Mixing
Probability of exactly one student failing?
I have the following problem:
If the probability that student A will fail a certain statis- tics
examination is 0.5, the probability that student B will fail the
examination is 0.2, and the probability that both student A and
student B will fail the examination is 0.1, what is the probability
and that exactly one of the two students will fail the examina- tion?
I came up with the following solution:
$$P(A) = 0.5; P(B) = 0.2$$
'exactly one' means either A only fails or B only fails.
Event $X_1$: 'A only fails'$$ P(X_1) = P(A) *P(B)^c = 0.5*0.8 = 0.4 $$
Event $X_2$: 'B only fails' $$P(X_2) = P(A)^c * P(B) = 0.5*0.2 = 0.1 $$
And therefore:
$$P(X_1lor X_2) = P(X_1)+P(X_2)-P(X_1land X_2) = 0.4+0.1-0 = 0.5 $$
My thoughts behind $P(X_1 land X_2) = 0$ were that it is not possible that both only happens at the same time. I'm not fully sure whether that is correct. Should these logical thought be correct is the value of $0.5$ correct?
probability
asked Nov 21 '18 at 15:50
thebilly
566
add a comment |
I have the following problem:
If the probability that student A will fail a certain statis- tics
examination is 0.5, the probability that student B will fail the
examination is 0.2, and the probability that both student A and
student B will fail the examination is 0.1, what is the probability
and that exactly one of the two students will fail the examina- tion?
I came up with the following solution:
$$P(A) = 0.5; P(B) = 0.2$$
'exactly one' means either A only fails or B only fails.
Event $X_1$: 'A only fails'$$ P(X_1) = P(A) *P(B)^c = 0.5*0.8 = 0.4 $$
Event $X_2$: 'B only fails' $$P(X_2) = P(A)^c * P(B) = 0.5*0.2 = 0.1 $$
And therefore:
$$P(X_1lor X_2) = P(X_1)+P(X_2)-P(X_1land X_2) = 0.4+0.1-0 = 0.5 $$
My thoughts behind $P(X_1 land X_2) = 0$ were that it is not possible that both only happens at the same time. I'm not fully sure whether that is correct. Should these logical thought be correct is the value of $0.5$ correct?
probability
asked Nov 21 '18 at 15:50
thebilly
566
Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures.
– Nutle
Nov 21 '18 at 15:54
It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.)
– littleO
Nov 21 '18 at 15:54
I see. Thank you both, I edited my question now.
– thebilly
Nov 21 '18 at 16:10
add a comment |
I have the following problem:
If the probability that student A will fail a certain statis- tics
examination is 0.5, the probability that student B will fail the
examination is 0.2, and the probability that both student A and
student B will fail the examination is 0.1, what is the probability
and that exactly one of the two students will fail the examina- tion?
I came up with the following solution:
$$P(A) = 0.5; P(B) = 0.2$$
'exactly one' means either A only fails or B only fails.
Event $X_1$: 'A only fails'$$ P(X_1) = P(A) *P(B)^c = 0.5*0.8 = 0.4 $$
Event $X_2$: 'B only fails' $$P(X_2) = P(A)^c * P(B) = 0.5*0.2 = 0.1 $$
And therefore:
$$P(X_1lor X_2) = P(X_1)+P(X_2)-P(X_1land X_2) = 0.4+0.1-0 = 0.5 $$
My thoughts behind $P(X_1 land X_2) = 0$ were that it is not possible that both only happens at the same time. I'm not fully sure whether that is correct. Should these logical thought be correct is the value of $0.5$ correct?
probability
asked Nov 21 '18 at 15:50
thebilly
566
I have the following problem:
If the probability that student A will fail a certain statis- tics
examination is 0.5, the probability that student B will fail the
examination is 0.2, and the probability that both student A and
student B will fail the examination is 0.1, what is the probability
and that exactly one of the two students will fail the examina- tion?
