This response includes 3 different probability questions based on the scenario of drawing two balls from a bag, and provides step-by-step solutions for each question.
B) Solve the probability question from A):
b) P(Drawing a blue ball and then drawing a red ball) = (3/5) * (2/4) = 3/10
c) P(Drawing a green ball and then drawing a yellow ball) = (4/10) * (6/9) = 4/15
d) P(Drawing one white ball and one black ball) = ((2/9) * (3/8)) + ((3/9) * (4/8)) + ((4/9) * (3/8)) = 19/36
Step-by-step explanation:
A) Create a probability question such that the compound event of drawing the two balls is:
Independent events: Suppose there are 3 blue balls and 2 red balls in the bag. What is the probability of drawing a blue ball and then drawing a red ball?
Dependent events: Suppose there are 4 green balls and 6 yellow balls in the bag. What is the probability of drawing a green ball, not replacing it, and then drawing a yellow ball?
Multiple cases: Suppose there are 2 white balls, 3 black balls, and 4 gray balls in the bag. What is the probability of drawing one white ball and one black ball?
B) Solve the probability question from A):
b) P(Drawing a blue ball and then drawing a red ball) = (3/5) * (2/4) = 3/10
c) P(Drawing a green ball and then drawing a yellow ball) = (4/10) * (6/9) = 4/15
d) P(Drawing one white ball and one black ball) = ((2/9) * (3/8)) + ((3/9) * (4/8)) + ((4/9) * (3/8)) = 19/36