bsahil wrote:
n is integer from 1-99:
1. Probabilty that n(n+1) is divisible by 3
2. Probabilty that n(n+1)(n+2) is divisible by 6
3. Probabilty that n(n+1)(n+2) is divisible by 8
Please help me... I am unable to get this one
1.
If n is between 1-99,
probability that n is divisible by 3 = 1/3 (every 3 rd number is divisible by 3)
probability that n+1 is divisible by 3 = 1/3
Divisibility of n and (n+1) are mutually exclusive. n(n+1) is divisible by 3, if either n or (n+1) is divisible by 3.
P[n(n+1)] = P[n divisible by 3 OR (n+1) divisible by 3] = P[n] + P[(n+1)] = 1/3 + 1/3 = 2/3
2. Probabilty that n(n+1)(n+2) is divisible by 6 - Product of 3 consecutive numbers will be always divisible by 6. Probability = 1
3. Probabilty that n(n+1)(n+2) is divisible by 8
n is even: (n+1) doesn't contribute to 8 divisibility, as this will be an odd number.
n(n+1)(n+2) - will be divisible by 8, if n(n+2) is divisible by 8.
n(n+2) will be divisible by 8, if one of them is divisible by 4. (8=4*2, Only one number can be divisible by 4, and we can get factor 2 from the other number)
for n between 1-99
P[n div. by 4] = 24/99 [had it been 1-100, we cud have got 1/4 - every 4th number)
P[(n+2) div. by 4] = 25/99 [from 3-101]
P[n(n+2) div. by 4 or 8] = 24/99 + 25/99 = 49/99
n is even, P[n(n+1)(n+2) div. by 8] = 49/99
n is Odd: n(n+1)(n+2) - will be divisible by 8, if n+1 is divisible by 8.
for n between 1-99, n+1 is between 2-100.
between 2-100, we have 12 multiples of 8.
P[(n+1) div. 8] = 12/99
n is odd, P[n(n+1)(n+2) div. 8] = 12/99
Summing up even and odd cases,
Probabilty that n(n+1)(n+2) is divisible by 8 = 49/99 + 12/99 = 61/99
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