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    A question asks: in a group of 60 girls and 40 boys, it is found that the probability of a person being left handed is 0.18. The probability of a girl being left handed is 0.1. What is the probability of a boy being right handed?

    I got this wrong and in the solution I'm not seeing what exactly the difference of being a girl and left handed P(AuB) = 0.1 and being left handed given one is a girl P(A|B)*

    Could somebody please explain the difference?

    *P(AuB) = P(A) * P(A|B)
    therefore, P(A|B) = 0.1 / 0.6 = 1/6.

    I can't intuit what exactly is being said here, when looking at u vs |
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    I think I just worked it out.

    A girl who is left handed is different to the probability of being left handed given she’s a girl.
    P(A|B) is dependant on the probability of being a girl first (.6) while P(AuB) assumes she is a girl (so is probability of being left instead of right handed in subgroup Girls)

    Is this correct?
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    Do you need to calculate it this way? Intuitively there are 18 left-handed people in the group, 6 of which are girls. This means that there are 12 left-handed boys, which means there are 28 right-handed boys. 28 / 40 = 7 / 10 = 70% chance of a boy being right-handed.
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    (Original post by eniet)
    Do you need to calculate it this way? Intuitively there are 18 left-handed people in the group, 6 of which are girls. This means that there are 12 left-handed boys, which means there are 28 right-handed boys. 28 / 40 = 7 / 10 = 70% chance of a boy being right-handed.
    The answer is 8/25 so no
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    (Original post by Mathstatician)
    A question asks: in a group of 60 girls and 40 boys, it is found that the probability of a person being left handed is 0.18. The probability of a girl being left handed is 0.1. What is the probability of a boy being right handed?

    I got this wrong and in the solution I'm not seeing what exactly the difference of being a girl and left handed P(AuB) = 0.1 and being left handed given one is a girl P(A|B)*

    Could somebody please explain the difference?

    *P(AuB) = P(A) * P(A|B)
    therefore, P(A|B) = 0.1 / 0.6 = 1/6.

    I can't intuit what exactly is being said here, when looking at u vs |
    \mathbb{P}(A \cup B) is the probability of either A or B occuring.
    \mathbb{P}(A|B) is the probability of A occurring given that B has occurred.
    You seem to be confusing \cup for \cap. The latter means "and", and \mathbb{P}(A \cap B)=\mathbb{P}(A|B)\mathbb{P}(B)

    For example, if you roll two dice, with A being the event that you roll a 3 and B being the event you roll a 4 then \mathbb{P}(A \cap B) = \mathbb{P}( \text{roll a 3 and a 4}) but \mathbb{P}(A \cup B) = \mathbb{P}( \text{roll a 3 or a 4}).
    Finally \mathbb{P}(A | B) = \mathbb{P}(\text{roll a 3, given one dice is a 4}).
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    (Original post by joostan)
    \mathbb{P}(A \cup B) is the probability of either A or B occuring.
    \mathbb{P}(A|B) is the probability of A occurring given that B has occurred.
    You seem to be confusing \cup for \cap. The latter means "and", and \mathbb{P}(A \cap B)=\mathbb{P}(A|B)\mathbb{P}(B)

    For example, if you roll two dice, with A being the event that you roll a 3 and B being the event you roll a 4 then \mathbb{P}(A \cap B) = \mathbb{P}( \text{roll a 3 and a 4}) but \mathbb{P}(A \cup B) = \mathbb{P}( \text{roll a 3 or a 4}).
    Finally \mathbb{P}(A | B) = \mathbb{P}(\text{roll a 3, given one dice is a 4}).
    I've not used latex before and I apologise for not using it in my post. This answer was very useful thank you.

    I think I realise now that in the question asked P(AnB) is P(Girl) * P(Left|Girl) which is different to P(B|A) or P(Left|Girl) in that the former is the probability of both events happening while the latter is saying if A happens then the probability of B happening is (x). Is that right?
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    (Original post by Mathstatician)
    I've not used latex before and I apologise for not using it in my post. This answer was very useful thank you.

    I think I realise now that in the question asked P(AnB) is P(Girl) * P(Left|Girl) which is different to P(B|A) or P(Left|Girl) in that the former is the probability of both events happening while the latter is saying if A happens then the probability of B happening is (x). Is that right?
    I'm not sure what your x is.
    Furthermore I would like to comment that my previous post is merely meant to explain notation and the method suggested by eniet is the way I'd go about it - the answer you've given/been given, of \frac{8}{25} must surely be incorrect.
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    (Original post by joostan)
    the answer you've given/been given, of \frac{8}{25} must surely be incorrect.
    Are you sure? Here's how I figure it is correct, given my current understanding:
    in a group of 60 girls and 40 boys, it is found that the probability of a person being left handed is 0.18. The probability of a girl being left handed is 0.1. What is the probability of a boy being right handed?
    Given:
    P(G) = 0.6
    P(B) = 0.4
    P(L) = 0.18
    P(GnL) = 0.1
    -
    Solving for the girls:
    P(GnL) = P(G) * P(G|L)
    Therefore, P(G|L) = 0.1 / 0.6 = 1/6
    And P(G|R) = 5/6
    -
    Solving for the boys:
    P(L) = P(GnL) + P(BnL)
    Therefore, P(L) = P(GnL) + P(B) + P(B|L)
    So 0.18 = 0.1 + 0.4 + P(B|L)
    0.08/0.4 = 1/5 = P(B|L)
    And P(B|R) = 4/5
    -
    Finally solving for what the question asks for:
    P(BnR) = P(B) * P(B|R)
    P(BnR) = 0.4 * 4/5
    P(BnR) = 8/25

