A 3 digit natural number is selected at random .Find the probability that it is divisible by 2 but not by 3?
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OpenStudy (anonymous):
y u delete the answer
OpenStudy (anonymous):
I think I missed something
OpenStudy (anonymous):
k
OpenStudy (anonymous):
so by solving it , we can get asnwer
OpenStudy (anonymous):
2/3 of (9*10*5)/(9*10*10)
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OpenStudy (anonymous):
So, it is 1/3
OpenStudy (anonymous):
final answer
OpenStudy (anonymous):
yep
OpenStudy (anonymous):
see we have 3 digit nos from 100 to 999
and we have to find prob , no should div by 2 not by 3
OpenStudy (anonymous):
i think something is missing
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OpenStudy (anonymous):
number should be divisible by 2 and not by 3
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
@ExperimentX
would u plse guide me
OpenStudy (anonymous):
So, there are 9*10*5, 3 digit numbers that are divisible by 2
OpenStudy (anonymous):
right?
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OpenStudy (anonymous):
@Shubhamsrg
OpenStudy (shubhamsrg):
nos divisible by 2= 100,102...998 ->450 nos.
nos divisible by 6(both 2 and 3) = 102,108..996 ->150 nos.
so total nos we have to take care of = 450 - 150 = 300
total nos. from 100 to 999 = 900
so our required probability = 300/900 = 1/3
as @sauravshakya rightly said.