Math, asked by modijanvi, 11 months ago

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 Anonymous
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 rajsingh24
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.

Similar questions
Math, 11 months ago