Question 1097765: In a factory it is known from past experience that the probability is 0.75 that a new worker who has attended the company training program will meet the production quota, and the corresponding probability for a new worker who has not attended the company training program is 0.50. If 60% of the new workers attend the training program then What is the probability that a new worker will meet the production quota? also thus, What is the probability that a new worker, who meets the production quota, attended the company’s training program?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i think your solution will be as follows:
p(M) = probability that the worker will meet the production quota
p(A) = probability that the worker attended the training program.
p(B) = probability that the worker did not attend the training program.
p(M given A) = probability that the worker met the quota given the worker attended the training program.
p(M given B) = probability that that the worker met the quota given the worker did not attend the training program.
p(A given M) = probability that the worker took the training program given that the worker met the quota.
p(A) = .6 which is given.
p(B) = .4 which is equal to 1 minus p(A) since either the worker attended the training program or the worker did not attend the training program.
p(M given A) = .75 which is given.
p(M given B) = .5 which is given.
p(M) is equal to the probability that any worker, chosen at random, met the quota.
since 60% attended the training program and 40% didn't, then:
p(M) = .6 * .75 + .4 * .5 = .65
this means the probability that a new worker met the production quota is .65.
therefore p(M) = .65
you want to know the probability that a worker attended the training program given that the worker met the quota.
this would be p(A given M) which is equal to p(A and M) / p(M).
so solve this, you need to find p (A and M).
p(M given A) is equal to p(A and M) / p(A).
you know p(M given A) is equal to .75 and you know that p(A) is equal to .6.
therefore .75 = p(A and M) / .6
solve for p(A and M) to get p(A and M) = .6 * .75 = .45
you now know p(A and M) = .45
p(A given M) = p(A and M) / p(M) becomes p(A given M) = .45 / .65 which is equal to .6923 rounded to 4 decimal digits.
you have p(M) = .65 and you have p(A given M) = .6923.
to test this out, assume 100 workers.
60 took the training program and 40 did not.
75% of the workers who took the training program met their quota and 50% of the workers who did not take the training program met their quota.
.75 * 60 + .50 * 40 = 45 + 20 = 65 workers met quota.
out of the 65 who met the quota, 45 took the training program.
probability of any worker meeting the quota is 65/100 = .65
probability that a worker attended the training program given that the worker met the quota is 45/65 = .6923.
solution appears to be good and appears to be confirmed if i did my testing right.
your solution should be:
probability that a new worker will meet the quota is .65.
probability that a new worker who met the quota attended the training program is .6923 rounded to 4 decimal places, or .69 rounded to 2 decimal places.
i'm pretty sure i'm right but i could be wrong so take it for what it is .
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