If a bag contains a set of 20 red balls, 10 green balls, and 15 black balls, then what is the possiblity of selecting (in this order) a red ball, a green ball, and a red ball. Without Replacement of the items selected.
Can you explain how you get the answer? Thank you
To find the probability of selecting a red ball, a green ball, and another red ball in that order without replacement, you need to calculate the probability of each step.
Step 1: Probability of selecting a red ball
The bag contains a total of 20 red balls, 10 green balls, and 15 black balls, for a total of 45 balls. Since the ball is not replaced after selection, there will be one less ball in the bag for each subsequent step.
The probability of selecting a red ball on the first draw is 20/45, since there are 20 red balls out of the total 45 balls.
Step 2: Probability of selecting a green ball
After the first red ball is selected, there are now 19 red balls and 44 total balls remaining in the bag. The probability of selecting a green ball on the second draw is 10/44 since there are 10 green balls remaining out of the total 44 balls.
Step 3: Probability of selecting another red ball
After the green ball is selected, there will be 9 green balls, 19 red balls, and 43 total balls remaining in the bag. The probability of selecting another red ball on the third draw is 19/43 since there are 19 red balls remaining out of the total 43 balls.
To find the combined probability of all three steps, you need to multiply the probabilities of each step together. Therefore, the probability of selecting a red ball, a green ball, and another red ball in that order without replacement is:
(20/45) * (10/44) * (19/43) = 0.078 or 7.8%.