There are 8 disks in a container that are numbered 23, 24

In order 27 to be the median of a set of 5 numbers it should be the middle term when arranged ascending/descending order. So, as you correctly noticed, we need to select 2 numbers from {23, 24, 25, 26} and 2 numbers from {28, 39, 30, 31}.

\(P=\frac{C^2_4*C^2_4}{C^4_8}=\frac{6*6}{70}=\frac{18}{35}\).

Answer: D.

Hope it's clear.
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Note also that, if we want to use alexpavlos reasoning, we must consider that there are two ways to choose from A or B, and then two ways to choose each number from their respective groups:

4/8 x 3/7 x 4/6 x 3/5 x (1/2) x (1/2) = 18/35

As usual Bunuel, your explanations to the problems are always great...

i have a question here about the question itself.. i always tend to get more confused on the probability questions.

"4 disks are randomly removed from the container, one after the other, without replacement. Then a list of 5 numbers containing the 4 numbers on the disks selected, together with the number 27 will be made. "

What is the meaning of the part of the sentence pasted from question above?

Does it mean that they took the numbers (4 numbers ) from the original list, now with 27 they are again placing the numbers back into the list?

This means that the set we are considering consists of 4 numbers selected plus number 27, so total of 5 numbers. For example if we select disks numbered 23, 24, 25, and 26 then the set would be {23, 24, 25, 26, 27}.

Hope it's clear.
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Someone kindly edit the answer choice.
There is a typo.
It should be 18/35 instead of 18/15.

Typo edited. Thank you.
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For 27 to be the median of the drawn numbers, we must have 2 numbers from the Group A and 2 numbers from Group B. There are 4!/(2!*2!) = 6 ways to arrange AABB.

The probability of drawing AABB is the same regardless of the order they were drawn in = (4/8)(3/7)(4/6)(3/5) = 3/35.

Thus the total probability is 6 * (3/35) = 18/35.

Answer: D

[wrapimg=]alex1233 wrote:
There are 8 disks in a container that are numbered 23, 24, 25, 26, 28, 29, 30 and 31. 4 disks are randomly removed from the container, one after the other, without replacement. Then a list of 5 numbers containing the 4 numbers on the disks selected, together with the number 27 will be made. What is the probability that 27 is the median of the list of 5 numbers?

A. 2/5
B. 17/35
C. 1/2
D. 18/35
E. 5/9

I split the numbers into two groups: Group A (23,24,25,26) and Group B (28,39,30,31). Then I saw that in order to have a median of 27 we need to have 2 from group and and 2 from group B. The probability of that is 4/8 x 3/7 x 4/6 x 3/5 = 3/35. However this is not one of the answers provided. What am I doing wrong?

Thanks!


For 27 to be the median of the drawn numbers, we must have 2 numbers from the Group A and 2 numbers from Group B. There are 4!/(2!*2!) = 6 ways to arrange AABB.

The probability of drawing AABB is the same regardless of the order they were drawn in = (4/8)(3/7)(4/6)(3/5) = 3/35.

Thus the total probability is 6 * (3/35) = 18/35.

Answer: D[/wrapimg]

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To elaborate Speedily's answer & to help those who are wondering why 3/35 is not the answer.

when u do.... 4/8*3/7*4/6*3/5 ...what we are doing here is we are doing permutation 4P2 * 4P2 / 8P4

But this problem doesn't require permutation for the arrangement doesn't matter here...we just need two nos smaller than 27 and two nos greater than 27..no matter how the numbers are arranged. Hence Bunuel method of using combinations 4C2 * 4C2 / 8C4 is the best method.

But if we use 4/8*3/7*4/6*3/5 which is a permutation, we need to change it to combination.

[A permutation is changed to combination by multiplying/dividing with Factorials]

what is that factorial?

In how many ways can the disc be chosen...

1. Choosing first disc between nos [23-26]1 ; second disc again between [23-26]2; third disc between [28-31]1 and fourth again bet [28-31]2 [I have written numbers 1 and 2 after the brackets to denote first & second disc from the same group] OR

2. Choosing first disc between nos [23-26]1; second bet [28-31]1 ; third between [28-31]2 & fourth bet [23-26]2.........

hence there in 4! ways this can be done [all choices have similar outcome only] but two numbers belong to the group [23-26] and the other two belong to the group [28-31] and therefore no. of ways = 4! / [2!*2!] = 6


Now multiply 3/35 with 6 to get the ans 18/35.

