Problem Solving

If 5 numbers are to be selected from integers from 1 to 20, inclusive, without replacement, what is the probability that both 5 and 15 will be selected?
(A) 1/57
(B) 1/19
(C) 3/19
(D) 4/57
(E) 2/19

What I thought of was:
(2C2 * 19C3)/ 21C5

# nos to select from = 20
Total # ways to select 5 out of 20 nos = 20C5
prob that both 5 and 15 is selected (among 5 selected numbers) = 2 (either 5,15 or 15,5)
Ways to select other 3 numbers = 18C3

Reqd prob = (2 * 18C3) / 20C5
= 2/19
Ans E

What's the OA?

Thanks anju@gurome.
I had taken 21C3 instead of 20C3, hence the error!

shashwats wrote:If 5 numbers are to be selected from integers from 1 to 20, inclusive, without replacement, what is the probability that both 5 and 15 will be selected?
(A) 1/57
(B) 1/19
(C) 3/19
(D) 4/57
(E) 2/19

An alternate approach:

Let Y = selecting 5 or 15 and N = selecting any number other than 5 or 15.

ONE WAY to select both 5 and 15: YYNNN
P(Y on the 1st pick) = 2/20. (Of the 20 numbers, either 5 or 15.)
P(Y on the 2nd pick) = 1/19. (Of the 19 remaining numbers, either 5 or 15, whichever was not selected on the first pick.)
P(N on the 3rd pick) = 1. (Since 5 and 15 have already been selected, the 3rd number selected must be a number other than 5 or 15.)
P(N on the 4th pick) = 1. (Since 5 and 15 have already been selected, the 4th number selected must be a number other than 5 or 15.)
P(N on the 5th pick) = 1. (Since 5 and 15 have already been selected, the 5th number selected must be a number other than 5 or 15.)
Since we want all of these outcomes to happen, we multiply the probabilities:
2/20 * 1/19 * 1 * 1 * 1 = 1/10 * 1/19.

TOTAL POSSIBLE WAYS:
YYNNN represents only ONE WAY to select both 5 and 15.
Now we must account for ALL OF THE WAYS to select both 5 and 15.
Any arrangement of the letters YYNNN represents a way to select both 5 and 15.
To illustrate:
NNNYY = selecting 5 and 15 on the 4th and 5th picks.
NYNYN = selecting 5 and 15 on the 2nd and 4th picks.
To account for ALL OF THE WAYS to select both 5 and 15, the product above must be multiplied by the number of ways to arrange YYNNN.

The number of ways to arrange 5 letters = 5!.
But the arrangement here includes IDENTICAL ELEMENTS.
When an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! (to account for the two identical Y's) and by 3! (to account for the three identical N's).
Thus:
Number of ways to arrange YYNNN = 5!/(2!3!) = 10.

Multiplying the two results, we get:
P(selecting both 5 and 15) = 1/10 * 1/19 * 10 = 1/19.

The correct answer is B.

Check here for a similar problem:
https://www.beatthegmat.com/select-exact ... 88786.html

experts please suggest

whats wrong with this

1/20 * 1/19 * 1 * 1 * 1 * 5(factorial) = 6/19

1/20 * 1/19 * 1 * 1 * 1 * 5(factorial) = 6/19

first i took 5 then 15 then for third we have 18 out of 18 options then forr fourth we have 17 out of 17 options and for fifth we have 16 out of 16 options hence

1/20 * 1/19 * 18/18 * 17/17 * 16/16
1/20 * 1/19 * 1 * 1 * 1
since all the 5 numbers are differant because we are talking abt without replacement

1/20 * 1/19 * 1 * 1 * 1 * 5(factorial) = 6/19

Hi Anju,

I understand the solution but i have a small question.

why when choosing 5 and 15, order doesn't matter?

we can choose 5 first or 15 first so this should double the probability.

for example u said it should be (2c2 x 18c3)/ 20c5

which is basically 18c3 / 20c5

well why did we not consider that fact that we might choose one element before the other and that should count in probability? i understand that if it is a combination question, the order doesn't count because 5,15 is the same as 15,5, but when tackling a probability question shouldn't we consider the probability of taking options?

Also when i solve it using the probability method i do the following:

2/20 x 1/19 x 10--> which is the number of ways to arrange the 2 elements in 5 places/choices

this answer yields to 1/19

so here i started with 2/20, meaning the first selection can include either 5 or 15.

please explain

Thanks

Anju@Gurome wrote:

srcc25anu wrote:# nos to select from = 20
Total # ways to select 5 out of 20 nos = 20C5
prob that both 5 and 15 is selected (among 5 selected numbers) = 2 (either 5,15 or 15,5)
Ways to select other 3 numbers = 18C3

Reqd prob = (2 * 18C3) / 20C5?

First of all it is not probability but number of ways to select both 5 and 15.
And secondly, it is selection, so we do not need to consider relative arrangements or the order in which 5 and 15 are selected.

For example, {5, 6, 7, 8, 15} and {15, 6, 7, 8, 5} are same selections.