Using Bayes rule
$begingroup$
There are two schools in a village: School A and School B
- School A has twice as many pupils as School B
- At School A, 70% of the pupils are girls
- At School B, 40% of the pupils are girls
i) You meet a pupil from the village. What is the probability that this pupil is a boy?
ii) Charlie is a boy who goes to either School A or School B. What is the probability that Charlie goes to School A?
My attempts:
i)
Since school A has twice has many pupils, I give it double weighting.
School A: (1 - 0.7)*2 = 0.6
School B: 1 - 0.4 = 0.6
Probability = average of A and B = 0.6
ii)
P(School A|Boy) = (P(Boy|School A) * P(School A)) / P(Boy)
P(Boy|School A) = 0.3 because probability of boy, given school A
P(School A) = 2/3 because twice as many pupils in school A
P(Boy) from part i
P(School A|Boy) = (0.3 * 2/3) / 0.6
P(School A|Boy) = 1/3
Have I got these right?
Thanks
probability
asked Jan 2 at 20:27
stripe123stripe123
11
$endgroup$
add a comment |
$begingroup$
There are two schools in a village: School A and School B
- School A has twice as many pupils as School B
- At School A, 70% of the pupils are girls
- At School B, 40% of the pupils are girls
i) You meet a pupil from the village. What is the probability that this pupil is a boy?
ii) Charlie is a boy who goes to either School A or School B. What is the probability that Charlie goes to School A?
My attempts:
i)
Since school A has twice has many pupils, I give it double weighting.
School A: (1 - 0.7)*2 = 0.6
School B: 1 - 0.4 = 0.6
Probability = average of A and B = 0.6
ii)
P(School A|Boy) = (P(Boy|School A) * P(School A)) / P(Boy)
P(Boy|School A) = 0.3 because probability of boy, given school A
P(School A) = 2/3 because twice as many pupils in school A
P(Boy) from part i
P(School A|Boy) = (0.3 * 2/3) / 0.6
P(School A|Boy) = 1/3
Have I got these right?
Thanks
probability
asked Jan 2 at 20:27
stripe123stripe123
11
$endgroup$
2
$begingroup$
It's sometimes helpful to think through numerical examples. Say $A$ has $30$ boys and $70$ girls, and $B$ has $30$ boys and $20$ girls. That fits your requirements, yes? Then there are $60$ boys all in all out of $150$ students. Does that fit your answer for (i)? Note too that each school has the same number of boys. Does that fit your answer for (ii)?
$endgroup$
– lulu
Jan 2 at 20:50
1
$begingroup$
Note: your notation is very confusing. I think (but am not sure) that you are using "$A$" to denote the event "a randomly observed child attends school $A$" but you are using "$B$" to denote the event "a randomly observed child is a boy". At the very least, you should define the variables you want to use.
$endgroup$
– lulu
Jan 2 at 20:57
1
$begingroup$
In $(a)$, you need to multiply $1-0.7$ by $2/3$, not $2$. Similarly, $1-0.4$ gets multiplied by $1/3$. The 'weighting' you are talking about is a probability.
$endgroup$
– Shubham Johri
Jan 2 at 21:01
$begingroup$
Thanks for the comments. Updated notation, hopefully less confusing now. So is the probability of being a boy 60/150 = 0.4? If so, would I be correct in saying the answer for part 2 is 0.5?
$endgroup$
– stripe123
Jan 2 at 21:15
add a comment |
$begingroup$
There are two schools in a village: School A and School B
- School A has twice as many pupils as School B
- At School A, 70% of the pupils are girls
- At School B, 40% of the pupils are girls
i) You meet a pupil from the village. What is the probability that this pupil is a boy?
ii) Charlie is a boy who goes to either School A or School B. What is the probability that Charlie goes to School A?
My attempts:
i)
Since school A has twice has many pupils, I give it double weighting.
