What is the probability that a two digit number selected at random will be a multiple of 3 and not 5?
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Answered by
62
There are 90 two digit numbers(99–9).
Out of this there are 30 numbers divisible by 3(33–3)
Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5.
Therefore, the favorable cases are 30–6=24.
Hence, the required probability is 24/90 = 4/15.
×××
hope it helps):-
Out of this there are 30 numbers divisible by 3(33–3)
Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5.
Therefore, the favorable cases are 30–6=24.
Hence, the required probability is 24/90 = 4/15.
×××
hope it helps):-
Answered by
23
>> There are 90 two digit numbers(99–9).
>> Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5.
>> Therefore, the favorable cases are 30–6=24.
>> Hence, the required probability is 24/90 = 4/15.
( hope u understand it !! )
>> Out of this there are 6 numbers divisible by 15(15, 30, 45, 60, 75, 90), which are also divisible by 5.
>> Therefore, the favorable cases are 30–6=24.
>> Hence, the required probability is 24/90 = 4/15.
( hope u understand it !! )
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