IIT JEE , AIEEE Forum - Mathematics , Algebra

abhishektripathi

given two no. 1,2.....100, & r multiplied . problity of no. divisiblle by 3

magiclko

the number obtained on multiplication will be divisible by 3, if any of the twon numbers we have taken is divisible by 3.... that means for every factor of 3 belonging to [1, 100]...we'll be having 99 possibilties, nd thr will be 33 such factors, thrfore total no of favourable cases = 99 X 33

thus probabilty = 99 X 33

¹⁰C₂

= 33/50

iitkgp_bipin

We have to pick two numbers from 1,2,3,......100 such that their product is divisble by 3.
It can happen when one of the selected numbers is divisible by 3.
No. of ways of selecting those nos. = 33
Hence total no. of ways = 99 X 33

No. of ways of choosing any 2 nos.= ¹⁰⁰C₂ = (100 X 99)/2 = 50 X 99

Hence probability = (99 X 33)/(50 X 99) = 33/50

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