    Is this incorrect?
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    (Original post by Mathstatician)
    Are you sure? Here's how I figure it is correct, given my current understanding:

    Given:
    P(G) = 0.6
    P(B) = 0.4
    P(L) = 0.18
    P(GnL) = 0.1
    -
    Solving for the girls:
    P(GnL) = P(G) * P(G|L)
    Therefore, P(G|L) = 0.1 / 0.6 = 1/6
    And P(G|R) = 5/6
    -
    Solving for the boys:
    P(L) = P(GnL) + P(BnL)
    Therefore, P(L) = P(GnL) + P(B) + P(B|L)
    So 0.18 = 0.1 + 0.4 + P(B|L)
    0.08/0.4 = 1/5 = P(B|L)
    And P(B|R) = 4/5
    -
    Finally solving for what the question asks for:
    P(BnR) = P(B) * P(B|R)
    P(BnR) = 0.4 * 4/5
    P(BnR) = 8/25

    Is this incorrect?
    I've not looked thoroughly at this, but you seem to have muddled some pluses and multiplications, moreover, how did you get \mathbb{P}(G|R)=\frac{5}{6} and \mathbb{P}(B|R)=\frac{4}{5}.
    I've actually got to get up early tomorrow, so I've gotta scoot, try and get Zacken or someone to help you.
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    (Original post by joostan)
    I've not looked thoroughly at this, but you seem to have muddled some pluses and multiplications, moreover, how did you get \mathbb{P}(G|R)=\frac{5}{6} and \mathbb{P}(B|R)=\frac{4}{5}.
    I've actually got to get up early tomorrow, so I've gotta scoot, try and get Zacken or someone to help you.
    Do those numbers not make sense? My understanding is the because we split at the initial probability Girls and Boys, We then split again at Left and Right, and therefore the 1/6 and 5/6 for girls and 1/5 and 4/5 for boys is perfectly acceptable, since the only requirement is that the totals are 1.

    I'm only a GCSE student. Maybe I have completely misunderstood. I appreciate your help thus far. Will wait on somebody else. Thank you.
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    (Original post by Mathstatician)
    Are you sure? Here's how I figure it is correct, given my current understanding:

    Given:
    P(G) = 0.6
    P(B) = 0.4
    P(L) = 0.18
    P(GnL) = 0.1
    The problems set in at the bolded step. The question states that "the probability of a girl being left handed is 0.1". In other words, if you choose a girl from the group of 100 students, the probability of her being left handed is 0.1. This is precisely P(Left handed | individual is a girl) (= P(L | G) ) and not P(individual is left handed and individual is a girl) (= P(L \cap G)).

    The solution joostan gave is correct.
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    PS Reviewer
    You can also see it covered by Exam Solutions: http://www.examsolutions.net/maths-r...examples-1.php
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    (Original post by Gregorius)
    The problems set in at the bolded step. The question states that "the probability of a girl being left handed is 0.1". In other words, if you choose a girl from the group of 100 students, the probability of her being left handed is 0.1. This is precisely P(Left handed | individual is a girl) (= P(L | G) ) and not P(individual is left handed and individual is a girl) (= <a rel="nofollow" rel="nofollow" href="javascript:void(0)"><img src="http://www.thestudentroom.co.uk/latexrender/pictures/50/50e7c1e074549840fb77d3a36abd0d6b .png" width="79" height="21" style="border: 0px; margin: 0px; padding: 0px; vertical-align: -5px;" alt="P(L \cap G)" title="P(L \cap G)" onclick="newWindow=window.open(' http://www.thestudentroom.co.uk/latexrender/latexcode.php?code=P%28L+%5Ccap+ G%29','latexCode','toolbar=no,lo cation=no,scrollbars=yes,resizab le=yes,status=no,width=460,heigh t=320,left=200,top=100';" /></a>.The solution @joostan gave is correct.
    (Original post by ByronicHero)
    You can also see it covered by Exam Solutions: http://www.examsolutions.net/maths-r...examples-1.php
    He (the guy in exam solutions) gets the same answer as me. Are we both wrong?
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    (Original post by Mathstatician)
    He (the guy in exam solutions) gets the same answer as me. Are we both wrong?
    Having had a look at the video, I think the issue is that the question is very poorly worded, and as Gregorius said, by the probability of a girl being left handed he has taken that to mean the probability that a girl is selected from the group and that said girl is left handed to be 0.1, rather than the more sensible inference in my opinion that this is the probability of being left handed, given you are a girl.
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    (Original post by joostan)
    Having had a look at the video, I think the issue is that the question is very poorly worded, and as Gregorius said, by the probability of a girl being left handed he has taken that to mean the probability that a girl is selected from the group and that said girl is left handed to be 0.1, rather than the more sensible inference in my opinion that this is the probability of being left handed, given you are a girl.
    I appreciate that you say this, since my initial answer was dependent on the same inference you made. The wording and answer just confused matters and for the given answer to make sense I had to reassess my stats understanding (not good!) Think I'll just ignore this as a poor example and move on.

    Thanks a lot everybody for your input. I think we should consider the matter resolved under: Poor Question.
 
 
 

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