An integer is chosen at random between 1 and 100 . find the probability that is 1 and 109.find probability that is divisible by 8 and not divisible by 8
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3
the pb of choosing 1=1/100
the pb of 108=0
divisble of 8=8,16,24,32.......96
the pb of divisible of 8=12/100
3/25
the pb of not divisible by 8=88/100
22/25
the pb of 108=0
divisble of 8=8,16,24,32.......96
the pb of divisible of 8=12/100
3/25
the pb of not divisible by 8=88/100
22/25
Answered by
0
Hey, your answer is --
Given, an integer is chosen at random
b/w 1 and 100
so, total number of outcomes = 100
now , probability of getting 1 is
P = 1/100
and probability of getting 109 is
P = 0/100 = 0
now , the possible outcomes that is divisible by 8 is -
8 , 16,24 ,36,40,48,56,64,72,80,88,96
therefore , favorable number of outcomes is 12
so, P(getting divisible by 8) = 12/100
= 3/25
probability of getting number is not divisible by 8 is
P = 100 - 12/100
= 88/100 = 22/25
hope u like
Given, an integer is chosen at random
b/w 1 and 100
so, total number of outcomes = 100
now , probability of getting 1 is
P = 1/100
and probability of getting 109 is
P = 0/100 = 0
now , the possible outcomes that is divisible by 8 is -
8 , 16,24 ,36,40,48,56,64,72,80,88,96
therefore , favorable number of outcomes is 12
so, P(getting divisible by 8) = 12/100
= 3/25
probability of getting number is not divisible by 8 is
P = 100 - 12/100
= 88/100 = 22/25
hope u like
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