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Question : A bag contains 40 tickets numbered $1, 2, 3, ...,40$ of which four are drawn at random and arranged order $(t_1 < t_2 < t_3 < t_4)$. Find the probabitlity of $t_3$ being $20$.

My solution was: If $t_3 = 20$, then the tickets $t_1$ and $t_2$ must come out from $19$ tickets(1 to 19). This can be done in $C(19, 2)$ ways. $t_4$ must come from $20$ tickets ($21$ to $40$), which can be done in $C(20, 1)$ ways. So Ans $ = C(19, 2) \times C(20, 1)$

But I checked the solution in book, it states that total number of cases$ = C(40, 4)$ ways I Agree

Therefore favorable number of cases(i.e. solution) is $(C(19, 2) \times C(20,1)) / (C(40, 4))$

I want to know why extra division by whole possible cases $C(40, 4)$ is done to get the solution of problem. And what number does my solution represents

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up vote 1 down vote accepted

If the problem had asked you in how many ways can $t_3$ be 20, you would not divide. You would simply compute the number of favorable outcomes, as you have done.

The problem asks, however, for the probability that $t_3$ is 20. In probability problems, you must divide by the total number of outcomes, since the probability of an event is defined to be the number of favorable outcomes divided by the total number of outcomes.

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It took me sometime to even understand your those two lines. But now I get the gist. Probability is the keyword here. I have to divide by total outcomes because probability of outcomes is asked not just outcomes. Thanks. Now this site can be my next next home. – Master Chief Nov 18 '12 at 6:46

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