You under counted the possibilities.
Any selection of four tickets can be arranged is ascending order.
There are such selections.
How of those contain the ticket numbered 20, two ticket numbered less than 20 and one numbered greater than 20?
Hi,
A bag has 40 tickets number from 1 to 40. 4 tickets namely t1,t2,t3,t4 have been taken and arranged as t1<t2<t3<t4.
Then what is the probability of having t3 as 20.
What i have tried is, t1 can range from 1 to 19. so the t1's probability is 19/40.
t2 can be remaining 18 numbers which are less than 20, and total tickets are 39 now. so 18/39.
And t3 is 1/38 and t4 can be any one of the remaining 37 tickets so 1/37.
The probability of the interested event is 19/40 * 18/39 * 1/38 * 20/37.
But, this is not matching with the answer.
Please correct if my approach is wrong.
thanks.
Hello, kumaran5555!
The answer would be: .
But, how should one interpret my approach.
I don't understand what i have tried there.
Now, I don't understand!
You wrote that answer . . . but don't know how or why?
There are possible outcomes.
. . That is the denominator.
Now consider the numerator. .(Plato already outlined it.)
We want "20" . . . There is one (1) way.
We want two numbers to be less than 20.
. . There are 19 numbers less than 20; select 2 of them.
.There are ways.
We want one number to be greater than 20.
. . There are 20 numbers greater than 20; select 1 of them.
. . There are ways.
The numerator is: .
Got it?
Thanks Soroban.
But I want to know the interpretation of my first approach, which I have showed in the first post of this thread.
I know, I have really messed up there. But to make myself clear, i need to know what i have done there.
So please if you could check.
As I have told you, I don’t know what you were thinking in that first post.
It really makes no sense.
Although this may help you: Order makes no difference in this question.
We make a selection of four, one of which is 20, two less than 20 and one more.
You may have been thinking order matters?