I came up with the following solution:
$$P(A) = 0.5; P(B) = 0.2$$
'exactly one' means either A only fails or B only fails.
Event $X_1$: 'A only fails'$$ P(X_1) = P(A) *P(B)^c = 0.5*0.8 = 0.4 $$
Event $X_2$: 'B only fails' $$P(X_2) = P(A)^c * P(B) = 0.5*0.2 = 0.1 $$
And therefore:
$$P(X_1lor X_2) = P(X_1)+P(X_2)-P(X_1land X_2) = 0.4+0.1-0 = 0.5 $$
My thoughts behind $P(X_1 land X_2) = 0$ were that it is not possible that both only happens at the same time. I'm not fully sure whether that is correct. Should these logical thought be correct is the value of $0.5$ correct?
probability
probability
asked Nov 21 '18 at 15:50
thebilly
566
asked Nov 21 '18 at 15:50
thebilly
566
edited Nov 21 '18 at 16:10
asked Nov 21 '18 at 15:50
thebilly
566
asked Nov 21 '18 at 15:50
thebilly
566
asked Nov 21 '18 at 15:50
thebilly
566
566
Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures.
– Nutle
Nov 21 '18 at 15:54
It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.)
– littleO
Nov 21 '18 at 15:54
I see. Thank you both, I edited my question now.
– thebilly
Nov 21 '18 at 16:10
add a comment |
Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures.
– Nutle
Nov 21 '18 at 15:54
It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.)
– littleO
Nov 21 '18 at 15:54
I see. Thank you both, I edited my question now.
– thebilly
Nov 21 '18 at 16:10
Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures.
– Nutle
Nov 21 '18 at 15:54
Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures.
– Nutle
Nov 21 '18 at 15:54
It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.)
– littleO
Nov 21 '18 at 15:54
It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.)
– littleO
Nov 21 '18 at 15:54
I see. Thank you both, I edited my question now.
– thebilly
Nov 21 '18 at 16:10
I see. Thank you both, I edited my question now.
– thebilly
Nov 21 '18 at 16:10
add a comment |
2 Answers
2
active
oldest
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You need to find $P(A, neg B) + P(neg A, B)$. We know $P(A)$, $P(B)$ and $P(A, B)$.
We also know that $P(A,B)+P(A,neg B)=P(A)$. You can get the value $P(A, neg B)$ from here. Likewise, we also know that $P(A, B)+P(neg A, B)=P(B)$ and thus you can also get the value $P(neg A, B)$.
answered Nov 21 '18 at 16:15
DavidPM
28118
add a comment |
Your $P(X_1)=0.4$ and $P(X_2)=0.1$ are right. But then it is asked for $P(X_2cap overline X_1)+P(X_1cap overline X_2)$
This is $[P(X_1)-P(X_1cap X_2)]+[P(X_2)-P(X_1cap X_2)]=P(X_1)+P(X_2)-2cdot P(X_1cap X_2)$
$=0.2+0.5-2cdot 0.1=0.5$
answered Nov 21 '18 at 16:16
callculus
17.9k31427
add a comment |
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2 Answers
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2 Answers
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You need to find $P(A, neg B) + P(neg A, B)$. We know $P(A)$, $P(B)$ and $P(A, B)$.
We also know that $P(A,B)+P(A,neg B)=P(A)$. You can get the value $P(A, neg B)$ from here. Likewise, we also know that $P(A, B)+P(neg A, B)=P(B)$ and thus you can also get the value $P(neg A, B)$.
answered Nov 21 '18 at 16:15
DavidPM
28118
add a comment |
You need to find $P(A, neg B) + P(neg A, B)$. We know $P(A)$, $P(B)$ and $P(A, B)$.
We also know that $P(A,B)+P(A,neg B)=P(A)$. You can get the value $P(A, neg B)$ from here. Likewise, we also know that $P(A, B)+P(neg A, B)=P(B)$ and thus you can also get the value $P(neg A, B)$.
answered Nov 21 '18 at 16:15
DavidPM
28118
add a comment |
You need to find $P(A, neg B) + P(neg A, B)$. We know $P(A)$, $P(B)$ and $P(A, B)$.
