SOLUTION: If you have 14 bad calculators and 24 good calculators and you pull 4 what is the probability that 1 is bad

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Question 477798: If you have 14 bad calculators and 24 good calculators and you pull 4 what is the probability that 1 is bad
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
probability that a calculator is bad is equal to the number of bad calculators divided by the total number of calculators.
the probability of getting a bad calculator on any 1 draw from the pile is 14/24.
p(b) is equal to 14/24
p(g) is equal to 10/24
p(b) is defined as the probability of getting a bad calculator.
p(g) is defined as the probability of getting a good calculator.
you pull 4 calculators out of the pile of 24.
the probability of exactly x out of the 4 calculators being bad is given by the following table.
n	p(b)^x	        p(g)^(n-x)	nCx	p(x)
0	1	        0.030140818	1	0.030140818
1	0.583333333	0.072337963	4	0.16878858
2	0.340277778	0.173611111	6	0.354456019
3	0.19849537	0.416666667	4	0.330825617
4	0.115788966	1	        1	0.115788966
sum of all probabilities >>>>>>>>>>>>>>>>>>>>   1

sum of all probabilities equals 1 as it should.
n is the number of bad calculators in each sample.
p(b)^x is the probability of getting x bad calculators out of 4.
p(g)^(n-x) is the probability of getting (n-x) good calculators out of 4.
nCx is the number of ways you can get x bad calculators out of 4.
here's how the formulas work.
the probability of getting exactly 1 bad calculator out of the 4 would be given by the formula:
p(b)^1 * p(g)^3 * nC1
nC1 is equal to 4! / (1!*3!) which is equal to 4
the probability of getting exactly 1 bad calculator out of a sample of 4 is therefore equal to:
(14/24)^1 * (10/24)^3 * 4 which is equal to .16878858
this agrees with the entry for n = 1 in the table, as it should because the formulas used to generate the table are the same formulas being shown here.
the probability of getting 4 bad calculators when you draw 4 calculators out of the pile of 24 would be calculated as follows:
(14/24)^4 * (10/24)^0 * 4C4 which equals:
(14/24)^4 * (10/24)^0 * 1 which equals .1215788966
this number agrees with the comparable number in the table for 4 bad calculators out of 4 as it should.
The answer to your question is:
The probability of getting exactly 1 bad calculator out of the 4 is equal to .16878858