Asked by Liz
Hi helper, I used permutation for the following question. How can i use combination to find the answer? Help me please!
There are 12 red checkers and 3 black checkers in a bag. Checkers are selected one at a time, with replacement. Each time, the color of the checker is recorded. Find the probability of selecting a red checker exactly 7 times in 10 selections. Show your work.
Answer:
P(success)=P(red)= 12/15= 0.8
n=7 r=5
10C7= 10!/(10-7)! = 10!/3! = 10*9*8*7*6*5*4*3!/ 3!= 10*9*8*7*6*5*4= 604,800
P(red7times)= 10C7 (0.8)^7(0.2)^3
= 604,800 (0.8)^7(0.2)^3
= 604,800 (0.2)(.008)
= 604,800(0.0016)
=967.68
There are 12 red checkers and 3 black checkers in a bag. Checkers are selected one at a time, with replacement. Each time, the color of the checker is recorded. Find the probability of selecting a red checker exactly 7 times in 10 selections. Show your work.
Answer:
P(success)=P(red)= 12/15= 0.8
n=7 r=5
10C7= 10!/(10-7)! = 10!/3! = 10*9*8*7*6*5*4*3!/ 3!= 10*9*8*7*6*5*4= 604,800
P(red7times)= 10C7 (0.8)^7(0.2)^3
= 604,800 (0.8)^7(0.2)^3
= 604,800 (0.2)(.008)
= 604,800(0.0016)
=967.68
Answers
Answered by
PsyDAG
How can the probability be greater than one?
The probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(12/15)^7 * (3/15)^3 = ?
The probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
(12/15)^7 * (3/15)^3 = ?
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