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Paritosh
CAT Preparation Community 9y ago
Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. What is the probability that the numbers on them will be in AP?
7
Subhadeep Chakraborty 9y ago
10/399 ??
ANKUR MAHIWAL 9y ago
ans -- 10/119 , maximum common difference can be 10 (tickes nos selected (1 , 11 , 21) ) and minimum common difference is 1 , comm difference ---- count of triplets with that common difference 1
common diff ---- count of triplets with that common diff
1 19
2 17
.......
10 1
this will form an AP
TOTAL = 1+3.+5+7...........17+19 = 100
possible ways -- 21C3
ans -- 10/119
Paritosh 9y ago
@ANKUR MAHIWAL Procedure is correct. But 100/(21C3) = 10/133.....
ANKUR MAHIWAL 9y ago
@paritosh , yes it should be 133 , i read 19 table wrongly ..
Wannabe JOKAR 9y ago
20/133
Rishi Kala 6y ago
100/1330
Anurag pant 6y ago
let the numbers be a,b,c so if a,b,c are in AP the condition is
b=a+c/2
it means that b is an integer ,so from here we can if a,b are even or a,b are odd because odd+odd=even or even+even=even
so consider case 1 where both a and b are odd numbers so the possible numbers of outcomes is 11C2 because there are 11 odd numbers between 1 to 21
now consider case no.2 where a and b are even numbers so total possible ways is 10C2 because there are 10 even numbers between 1 to 21
if we get a and b then we can easily calculate the b so we dont need to calculate the total ways of selecting b
and the total no. of outcomes are 21C3
so probibality =[10C2 + 11C2]/21C3=10/133
