If 5 integers are to be selected at random and without replacement

 

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

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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Bunuel

Thank you for the question. Can you explain further? I am struggling to understand the 5!/3!

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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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 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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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.

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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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­I think in the case 20 there is a typo and should be this
Case 20: {any, any, any, 20, 10}. P = 18/20 * 17/19 * 16/18 * 1/17 * 1/16
 

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Right. Edited. Thank you for noticing. 
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Bunuel, i am not clear on why we are diving 5! by 3! in the probability approach. Since the 3 values of any,any,any are technically 3 different options, why can't we just account for the arrangement of 5 terms by 5! ? If they were the same item, I would divide by 3! as well. Thanks!