We also know that $P(A,B)+P(A,neg B)=P(A)$. You can get the value $P(A, neg B)$ from here. Likewise, we also know that $P(A, B)+P(neg A, B)=P(B)$ and thus you can also get the value $P(neg A, B)$.
answered Nov 21 '18 at 16:15
DavidPM
28118
You need to find $P(A, neg B) + P(neg A, B)$. We know $P(A)$, $P(B)$ and $P(A, B)$.
We also know that $P(A,B)+P(A,neg B)=P(A)$. You can get the value $P(A, neg B)$ from here. Likewise, we also know that $P(A, B)+P(neg A, B)=P(B)$ and thus you can also get the value $P(neg A, B)$.
answered Nov 21 '18 at 16:15
DavidPM
28118
answered Nov 21 '18 at 16:15
DavidPM
28118
answered Nov 21 '18 at 16:15
DavidPM
28118
answered Nov 21 '18 at 16:15
DavidPM
28118
28118
add a comment |
add a comment |
Your $P(X_1)=0.4$ and $P(X_2)=0.1$ are right. But then it is asked for $P(X_2cap overline X_1)+P(X_1cap overline X_2)$
This is $[P(X_1)-P(X_1cap X_2)]+[P(X_2)-P(X_1cap X_2)]=P(X_1)+P(X_2)-2cdot P(X_1cap X_2)$
$=0.2+0.5-2cdot 0.1=0.5$
answered Nov 21 '18 at 16:16
callculus
17.9k31427
add a comment |
Your $P(X_1)=0.4$ and $P(X_2)=0.1$ are right. But then it is asked for $P(X_2cap overline X_1)+P(X_1cap overline X_2)$
This is $[P(X_1)-P(X_1cap X_2)]+[P(X_2)-P(X_1cap X_2)]=P(X_1)+P(X_2)-2cdot P(X_1cap X_2)$
$=0.2+0.5-2cdot 0.1=0.5$
answered Nov 21 '18 at 16:16
callculus
17.9k31427
add a comment |
Your $P(X_1)=0.4$ and $P(X_2)=0.1$ are right. But then it is asked for $P(X_2cap overline X_1)+P(X_1cap overline X_2)$
This is $[P(X_1)-P(X_1cap X_2)]+[P(X_2)-P(X_1cap X_2)]=P(X_1)+P(X_2)-2cdot P(X_1cap X_2)$
$=0.2+0.5-2cdot 0.1=0.5$
answered Nov 21 '18 at 16:16
callculus
17.9k31427
Your $P(X_1)=0.4$ and $P(X_2)=0.1$ are right. But then it is asked for $P(X_2cap overline X_1)+P(X_1cap overline X_2)$
This is $[P(X_1)-P(X_1cap X_2)]+[P(X_2)-P(X_1cap X_2)]=P(X_1)+P(X_2)-2cdot P(X_1cap X_2)$
$=0.2+0.5-2cdot 0.1=0.5$
answered Nov 21 '18 at 16:16
callculus
17.9k31427
answered Nov 21 '18 at 16:16
callculus
17.9k31427
answered Nov 21 '18 at 16:16
callculus
17.9k31427
answered Nov 21 '18 at 16:16
callculus
17.9k31427
17.9k31427
add a comment |
add a comment |
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Welcome to math.SE! Asking for correctness here is OK (and it is encouraged to provide your own ideas for every question, what did you try and what didn't work), but it would also be good in the future to write your solution using MathJax, instead of uploading pictures.
– Nutle
Nov 21 '18 at 15:54
It'd be better to write your solution in the body of the question rather than linking to an image. (Also, the image should be rotated to make it easier to read.)
– littleO
Nov 21 '18 at 15:54
I see. Thank you both, I edited my question now.
– thebilly
Nov 21 '18 at 16:10