11yes... i got it...
nishant sir, is the ans 12/ 30C3 ??
OUT OF 30 TICKETS MARKED WITH NUMBERS 1 TO 30 , 3 ARE DRAWN AT RANDOM ;
FIND THE PROBABILITY THAT THESE ARE IN GEOMETRIC PROGRESSION.
PLS. HELP ME WITH THESE SUM.
pls... tell how the answer comes...
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5 Answers
62first thing is that if you chose 3 tickets.. the number of ways is 30C3
you have to chose 3 that are in GP...
YOu have different options...
the largest number is ar^2
so ar^2 has to be less than equal to 30
it is clear that r can be atmost 5... otherwise r^2>30
now we have a very few choices for a, r .. so we can find the number of favorable cases..
can you solve it from here?
11@ Nishant bhaiya, you forgot probably that in this problem, r can have rational values as well.
Like 8,12,18.
What abt them?
I don't think the upper limit of r helps much.
11i got a soln. out...
we can analyse it in this way..
as they are in g.p.
then b2 = a.c
therefore analysing the squares, we get a no. of ordered pairs..
like
22 = (1,4)
32 = (1,9)
42 = (2,8),(1,16)
52 = (1,25)
62 = (2,18),(3,12),(4,9)
72 = can not provide any..
82 = (4,16)
92 = (3,27)
102 = (4,25),(5,20)
112 = no cases..
122 = (6,24),(8,18),(9,16)
132 = can not be obtained..
142 = no cases..
152 = (9,25)
162 = no cases..
172 = no cases.
182 = (12,27)
192 = no cases..
202 = (25,16)
.
.
.
we dont need to go further as the squares are big enough to give the combination of nos between 1 to 30.
hence the ans is 18/30C3.