An integer is chosen between 0 and 100. what is the probability of getting not divisible by 9?
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Answered by
10
sample space divisible by 9
( 9, 18,27,36,45,54,63,72,81,90,99)
therefore there would be only 11 number divisible by nine in numbers 1 to 100
so rather than these eleven would be
100-11
=89
so probability of getting a number not divisible by 9 would be =89/100
= 0.89
( 9, 18,27,36,45,54,63,72,81,90,99)
therefore there would be only 11 number divisible by nine in numbers 1 to 100
so rather than these eleven would be
100-11
=89
so probability of getting a number not divisible by 9 would be =89/100
= 0.89
Answered by
2
the numbers divisible by 9 are (9,18,27,36,45,54,63,72,81,90) =10
the no. of possible outcomes =100
the no. of favourable outcomes =100-10
=90
probability =90/100
=9/10
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the no. of possible outcomes =100
the no. of favourable outcomes =100-10
=90
probability =90/100
=9/10
@@HOPEIT HELPS U N PLZ MARK ME AS BRAINLIEST ☆☆☆☆
Bansarikikz:
your answer is wrong it's 11 number of sample space not 10
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