19. What is the probability of choosing 3 distinct
numbers randomly from (1, 2, 3, ...,100) such that
all are divisible by both 2 and 3?
Answers
Answered by
49
Numbers which are divisible by both 2 and 3 = multiples of 6 =
6,12,18....96.
a = 6 , d = 6
a(n) = a+(n-1)d
96 = 6+(n-1)6
90 = 6(n-1)
15= n-1
n = 16 .
So numbers between 1-100 which are divisible by both 2 and 3 = 16 .
Probability = Required no. of events/Total no. of events
= 16/100 = 8/50 = 4/25
Answered by
62
QUESTION :-
What is the probability of choosing 3 distinct
numbers randomly from (1, 2, 3, ...,100) such that
all are divisible by both 2 and 3?
ANSWER:-
the Numbers is are divisible by both 2 and 3 = multiples of 6 =
a = 6 , d = 6
a(n) = a+(n-1)d. [FORMULA]
put values,
96 = 6+(n-1)6
90 = 6(n-1)
15= n-1
n = 16 .
Number is 16.
Probability = Required no. of events/Total no. of events
= 16/100 = 4/25
hence,
Probability is 4/25.
THANKS.
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