Probability of two different groups of students passing an exam
$begingroup$
A group of $60$ students pass their exam with a probability of $0.3$. Another group of $70$ students pass their exam with a probability of $0.5$. What is the expected number of students who will pass their exams? What is the probability of one student passing her/his exam?
So far the only way to approach the problem I can think of is $0.3 cdot 60 + 0.5 cdot 70 = 53$ will pass their exams. Therefore, $frac{53}{(60+70)} = 0.4$ is the probability of one student passing her/his exam.
However, the groups have different number of people. I am not sure if this makes one of the group a more reliable prediction of the outcome.
probability statistics
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
$endgroup$
add a comment |
$begingroup$
A group of $60$ students pass their exam with a probability of $0.3$. Another group of $70$ students pass their exam with a probability of $0.5$. What is the expected number of students who will pass their exams? What is the probability of one student passing her/his exam?
So far the only way to approach the problem I can think of is $0.3 cdot 60 + 0.5 cdot 70 = 53$ will pass their exams. Therefore, $frac{53}{(60+70)} = 0.4$ is the probability of one student passing her/his exam.
However, the groups have different number of people. I am not sure if this makes one of the group a more reliable prediction of the outcome.
probability statistics
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
$endgroup$
6
$begingroup$
As another approach, note that a randomly selected student has a $frac {60}{130}$ chance of being from the first group, and a $frac {70}{130}$ chance of being from the second. Thus the probability that this randomly selected student passes is $frac {60}{130}times .3+frac {70}{130}times .5=.408$. This is really the same as the calculation you did, but perhaps it is more intuitive.
$endgroup$
– lulu
Jan 22 at 15:30
add a comment |
$begingroup$
A group of $60$ students pass their exam with a probability of $0.3$. Another group of $70$ students pass their exam with a probability of $0.5$. What is the expected number of students who will pass their exams? What is the probability of one student passing her/his exam?
So far the only way to approach the problem I can think of is $0.3 cdot 60 + 0.5 cdot 70 = 53$ will pass their exams. Therefore, $frac{53}{(60+70)} = 0.4$ is the probability of one student passing her/his exam.
However, the groups have different number of people. I am not sure if this makes one of the group a more reliable prediction of the outcome.
probability statistics
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
$endgroup$
A group of $60$ students pass their exam with a probability of $0.3$. Another group of $70$ students pass their exam with a probability of $0.5$. What is the expected number of students who will pass their exams? What is the probability of one student passing her/his exam?
So far the only way to approach the problem I can think of is $0.3 cdot 60 + 0.5 cdot 70 = 53$ will pass their exams. Therefore, $frac{53}{(60+70)} = 0.4$ is the probability of one student passing her/his exam.
However, the groups have different number of people. I am not sure if this makes one of the group a more reliable prediction of the outcome.
probability statistics
probability statistics
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
edited Jan 22 at 15:53
N. F. Taussig
44.5k103357
44.5k103357
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
asked Jan 22 at 15:24
Radoslav BonevRadoslav Bonev
414
414
6
$begingroup$
As another approach, note that a randomly selected student has a $frac {60}{130}$ chance of being from the first group, and a $frac {70}{130}$ chance of being from the second. Thus the probability that this randomly selected student passes is $frac {60}{130}times .3+frac {70}{130}times .5=.408$. This is really the same as the calculation you did, but perhaps it is more intuitive.
$endgroup$
– lulu
Jan 22 at 15:30
add a comment |
6
$begingroup$
As another approach, note that a randomly selected student has a $frac {60}{130}$ chance of being from the first group, and a $frac {70}{130}$ chance of being from the second. Thus the probability that this randomly selected student passes is $frac {60}{130}times .3+frac {70}{130}times .5=.408$. This is really the same as the calculation you did, but perhaps it is more intuitive.
$endgroup$
– lulu
Jan 22 at 15:30
6
6
$begingroup$
As another approach, note that a randomly selected student has a $frac {60}{130}$ chance of being from the first group, and a $frac {70}{130}$ chance of being from the second. Thus the probability that this randomly selected student passes is $frac {60}{130}times .3+frac {70}{130}times .5=.408$. This is really the same as the calculation you did, but perhaps it is more intuitive.
$endgroup$
– lulu
Jan 22 at 15:30
$begingroup$
As another approach, note that a randomly selected student has a $frac {60}{130}$ chance of being from the first group, and a $frac {70}{130}$ chance of being from the second. Thus the probability that this randomly selected student passes is $frac {60}{130}times .3+frac {70}{130}times .5=.408$. This is really the same as the calculation you did, but perhaps it is more intuitive.
$endgroup$
– lulu
Jan 22 at 15:30
add a comment |
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6
$begingroup$
As another approach, note that a randomly selected student has a $frac {60}{130}$ chance of being from the first group, and a $frac {70}{130}$ chance of being from the second. Thus the probability that this randomly selected student passes is $frac {60}{130}times .3+frac {70}{130}times .5=.408$. This is really the same as the calculation you did, but perhaps it is more intuitive.
$endgroup$
– lulu
Jan 22 at 15:30