SOLUTION: If you have 14 bad calculators and 24 good calculators and you pull 4 what is the probability that 1 is bad
Algebra ->
Probability-and-statistics
-> SOLUTION: If you have 14 bad calculators and 24 good calculators and you pull 4 what is the probability that 1 is bad
Log On
Question 477798: If you have 14 bad calculators and 24 good calculators and you pull 4 what is the probability that 1 is bad Found 2 solutions by Theo, ikleyn:Answer by Theo(13342) (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
You can put this solution on YOUR website! .
If you have 14 bad calculators and 24 good calculators and you pull 4,
what is the probability that 1 is bad ?
~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @Theo is incorrect.
It is because @Theo used the Binomial distribution model in his solution,
but this model does not work in this case.
See my correct solution below.
We should calculate the total number of all possible different quadruples,
the number of all possible different quadruples containing 1 bad and 3 good calculators
and relate the second quantity to the first quantity.
The total number of all calculators in the problem is 14+24 = 38.
The total number of all possible quadruples of 38 calculators is the number
of combinations C(38,4) = = 73815.
The total number of all possible quadruples containing 1 bad and 3 good calculators
is this product
C(14,1)*C(24,3) = 14*2024 = 28336.
Finally, the probability under the problem's question is
P = = 0.3839 (rounded). ANSWER
