Probability
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Q.

$x_1,x_2,x_3,x_4.......x_{50}$ are are fifty real numbers such that $x_r

The probability that the five numbers have $x_{20}$ as the middle is:

 A.

$\dfrac{^{20}C_2\times ^{30}C_2}{^{50}C_5}$

 B.

$\dfrac{^{19}C_2\times ^{30}C_2}{^{50}C_5}$

 C.

$\dfrac{^{19}C_2\times ^{31}C_2}{^{50}C_5}$

 D.

None of these

 Hide Ans

Solution:
Option(B) is correct

As $x_r<x_{r+1}$.This means $x_1<x_2<x_3......<x_{50}$

Hence, each of the 19 numbers are less than the number $x_{20}$

And each of the 30 numbers $x_1,x_2,x_3,x_4.......x_{50}$ is greater than $x_{20}$

Out of the five numbers that are randomly picked from the set, when two numbers are picked from the set

{$x_{1},x_{2}.....x_{19}$} and two others picked from the set {$x_1,x_2,x_3,x_4.......x_{50}$} then the number

$x_{20}$ will always be in the middle.

When five numbers are arranged in an order. 

Total number of ways of selecting such five numbers is $^{19}C_2\times ^{30}C_2$

As a total number of ways of selecting a set of any five number out of 50 is $^{50}C_5$. Thus, the required probability is:

$\dfrac{^{19}C_2\times ^{30}C_2}{^{50}C_5}$

Edit: Thank you shobhit for pointing out, corrected the question.


(10) Comment(s)


Pearl
 ()

Denominator not correct


Kriti
 ()

What is wrong with the denominator?


Ayush
 ()

There are more ways to this

1)

Fix $X_{20}$ as middle number then

$^{49}C_4$ ways to select other 4 numbers

$4!$ Different arrangements are possible

So $^{49}C_4 \times 4! = 5,085,024$ ways

Sample space is $^{50}C_5 \times 5!$

So probability is $\dfrac{^{49}C_4 \times 4!}{^{50}C_5 \times 5!} =0.02$

Please think on this,

Because so many different arrangements and selection of numbers from 50 different real nose is possible

So 0.02 probability may be correct


Surabhi
 ()

I didn't understand your solution properly, but even if I follow your approach how do I arrive at the correct option?

Ayush
 ()

To "surabhi"

My answer is not given in options

Ayush
 ()

Question is saying FIVE NUMBERS PICKED AT RANDOM

So we CAN'T ASSUME that 2 numbers before $X_{20}$ are picked from $X_1$ to $X_{19}$

And 2 are from $X_{21}$ to $X_{50}$

We can also pick $X_{23}, X_{31}, X_{34}, X_{44}$ (just one example)

It is very difficult for me to explain problem and solution by writing here

But I tried my best. See above

Surabhi
 ()

No, you are mistaken, the question does say that FIVE NUMBERS ARE PICKED AT RANDOM but it continues to ask NOW WHAT IS THE PROBABILITY THAT $X_{20}$ IS THE MIDDLE OF SELECTED FIVE NUMBERS.

For this, we need to make two groups such that $X_{20}$ lies in the middle of two groups. Thus two groups are numbers from $X_1$ to $X_{20}$ ( 19 numbers) and $X_{21}$ to $X_{50}$ (30 numbers).

The fact that 5 numbers have been selected in total forms the TOTAL AVAILABLE CHOICES and the term $^{50}C_5$ thus goes into the denominator.


Shobhit
 ()

Please, correct the denominator in the answer option.


Deepak
 ()

Thank you Shobhit for letting us know the anomaly. Corrected.


Veeraraghavan.K
 ()

With 5 number selection, number of ways is given as ${^{19}C_2} \times {^{30}C_2}$; how it is possible?

It is assumed that the 5th number is always 20.

Not necessary. Average of 25,50,40,2,3 is also 20. Such are not taken into account.