When should you multiply probabilities?
$begingroup$
In an entrance test that is graded on the basis of two examinations,
the probability of a randomly chosen student passing the first
examination is 0.8 and the probability of passing the second
examination is 0.7. The probability of passing atleast one of them is
0.95. What is the probability of passing both?
If the probability of passing the first examination is .8 and the probability of passing the second is 0.7, why can't I multiply the two probabilities to get the probability that someone would pass both the examinations? My textbooks gives the answer as .55, which is different from what you'd get if you multiplied the probabilities (.56).
I understand how my textbook arrives at .55 (you use $ P(A cup B) = P(A) + P(B) - P(A cap B) $), but I want to understand why I can't simply multiply the probabilities. The two answers are also very close to each other (.55 vs .56).
probability
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
$endgroup$
add a comment |
$begingroup$
In an entrance test that is graded on the basis of two examinations,
the probability of a randomly chosen student passing the first
examination is 0.8 and the probability of passing the second
examination is 0.7. The probability of passing atleast one of them is
0.95. What is the probability of passing both?
If the probability of passing the first examination is .8 and the probability of passing the second is 0.7, why can't I multiply the two probabilities to get the probability that someone would pass both the examinations? My textbooks gives the answer as .55, which is different from what you'd get if you multiplied the probabilities (.56).
I understand how my textbook arrives at .55 (you use $ P(A cup B) = P(A) + P(B) - P(A cap B) $), but I want to understand why I can't simply multiply the probabilities. The two answers are also very close to each other (.55 vs .56).
probability
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
$endgroup$
$begingroup$
Because they are not independent. This is the simple answer. If they are independent then $P[Acap B] = P[A]P[B]$. Otherwise, you use th formula you mentioned.
$endgroup$
– BlackMath
Dec 6 '18 at 23:54
add a comment |
$begingroup$
In an entrance test that is graded on the basis of two examinations,
the probability of a randomly chosen student passing the first
examination is 0.8 and the probability of passing the second
examination is 0.7. The probability of passing atleast one of them is
0.95. What is the probability of passing both?
If the probability of passing the first examination is .8 and the probability of passing the second is 0.7, why can't I multiply the two probabilities to get the probability that someone would pass both the examinations? My textbooks gives the answer as .55, which is different from what you'd get if you multiplied the probabilities (.56).
I understand how my textbook arrives at .55 (you use $ P(A cup B) = P(A) + P(B) - P(A cap B) $), but I want to understand why I can't simply multiply the probabilities. The two answers are also very close to each other (.55 vs .56).
probability
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
$endgroup$
In an entrance test that is graded on the basis of two examinations,
the probability of a randomly chosen student passing the first
examination is 0.8 and the probability of passing the second
examination is 0.7. The probability of passing atleast one of them is
0.95. What is the probability of passing both?
If the probability of passing the first examination is .8 and the probability of passing the second is 0.7, why can't I multiply the two probabilities to get the probability that someone would pass both the examinations? My textbooks gives the answer as .55, which is different from what you'd get if you multiplied the probabilities (.56).
I understand how my textbook arrives at .55 (you use $ P(A cup B) = P(A) + P(B) - P(A cap B) $), but I want to understand why I can't simply multiply the probabilities. The two answers are also very close to each other (.55 vs .56).
probability
probability
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
edited Dec 6 '18 at 23:50
WorldGov
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
asked Dec 6 '18 at 23:33
WorldGovWorldGov
305111
305111
$begingroup$
Because they are not independent. This is the simple answer. If they are independent then $P[Acap B] = P[A]P[B]$. Otherwise, you use th formula you mentioned.
$endgroup$
– BlackMath
Dec 6 '18 at 23:54
add a comment |
$begingroup$
Because they are not independent. This is the simple answer. If they are independent then $P[Acap B] = P[A]P[B]$. Otherwise, you use th formula you mentioned.
$endgroup$
– BlackMath
Dec 6 '18 at 23:54
$begingroup$
Because they are not independent. This is the simple answer. If they are independent then $P[Acap B] = P[A]P[B]$. Otherwise, you use th formula you mentioned.
$endgroup$
– BlackMath
Dec 6 '18 at 23:54
$begingroup$
Because they are not independent. This is the simple answer. If they are independent then $P[Acap B] = P[A]P[B]$. Otherwise, you use th formula you mentioned.
$endgroup$
– BlackMath
Dec 6 '18 at 23:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can only multiply probabilities if the two events are INDEPENDENT of each other. In other words, the result of one thing being true has no effect on the other one.
In this case, someone passing a test means they are likely a better student, and thus likelier to pass the second test.
Let's say 9 out of 10 students passed test 1, and 8 out of ten students passed test 2. In particular, in this case, lets say that 8 out of 8 students passed both tests. So we have one student who passed just the first test, and one student who failed both.
So if you tried to get the probability that a random student failed both tests, in this case, it's clearly 1 out of 10. If you tried by multiplication, you would get $.2$ times $.1$, which is $.02$, or 1 out of 50.