Thanks for the clarification Right, it might be easier to think about it the opposite way of how I explained it above:

We need two numbers from each group. AABB. Notice we can choose these in any order, we just need two from the group LEFT of the median (27) and two from the group RIGHT of the median.

The probability of this is (4/8)(3/7)(4/6)(3/5) = 3/35.

Notice we get the same probability if we had calculated drawing ABAB = (4/8)(4/7)(3/6)(3/5) = 3/35.

Now we just need to think of the total number of ways we can draw 2 from the left and 2 from the right. This is the number of permutations of AABB = 4C2 or (4!)/(2!)(2!) = 6.

Therefore the total probability is 3/35* 6 = 18/35.

Hopefully this makes sense

If 27 is the median of the 5 numbers, then 2 numbers must be less than 27 and the other 2 numbers must be greater than 27. Let’s denote a number less than 27 by L and a number greater than 27 by G. So,

P(LLGG in this order) = 4/8 x 3/7 x 4/6 x 3/5 = 1/2 x 3/7 x 2/3 x 3/5 = 1/1 x 1/7 x 1/1 x 3/5 = 3/35

We see that there are 4!/(2!2!) = 24/4 = 6 ways to arrange 2 L’s and 2 G’s and each of these has a probability of 3/35. Therefore, the overall probability is 6 x 3/35 = 18/35.

Answer: D
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I would use one of the approaches already suggested.
That said, below is another approach -- one that might prove useful for other probability problems.

good cases = all possible cases - bad cases

Let Group A represent the 4 values above 27:
28, 29, 30, 31
Let Group B represent the 4 values below 27:
23, 24. 25, 26

All possible cases:
From 8 integers, the number of ways to select 4 = 8C4 \(= \frac{8*7*6*5}{4*3*2*1} = 70\)

Bad case 1: 4 integers are selected from Group A or from Group B, with the result that 27 will not be the median of the resulting 5 numbers
Number of options for the selected group = 2 (A or B)
Number of ways to select 4 integers from this group = 4C4 \(= \frac{4*3*2*1}{4*3*2*1} = 1\)
To combine these options, we multiply:
2*1 = 2

Bad case 2: 3 integers are selected from Group A or from Group B, with the result that 27 will not be the median of the resulting 5 numbers
Number of options for the group from which 3 integers are selected = 2 (A or B)
Number of ways to select 3 integers from this group = 4C3 \(= \frac{4*3*2}{3*2*1} = 4\)
Number of ways to select 1 integer from the other group = 4C1 = 4
To combine these options, we multiply:
2*4*4 = 32

Good cases:
all possible cases - bad case 1 - bad case 2 = 70 - 2 - 32 = 36

Probability that 27 is the median:
\(\frac{good-cases}{all-possible-cases }= \frac{36}{70} = \frac{18}{35}\)


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Hi

Cant I make a set of the numbers required 26,26,28 and 29 and solve?
Ans came out to be 12/35 though.

Hi antdmcd
Could you help me understand this?
My approach:
[(4/8)(4/7)(3/6)(2/5)] + [(4/8)(3/7)(4/6)(3/5)] = 5/35 = 1/7
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Since the 4 selections are without replacement, using a combinatorics approach to finding the favorable and total possible outcomes might be the quickest way to solve.


23—24–25–26————28–29–30–31


We are picking 4 random numbers from the above 8 numbers.

In order for our favorable outcome to occur (27 is the median of the set with 4 number chosen and 27), we must successfully select 2 disks with numbers LESS THAN 27 and 2 disks with numbers GREATER THAN 27

Only then will the 4 numbers selected, when combined with 27, result in the Median being 27

Out of the 8 total numbers on the disks: 4 are less than 27 and 4 are greater than 27


P(27 is median) = P(A)

# or favorable outcome = (# of ways to choose 2 numbers less than 27) * (# of ways to choose 2 numbers greater than 27) =

(4 c 2) * (4 c 2) = 4!/(2! 2!) * 4!/(2! 2!) = 6 * 6 = 36 favorable outcomes


Total Possible Outcomes = Total number of ways to select Any 4 numbers from the 8 total numbers = 8 c 4 = 8!/(4! 4!) = 70 total outcomes


P(A) = 36/70 =

18/35

Posted from my mobile device

HI Bunuel, why have you divided it by 8C4? Didn't get that.

The probability of an event is the number of favorable outcomes divided by the total number of outcomes possible. There the numerator is the number of favorable outcomes and the denominator is the total number of outcomes possible.
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