7. A bag contains 5 red marbles, 6 green marbles, and 9 blue marbles. Suppose 3 marbles are chosen without replacement. What is the probability of choosing a red, a blue, and a red in that order?:
A. 9/320
B. 5/152
C. 1/1280
D. 1/38
Please explain answer. Thanks
Find the probability of picking one of the colors. Since there is no replacement, the probability for the second would have a divisor of n-1 and the third would have n-2. The probability of all events occurring is found by multiplying the probabilities ofthe individual events.
I hope this helps. Thanks for asking.
To find the probability of choosing a red, a blue, and a red in that order, we need to calculate the probability of each event happening and then multiply those probabilities together.
First, let's consider the probability of selecting a red marble on the first draw. There are a total of 5 red marbles out of 20 marbles in the bag, so the probability of choosing a red marble on the first draw is 5/20.
After the first draw, there are now 19 marbles left in the bag, with 4 of them being red. So, the probability of selecting a blue marble on the second draw is 9/19.
Finally, for the third draw, there is only 1 red marble left in the bag out of the remaining 18 marbles. Therefore, the probability of selecting a red marble again on the third draw is 1/18.
To find the overall probability, we multiply the probabilities of each event happening: (5/20) * (9/19) * (1/18) = 45/6840.
Simplifying the fraction gives us 1/152.
Therefore, the correct answer is B. 5/152.