Ben twice chooses a random integer between 1 and 50, inclusive (and he may choose the same integer both times). What is the probability that at least one of the numbers Ben chooses is a multiple of 8?
+66 There are 6 multiples of 8 below 50. 8, 16, 24, 32, 40, 48. There is a 44/50 chance that he doesn't choose 8 the first time. The second time is also a 44/50 chance he doesn't choose 8. so the probability that he doesn't choose at least 1 multiple of 8 is (44/50)^2=0.7744. so the probability that he chooses at least 1 multiple of 8 is 1-0.7744=\(\boxed{0.2256}\)
+2666 There are \(50^2 = 2500\) cases
To solve this problem, we will count the ways of getting at least 1 multiple of 8.
1 multiple of 8: 6 ways for the multiple of 8, 44 ways for the non-multiple, and 2 ways to order multiple and non-multiple, so \(44 \times 6 \times 2 = 528\)
2 multiples of 8: 6 ways for the first number, 6 ways for the second (we don't order because they are the same case), so \(6 \times 6 = 36\)
Thus, the total probability is \({{528+36} \over 2500 }= \color{brown}\boxed{0.2256}\)
