WAEC - Mathematics (1988 - No. 21)
What is the probability of having an odd number in a single toss of a fair die?
1/6
1/3
1/2
2/3
5/6
Тайлбар
n(S) = 6(odd) = {1, 3, 5}
n(odd) = 3
pr(odd) = \(\frac{n(odd)}{n(S)}\) 3/6 =1/2
n(odd) = 3
pr(odd) = \(\frac{n(odd)}{n(S)}\) 3/6 =1/2