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
$endgroup$
add a comment |
$begingroup$
The reason is that both events (passing the first exam and passing the second exam) are not independent. Only then is the rule of multiplying the individual probabilities correct.
It seems that the values for this problem were chosen such that the events are only slightly correlated (that means their dependence one another is not very big), such that the incorrect product formula gives an almost correct result. But that is just due to the values of the probabilities, it might be much different in another context with more differing result.
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You can only multiply probabilities if the two events are INDEPENDENT of each other. In other words, the result of one thing being true has no effect on the other one.
In this case, someone passing a test means they are likely a better student, and thus likelier to pass the second test.
Let's say 9 out of 10 students passed test 1, and 8 out of ten students passed test 2. In particular, in this case, lets say that 8 out of 8 students passed both tests. So we have one student who passed just the first test, and one student who failed both.
So if you tried to get the probability that a random student failed both tests, in this case, it's clearly 1 out of 10. If you tried by multiplication, you would get $.2$ times $.1$, which is $.02$, or 1 out of 50.
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
$endgroup$
add a comment |
$begingroup$
You can only multiply probabilities if the two events are INDEPENDENT of each other. In other words, the result of one thing being true has no effect on the other one.
In this case, someone passing a test means they are likely a better student, and thus likelier to pass the second test.
Let's say 9 out of 10 students passed test 1, and 8 out of ten students passed test 2. In particular, in this case, lets say that 8 out of 8 students passed both tests. So we have one student who passed just the first test, and one student who failed both.
So if you tried to get the probability that a random student failed both tests, in this case, it's clearly 1 out of 10. If you tried by multiplication, you would get $.2$ times $.1$, which is $.02$, or 1 out of 50.
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
$endgroup$
add a comment |
$begingroup$
You can only multiply probabilities if the two events are INDEPENDENT of each other. In other words, the result of one thing being true has no effect on the other one.
In this case, someone passing a test means they are likely a better student, and thus likelier to pass the second test.
Let's say 9 out of 10 students passed test 1, and 8 out of ten students passed test 2. In particular, in this case, lets say that 8 out of 8 students passed both tests. So we have one student who passed just the first test, and one student who failed both.
So if you tried to get the probability that a random student failed both tests, in this case, it's clearly 1 out of 10. If you tried by multiplication, you would get $.2$ times $.1$, which is $.02$, or 1 out of 50.
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
$endgroup$
You can only multiply probabilities if the two events are INDEPENDENT of each other. In other words, the result of one thing being true has no effect on the other one.
In this case, someone passing a test means they are likely a better student, and thus likelier to pass the second test.
Let's say 9 out of 10 students passed test 1, and 8 out of ten students passed test 2. In particular, in this case, lets say that 8 out of 8 students passed both tests. So we have one student who passed just the first test, and one student who failed both.
So if you tried to get the probability that a random student failed both tests, in this case, it's clearly 1 out of 10. If you tried by multiplication, you would get $.2$ times $.1$, which is $.02$, or 1 out of 50.
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
answered Dec 6 '18 at 23:43
AlanAlan
8,60221636
8,60221636
add a comment |
add a comment |
$begingroup$
The reason is that both events (passing the first exam and passing the second exam) are not independent. Only then is the rule of multiplying the individual probabilities correct.
It seems that the values for this problem were chosen such that the events are only slightly correlated (that means their dependence one another is not very big), such that the incorrect product formula gives an almost correct result. But that is just due to the values of the probabilities, it might be much different in another context with more differing result.
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
$endgroup$
add a comment |
$begingroup$
The reason is that both events (passing the first exam and passing the second exam) are not independent. Only then is the rule of multiplying the individual probabilities correct.
It seems that the values for this problem were chosen such that the events are only slightly correlated (that means their dependence one another is not very big), such that the incorrect product formula gives an almost correct result. But that is just due to the values of the probabilities, it might be much different in another context with more differing result.
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
$endgroup$
add a comment |
$begingroup$
The reason is that both events (passing the first exam and passing the second exam) are not independent. Only then is the rule of multiplying the individual probabilities correct.
It seems that the values for this problem were chosen such that the events are only slightly correlated (that means their dependence one another is not very big), such that the incorrect product formula gives an almost correct result. But that is just due to the values of the probabilities, it might be much different in another context with more differing result.
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
$endgroup$
The reason is that both events (passing the first exam and passing the second exam) are not independent. Only then is the rule of multiplying the individual probabilities correct.
It seems that the values for this problem were chosen such that the events are only slightly correlated (that means their dependence one another is not very big), such that the incorrect product formula gives an almost correct result. But that is just due to the values of the probabilities, it might be much different in another context with more differing result.
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
answered Dec 6 '18 at 23:44
IngixIngix
3,869146
3,869146
add a comment |
add a comment |
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$begingroup$
Because they are not independent. This is the simple answer. If they are independent then $P[Acap B] = P[A]P[B]$. Otherwise, you use th formula you mentioned.
$endgroup$
– BlackMath
Dec 6 '18 at 23:54