IIT JEE , AIEEE Forum - Mathematics , Algebra

kishan12

x1,x2,x3...........x50 are fifty real numbers such that x(r)<x(r+1) [here (r) and (r+1) are in subscript] for r=1,2,3.......49.Five numbers out of these are picked up at random.The probability that the five numbers have x20 as the middle number is..........

Ans:[20C2 * 19C2]/50C5

akhil_o

2 numbers are less than 20(ie from 1-19), 2 more than 20 such that middle is feixed at 20
ie(21,50)
so no of ways of choosing
=19C2*30C2
so P=19C2*30C2/total=19C2*30C2/50C5

computer001

else v can also say like:
prob of gettin x20 in 1st pick is 1/50
for gettin 2 nos from 1st 19 19/49,18/48
for gettin 2 nos from last 30: 30/47,29/46
but a diff 1 may b picked 1st so no of ways is 5! but among 2 from 19 there will b rep so /2! and same for last 30 so another /2!
so v get
(1/50)*(19/49)*(18/48)*(30/47)*(29/46 )*5!/(2!*2!)

computer001

but i think u shud rather do by total cases...thts easier to think of...
i did this way cuz im fond of it n regularly do this way....if u suddenly do the entire thing this way it will become quite confusin

akshay.khare91

if 20 is at middle then the starting two nos. should be picked from
1-19 therefore no. of ways of selecting are 19C2
and the last two nos. should be picked from (21-50)
therefore no. of ways of selecting = 30C2 .

therefore favorable cases= 19C2*30C2

total ways = 50C5

therefore probability = 19C2*30C2 / 50C5

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