Question

An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n + 1) will be divisible by 3 ? \small \frac{1}{9} \small \frac{1}{3} \small \frac{1}{2} \small \frac{2}{3} \small \frac{5}{6}

Answer 1

Answer:

66.666%

Step-by-step explanation:

For all multiple of 3 and all multiple of 3 - 1

Will produce a number divisible by 3 in the function n(n+1)

33 numbers are multiple of 3

33 numbers are multiple of 3 - 1

Sample 99

Event 66

So probability = 66/99 *100

= 66.666%