Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
A. 8/28
B. 9/28
C. 10/28
D. 10/18
E. 11/18
OA is b
How can I get the correct answer to this question and which formula did you use here? Thank you so much for your response.
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Total: There are 8 balls total. If we want to select 5, there are 8C5 = 8*7*6*5*4/5*4*3*2 = 56 total possible selections.Roland2rule wrote:Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
A. 8/28
B. 9/28
C. 10/28
D. 10/18
E. 11/18
OA is b
How can I get the correct answer to this question and which formula did you use here? Thank you so much for your response.
Desired: We want to select 1 red ball from 2, so there's 2C1 = 2 ways to select a red ball.
We want to select 2 green balls from 3, so there's 3C2 = 3 ways to select a green ball
We want to select 2 blue balls from 3, so there's 3C2 = 3 ways to select a blue ball.
# Desired = 2 * 3 * 3 = 18
Desired/Total = 18/56 = 9/28. The answer is B.
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The number of ways to select 1 red ball is 2C1 = 2.Roland2rule wrote:Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
A. 8/28
B. 9/28
C. 10/28
D. 10/18
E. 11/18
The number of ways to select 2 green balls is 3C2 = 3.
The number of ways to select 2 blue balls is 3C2 = 3.
So, the number of ways to select 1 red, 2 green, and 2 blue balls is 2 x 3 x 3 = 18.
The number of ways to select 5 balls from 8 is 8C5:
8!/5![(8-5)!] = 8!/(5!x3!) = (8 x 7 x 6 x 5 x 4)/5! = (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2 x 1) = 7 x 2 x 4 = 56
Thus, the probability is 18/56 = 9/28.
Answer: B
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