School A: (1 - 0.7)*2 = 0.6
School B: 1 - 0.4 = 0.6
Probability = average of A and B = 0.6
ii)
P(School A|Boy) = (P(Boy|School A) * P(School A)) / P(Boy)
P(Boy|School A) = 0.3 because probability of boy, given school A
P(School A) = 2/3 because twice as many pupils in school A
P(Boy) from part i
P(School A|Boy) = (0.3 * 2/3) / 0.6
P(School A|Boy) = 1/3
Have I got these right?
Thanks
probability
asked Jan 2 at 20:27
stripe123stripe123
11
$endgroup$
There are two schools in a village: School A and School B
- School A has twice as many pupils as School B
- At School A, 70% of the pupils are girls
- At School B, 40% of the pupils are girls
i) You meet a pupil from the village. What is the probability that this pupil is a boy?
ii) Charlie is a boy who goes to either School A or School B. What is the probability that Charlie goes to School A?
My attempts:
i)
Since school A has twice has many pupils, I give it double weighting.
School A: (1 - 0.7)*2 = 0.6
School B: 1 - 0.4 = 0.6
Probability = average of A and B = 0.6
ii)
P(School A|Boy) = (P(Boy|School A) * P(School A)) / P(Boy)
P(Boy|School A) = 0.3 because probability of boy, given school A
P(School A) = 2/3 because twice as many pupils in school A
P(Boy) from part i
P(School A|Boy) = (0.3 * 2/3) / 0.6
P(School A|Boy) = 1/3
Have I got these right?
Thanks
probability
probability
asked Jan 2 at 20:27
stripe123stripe123
11
asked Jan 2 at 20:27
stripe123stripe123
11
edited Jan 2 at 21:04
stripe123
asked Jan 2 at 20:27
stripe123stripe123
11
asked Jan 2 at 20:27
stripe123stripe123
11
asked Jan 2 at 20:27
stripe123stripe123
11
11
2
$begingroup$
It's sometimes helpful to think through numerical examples. Say $A$ has $30$ boys and $70$ girls, and $B$ has $30$ boys and $20$ girls. That fits your requirements, yes? Then there are $60$ boys all in all out of $150$ students. Does that fit your answer for (i)? Note too that each school has the same number of boys. Does that fit your answer for (ii)?
$endgroup$
– lulu
Jan 2 at 20:50
1
$begingroup$
Note: your notation is very confusing. I think (but am not sure) that you are using "$A$" to denote the event "a randomly observed child attends school $A$" but you are using "$B$" to denote the event "a randomly observed child is a boy". At the very least, you should define the variables you want to use.
$endgroup$
– lulu
Jan 2 at 20:57
1
$begingroup$
In $(a)$, you need to multiply $1-0.7$ by $2/3$, not $2$. Similarly, $1-0.4$ gets multiplied by $1/3$. The 'weighting' you are talking about is a probability.
$endgroup$
– Shubham Johri
Jan 2 at 21:01
$begingroup$
Thanks for the comments. Updated notation, hopefully less confusing now. So is the probability of being a boy 60/150 = 0.4? If so, would I be correct in saying the answer for part 2 is 0.5?
$endgroup$
– stripe123
Jan 2 at 21:15
add a comment |
2
$begingroup$
It's sometimes helpful to think through numerical examples. Say $A$ has $30$ boys and $70$ girls, and $B$ has $30$ boys and $20$ girls. That fits your requirements, yes? Then there are $60$ boys all in all out of $150$ students. Does that fit your answer for (i)? Note too that each school has the same number of boys. Does that fit your answer for (ii)?
$endgroup$
– lulu
Jan 2 at 20:50
1
$begingroup$
Note: your notation is very confusing. I think (but am not sure) that you are using "$A$" to denote the event "a randomly observed child attends school $A$" but you are using "$B$" to denote the event "a randomly observed child is a boy". At the very least, you should define the variables you want to use.
$endgroup$
– lulu
Jan 2 at 20:57
1
$begingroup$
In $(a)$, you need to multiply $1-0.7$ by $2/3$, not $2$. Similarly, $1-0.4$ gets multiplied by $1/3$. The 'weighting' you are talking about is a probability.
