an integer is chosen at random between 1 and 100. Find the probability that is
(i) divisible by 8
(ii) not divisible by 8
Answers
so, sample space { or total outcomes } = 98
So, you can say total number of sample space , n(S) = 98
Now , numbers , which are divisible by 8 are : 8 , 16 , 24 , 32, 40 , 48 , 56, 64, 72 , 80, 88 , 96
So, total number which are divisible by 8 = 12
So, number of possible event , n(E ) = 12
Now, probability that it is divisible by 8 , P(E) = n(E)/n(S) = 12/98 = 6/49
P(divisible by 8) = 6/49
Probability that it is not divisible by 8 , P(E') = 1 - probability that is is divisible by 8,P(E)
E.g., P( not divisible by 8) = 1 - 6/49 = (49 - 6)/49 = 43/49
To find the possibility of the number not divisible by 8,
Probability of an integer it is not divisible by 8,
By formula,
P(E') = 1 - P(E)
1 - 6/49 = (49 - 6)/49
Hence, P(integer not divisible by 8) = 43/49Given:
An integer is chosen between 1 and 100.
Numbers that are divisible by 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96.
The rest are not divisible by 8.
Calculation:
The integers are 2, 3, 4, 5, 6,..., 99.
The number of possible outcomes = 98 (Excluding 1 and 100)
Therefore,
Sample space, n(S) = 98.
So,
The possibility of the number divisible by 8, n(E) = 12.
To find probability,
By formula,
P(E) = n(E)/n(S)
=12/98
= 6/49
Hence, P(integer divisible by 8) = 6/49
To find the possibility of the number not divisible by 8,
Probability of an integer it is not divisible by 8,
By formula,
P(E') = 1 - P(E)
1 - 6/49 = (49 - 6)/49
Hence, P(integer not divisible by 8) = 43/49
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