Final answer:
The probability that all three balls drawn without replacement are yellow is 1/120.
The probability that at least one of the balls is green is 119/120.
Step-by-step explanation:
The student's question pertains to probability and involves calculating the likelihood of drawing certain colored balls from a bag without replacement. To find the probability of all three balls being yellow, you need to calculate the likelihood of each draw, one after the other, considering that the total number of balls reduces by one each time a ball is drawn and is not returned.
The probability for the first yellow ball is 3/10, because there are 3 yellow balls out of 10 total balls. When one yellow ball is taken, there are now 2 yellow balls and 9 total balls left, making the second draw's probability 2/9. For the third draw, the probability is 1/8, because only 1 yellow ball and 8 balls in total remain. The combined probability of drawing three yellow balls in a row is calculated by multiplying the probabilities of each draw:
(3/10) × (2/9) × (1/8) = (3 × 2 × 1)/(10 × 9 × 8) = 6/720 = 1/120.
To calculate the probability of at least one ball being green, it is easier to use the complement rule. This involves finding the probability that no green balls are drawn (which is the same as all balls being yellow) and subtracting this from 1. The probability that at least one ball is green is therefore:
1 - (probability all three balls are yellow) = 1 - (1/120) = 119/120.