# Finding the probability by combination?

Question as follows A bag contains 50 tickets numbered 1,2,3,....,50 of which five are drawn at random and arranged in ascending order of the number appearing on the tickets ( x1 < x2 < x3 < x4 < x5 ). Find the probability that x3 = 30.

Total number of elementary events =50C5 Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 =29C2 ways and remaining two in 20C2 ways.

Therfore,favourable number of events = 29C2×20C2

Hence,required probability = (29C2×20C2)/50C5 =551 / 15134

Answer given is 551/15134 but in the solution they have taken the case where x2 will be less than x1. Is my assumption correct? if it is not then what could be the correct answer.

• You need to provide the full method, not just the numeric bottom line, otherwise your question is practically unreadable. P.S., I did not down-vote you, because I see good intentions and effort made in the question. Nevertheless, you need to clarify things here if you're hoping to get an answer. Jan 4, 2017 at 10:37
• The two numbers less than 30 you refer to are chosen, and "arranged in ascending order.." Jan 4, 2017 at 10:37
• the number of ways you can select to satisfy the condition is (any 2 from 1 - 29) x (any 1 from 30) x (any 2 from 31 - 50) = 29C2 x 1C1 x 20C2 ......... for the condition to be true, 30 had to be selected, and two that were smaller and two that were larger, the actual ordering leads to that condition, but is not part of the calculation
– Cato
Jan 4, 2017 at 10:41
• You are taking the question as wrong. Suppose you have option to pick 2 numbers out of 29. You picked 23, 12. In C(29,2) ways. Then you have to arrange them in ascending order. That is 12, 23. Only 1 way. So we have 1 * C(29,2). Similarly for last 2 numbers. Jan 4, 2017 at 10:47
• the order they are drawn doesn't matter, there will always be a lowest number, then a 2nd ranked number, then a 3rd ranked number. Can you see that if 30 was drawn, then there has to be exactly 2 drawn from 1-29? If exactly 1 was drawn from 1-29, then the condition would not be met. 1,2,30,31,32 is a set of tickets that meets the condition, I'm only counting that once
– Cato
Jan 4, 2017 at 10:49

Required probability = $\frac{\binom{29}{2} \times \binom{20}{2}}{\binom{50}{5}}$
Because $x_3$=30 is fixed. You pick two numbers from first 29 and arrange them in ascending order. Similarly 2 numbers from last 20 and arrange them in ascending.