Probability - defective boxes
Out of 12 boxes, 2 were defective. If you randomly select two of the 12 boxes, what is the probability that one of them is defective? Would the probability be 1/12=0,08? Thank you.
probability
asked Nov 25 '18 at 8:59
BreatheEasy
83
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Out of 12 boxes, 2 were defective. If you randomly select two of the 12 boxes, what is the probability that one of them is defective? Would the probability be 1/12=0,08? Thank you.
probability
asked Nov 25 '18 at 8:59
BreatheEasy
83
add a comment |
Out of 12 boxes, 2 were defective. If you randomly select two of the 12 boxes, what is the probability that one of them is defective? Would the probability be 1/12=0,08? Thank you.
probability
asked Nov 25 '18 at 8:59
BreatheEasy
83
Out of 12 boxes, 2 were defective. If you randomly select two of the 12 boxes, what is the probability that one of them is defective? Would the probability be 1/12=0,08? Thank you.
probability
probability
asked Nov 25 '18 at 8:59
BreatheEasy
83
asked Nov 25 '18 at 8:59
BreatheEasy
83
asked Nov 25 '18 at 8:59
BreatheEasy
83
asked Nov 25 '18 at 8:59
BreatheEasy
83
asked Nov 25 '18 at 8:59
BreatheEasy
83
83
add a comment |
add a comment |
1 Answer
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We have $binom{12}{2} = 66$ ways to choose the two boxes randomly, all equally likely. We have to pick one non-defective one and one defective one in $10times 2= 20$ ways. So the asked-for probability is $frac{20}{66}= frac{10}{33}$.
Alternatively, we could compute $$P(text{box 1 defective and box 2 OK})= P(text{box 1 defective})P(text{box 2 OK}|text{box 1 defective})= frac{2}{12}cdot frac{10}{11}$$ (the second draw is from $11$ remaining boxes, with $1$ defective box left)
and $$P(text{box 1 OK and box 2 defective})=P(text{box 1 OK})P(text{box 2 defective}|text{box 1 OK})= frac{10}{12}cdot frac{2}{11}$$
(the second draw is from $11$ boxes of which $2$ are defective).
And add them to get the same result. The first way scales better with more draws, I think, and is a special case of the hypergeometric distribution.
But the second one is easier to come up with, maybe.
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
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We have $binom{12}{2} = 66$ ways to choose the two boxes randomly, all equally likely. We have to pick one non-defective one and one defective one in $10times 2= 20$ ways. So the asked-for probability is $frac{20}{66}= frac{10}{33}$.
Alternatively, we could compute $$P(text{box 1 defective and box 2 OK})= P(text{box 1 defective})P(text{box 2 OK}|text{box 1 defective})= frac{2}{12}cdot frac{10}{11}$$ (the second draw is from $11$ remaining boxes, with $1$ defective box left)
and $$P(text{box 1 OK and box 2 defective})=P(text{box 1 OK})P(text{box 2 defective}|text{box 1 OK})= frac{10}{12}cdot frac{2}{11}$$
(the second draw is from $11$ boxes of which $2$ are defective).
And add them to get the same result. The first way scales better with more draws, I think, and is a special case of the hypergeometric distribution.
But the second one is easier to come up with, maybe.
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
add a comment |
We have $binom{12}{2} = 66$ ways to choose the two boxes randomly, all equally likely. We have to pick one non-defective one and one defective one in $10times 2= 20$ ways. So the asked-for probability is $frac{20}{66}= frac{10}{33}$.
Alternatively, we could compute $$P(text{box 1 defective and box 2 OK})= P(text{box 1 defective})P(text{box 2 OK}|text{box 1 defective})= frac{2}{12}cdot frac{10}{11}$$ (the second draw is from $11$ remaining boxes, with $1$ defective box left)
and $$P(text{box 1 OK and box 2 defective})=P(text{box 1 OK})P(text{box 2 defective}|text{box 1 OK})= frac{10}{12}cdot frac{2}{11}$$
(the second draw is from $11$ boxes of which $2$ are defective).
And add them to get the same result. The first way scales better with more draws, I think, and is a special case of the hypergeometric distribution.
But the second one is easier to come up with, maybe.
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
add a comment |
We have $binom{12}{2} = 66$ ways to choose the two boxes randomly, all equally likely. We have to pick one non-defective one and one defective one in $10times 2= 20$ ways. So the asked-for probability is $frac{20}{66}= frac{10}{33}$.
Alternatively, we could compute $$P(text{box 1 defective and box 2 OK})= P(text{box 1 defective})P(text{box 2 OK}|text{box 1 defective})= frac{2}{12}cdot frac{10}{11}$$ (the second draw is from $11$ remaining boxes, with $1$ defective box left)
and $$P(text{box 1 OK and box 2 defective})=P(text{box 1 OK})P(text{box 2 defective}|text{box 1 OK})= frac{10}{12}cdot frac{2}{11}$$
(the second draw is from $11$ boxes of which $2$ are defective).
And add them to get the same result. The first way scales better with more draws, I think, and is a special case of the hypergeometric distribution.
But the second one is easier to come up with, maybe.
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
We have $binom{12}{2} = 66$ ways to choose the two boxes randomly, all equally likely. We have to pick one non-defective one and one defective one in $10times 2= 20$ ways. So the asked-for probability is $frac{20}{66}= frac{10}{33}$.
Alternatively, we could compute $$P(text{box 1 defective and box 2 OK})= P(text{box 1 defective})P(text{box 2 OK}|text{box 1 defective})= frac{2}{12}cdot frac{10}{11}$$ (the second draw is from $11$ remaining boxes, with $1$ defective box left)
and $$P(text{box 1 OK and box 2 defective})=P(text{box 1 OK})P(text{box 2 defective}|text{box 1 OK})= frac{10}{12}cdot frac{2}{11}$$
(the second draw is from $11$ boxes of which $2$ are defective).
And add them to get the same result. The first way scales better with more draws, I think, and is a special case of the hypergeometric distribution.
But the second one is easier to come up with, maybe.
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
edited Nov 25 '18 at 9:13
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
answered Nov 25 '18 at 9:05
Henno Brandsma
105k346114
105k346114
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