If three integers between 1 and 1000 are selected at random, what is the probability that all of them are divisible by 3 and product of the numbers if even.
- \(\text{Multiples of 3: } \left\lfloor \frac{1000}{3} \right\rfloor = 333\)
- Product is even if at least one of the three numbers is even
- 3 cases
- \(i. 3 \times 6 \times 9 = 162 \rightarrow Even\)
- \(ii. 3 \times 6 \times 12 = 212 \rightarrow Even\)
- \(iii. 3 \times 9 \times 15 = 405 \rightarrow Odd\)
- So favorable = all triples of multiples of 3 minus those with all three odd multiples of 3 \(\rightarrow \frac{\binom{333}{3} - \binom{167}{3}}{\binom{1000}{3}}\)
If three integers between 1 and 1000 are selected at random what is the probability that all of them are even and divisible by 3.
- So divisible by 6
- \(\text{Multiples of 6: } \left\lfloor \frac{1000}{6} \right\rfloor = 166\)
- \(\boxed{\frac{\binom{166}{3}}{\binom{1000}{3}}}\)