If we draw a ball from an urn containing 10 numbered balls, with seven red balls that are numbered 1, 2, 3, 4, 5, 6, 7 and three that are green and are numbered 8, 9, 10, what is the probability of drawing a ball that is red and odd?

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to