20, do we consider the (10&20) as a box and the rest of the 3 numbers as movable? Also why is the probability of choosing these numbers as 1?
Bunuel

Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

Probability approach:

We need the probability of {10, 20, any, any, any}.

P(10, 20, any, any, any) = 1/20 * 1/19 * 1 * 1 * 1 * 5!/3! = 1/19. We multiply by 5!/3! to account for all permutations of {10, 20, any, any, any} (such as {10, 20, any, any, any}, {10, any, 20, any, any}, {10, any, any, 20, any}, ...).

Answer: D.­

To elaborate more:
  • Case 1: {10, 20, any, any, any}. P = 1/20 * 1/19 * 1 * 1 * 1
  • Case 2: {10, any, 20, any, any}. P = 1/20 * 18/19 * 1/18 * 1 * 1
  • Case 3: {10, any, any, 20, any}. P = 1/20 * 18/19 * 17/18 * 1/17 * 1
  • ...
  • Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16

Here, when we say "any," it denotes "any number other than 10 or 20". It's important to note that the probability for each case is essentially 1/20 * 1/19. Given that there are a total of 20 cases, the overall probability becomes 1/20 * 1/19 * 20 = 1/19.

Combination approach:

The total number of ways to select 5 different numbers from 20 is 20C5.

The number of ways to select 10 and 20 is 1, and the total number of ways to select the remaining 3 numbers from 18 is 18C3.

Hence, the probability is:

\(\frac{18C3}{20C5} = \frac{(\frac{18!}{15!3!})}{(\frac{20!}{15!5!})} = \frac{18!}{15!3!}*\frac{15!5!}{20!}=\frac{1}{19}\)

­Answer: D.­
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We multiply by 5!/3! because we're fixing two numbers (10 and 20) and choosing three others from the remaining 18. The 5!/3! accounts for the number of ways to arrange 10, 20, and the 3 other numbers within the 5 slots.

As for the "probability of choosing these numbers is 1" part, once 10 and 20 are chosen, the remaining 3 can be any numbers from the other 18, so each of those has probability 1. The randomness is only in choosing 10 and 20; the rest are unrestricted.
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If 5 integers are to be selected at random and without replacement 27 Jul 2025, 07:06
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Hi Bunuel, i got the answer as 1/19, but please confirm if my approach is correct. for favorable outcomes, out of the 5 places I have to pick 2 places for both 10 & 20 so i wrote 5C2 and for favorable outcomes i did the same as out of 20 places I am choosing 2 for so 20C2 - F(O)/T(O) = 5C2/20C2 = 1/19

Please confirm
Bunuel
HammeredDeepu97
Hey Bunuel,

Why is it 5! / 3! for distribution? The 3 numbers that we choose apart from 10 and 20, do we consider the (10&20) as a box and the rest of the 3 numbers as movable? Also why is the probability of choosing these numbers as 1?
We multiply by 5!/3! because we're fixing two numbers (10 and 20) and choosing three others from the remaining 18. The 5!/3! accounts for the number of ways to arrange 10, 20, and the 3 other numbers within the 5 slots.

As for the "probability of choosing these numbers is 1" part, once 10 and 20 are chosen, the remaining 3 can be any numbers from the other 18, so each of those has probability 1. The randomness is only in choosing 10 and 20; the rest are unrestricted.
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