$endgroup$
– Shubham Johri
Jan 2 at 21:01
$begingroup$
Thanks for the comments. Updated notation, hopefully less confusing now. So is the probability of being a boy 60/150 = 0.4? If so, would I be correct in saying the answer for part 2 is 0.5?
$endgroup$
– stripe123
Jan 2 at 21:15
2
2
$begingroup$
It's sometimes helpful to think through numerical examples. Say $A$ has $30$ boys and $70$ girls, and $B$ has $30$ boys and $20$ girls. That fits your requirements, yes? Then there are $60$ boys all in all out of $150$ students. Does that fit your answer for (i)? Note too that each school has the same number of boys. Does that fit your answer for (ii)?
$endgroup$
– lulu
Jan 2 at 20:50
$begingroup$
It's sometimes helpful to think through numerical examples. Say $A$ has $30$ boys and $70$ girls, and $B$ has $30$ boys and $20$ girls. That fits your requirements, yes? Then there are $60$ boys all in all out of $150$ students. Does that fit your answer for (i)? Note too that each school has the same number of boys. Does that fit your answer for (ii)?
$endgroup$
– lulu
Jan 2 at 20:50
1
1
$begingroup$
Note: your notation is very confusing. I think (but am not sure) that you are using "$A$" to denote the event "a randomly observed child attends school $A$" but you are using "$B$" to denote the event "a randomly observed child is a boy". At the very least, you should define the variables you want to use.
$endgroup$
– lulu
Jan 2 at 20:57
$begingroup$
Note: your notation is very confusing. I think (but am not sure) that you are using "$A$" to denote the event "a randomly observed child attends school $A$" but you are using "$B$" to denote the event "a randomly observed child is a boy". At the very least, you should define the variables you want to use.
$endgroup$
– lulu
Jan 2 at 20:57
1
1
$begingroup$
In $(a)$, you need to multiply $1-0.7$ by $2/3$, not $2$. Similarly, $1-0.4$ gets multiplied by $1/3$. The 'weighting' you are talking about is a probability.
$endgroup$
– Shubham Johri
Jan 2 at 21:01
$begingroup$
In $(a)$, you need to multiply $1-0.7$ by $2/3$, not $2$. Similarly, $1-0.4$ gets multiplied by $1/3$. The 'weighting' you are talking about is a probability.
$endgroup$
– Shubham Johri
Jan 2 at 21:01
$begingroup$
Thanks for the comments. Updated notation, hopefully less confusing now. So is the probability of being a boy 60/150 = 0.4? If so, would I be correct in saying the answer for part 2 is 0.5?
$endgroup$
– stripe123
Jan 2 at 21:15
$begingroup$
Thanks for the comments. Updated notation, hopefully less confusing now. So is the probability of being a boy 60/150 = 0.4? If so, would I be correct in saying the answer for part 2 is 0.5?
$endgroup$
– stripe123
Jan 2 at 21:15
add a comment |
0
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2
$begingroup$
It's sometimes helpful to think through numerical examples. Say $A$ has $30$ boys and $70$ girls, and $B$ has $30$ boys and $20$ girls. That fits your requirements, yes? Then there are $60$ boys all in all out of $150$ students. Does that fit your answer for (i)? Note too that each school has the same number of boys. Does that fit your answer for (ii)?
$endgroup$
– lulu
Jan 2 at 20:50
1
$begingroup$
Note: your notation is very confusing. I think (but am not sure) that you are using "$A$" to denote the event "a randomly observed child attends school $A$" but you are using "$B$" to denote the event "a randomly observed child is a boy". At the very least, you should define the variables you want to use.
$endgroup$
– lulu
Jan 2 at 20:57
1
$begingroup$
In $(a)$, you need to multiply $1-0.7$ by $2/3$, not $2$. Similarly, $1-0.4$ gets multiplied by $1/3$. The 'weighting' you are talking about is a probability.
$endgroup$
– Shubham Johri
Jan 2 at 21:01
$begingroup$
Thanks for the comments. Updated notation, hopefully less confusing now. So is the probability of being a boy 60/150 = 0.4? If so, would I be correct in saying the answer for part 2 is 0.5?
$endgroup$
– stripe123
Jan 2 at 21:15