If three consecutive numbers are selected randomly from the first 100 natural numbers, then what is the probability that their product of the three numbers is divisible by 24?
If three consecutive numbers are selected randomly from the first 100 natural numbers, then what is the probability that their product of the three numbers is divisible by 24?
The correct option is C
No. of ways of selecting 'r' consecutive things from 'n' consecutive things = n - r + 1
So, you can select 3 consecutive numbers in 100 - 3 + 1 = 98 ways.
Now, out of 98 selections: you can get either of the two combinations:
i) even - odd - even: If out of numbers chosen two are even and one is odd, then certainly the product of these numbers will be divisible by 24. No. of ways we can select even-odd-even is 982 = 49
ii) odd - even - odd: The product of two odd numbers and one even number will be divisible by 12 if the even number is a multiple of 8. We have 12 even multiples of 8 within 100. Hence we can have 12 combinations of odd- even - odd.
So, total number of favourable cases: 49 + 12 = 61
So, probability = 6198. Hence option (c)
No. of ways of selecting 'r' consecutive things from 'n' consecutive things = n - r + 1
So, you can select 3 consecutive numbers in 100 - 3 + 1 = 98 ways.
Now, out of 98 selections: you can get either of the two combinations:
i) even - odd - even: If out of numbers chosen two are even and one is odd, then certainly the product of these numbers will be divisible by 24. No. of ways we can select even-odd-even is 982 = 49
ii) odd - even - odd: The product of two odd numbers and one even number will be divisible by 12 if the even number is a multiple of 8. We have 12 even multiples of 8 within 100. Hence we can have 12 combinations of odd- even - odd.
So, total number of favourable cases: 49 + 12 = 61
So, probability = 6198. Hence option (c)