# probability of a selection

• November 27th 2010, 03:41 AM
kumaran5555
probability of a selection
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.
• November 27th 2010, 04:18 AM
Plato
You under counted the possibilities.
Any selection of four tickets can be arranged is ascending order.
There are $\binom{40}{4}$ such selections.
How of those contain the ticket numbered 20, two ticket numbered less than 20 and one numbered greater than 20?
• November 27th 2010, 04:26 AM
kumaran5555
Thanks.

the answer would be, (19C2 * 20C1) / 40C4.

But, how should one interpret my approach. I don't understand what i have tried there.

I appreciate, if you could explain.
• November 27th 2010, 04:31 AM
Plato
Quote:

Originally Posted by kumaran5555
But, how should one interpret my approach. I don't understand what i have tried there.

I don't understand that approach either.
• November 27th 2010, 08:08 AM
Soroban
Hello, kumaran5555!

Quote:

The answer would be: . $\displaystyle \frac{(_{19}C_2)(_{20}C_1)}{_{40}C_4}$

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 $_{40}C_4$ 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 $_{19}C_2$ ways.

We want one number to be greater than 20.
. . There are 20 numbers greater than 20; select 1 of them.
. . There are $_{20}C_1$ ways.

The numerator is: . $1\cdot(_{19}C_2)\cdot(_{20}C_1)$

Got it?

• November 29th 2010, 03:57 AM
kumaran5555
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.
• November 29th 2010, 04:30 AM
Plato
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?