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Math Help - Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawshaw

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    Exclamation Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawshaw

    Exercise 3d Tree Diagrams, Section B, Question 2. (Page 201)


    A box contains 20 chocolates, of which 15 have Soft centres and 5 have Hard centres. Two chocolates are taken at random, one after the other.


    Calculate the probability that
    (a) both chocolates have Soft centres,
    [BOOK ANSWER: \frac{21}{38}]

    (b) one of each sort of chocolates is taken, [BOOK ANSWER: \frac{15}{38}]

    (c) both chocolates have Hard centres, given that the second chocolates has a Hard centres. [BOOK ANSWER: \frac{20}{83}]


    ************************************************** ************************************************** *******************


    I manage as shown below, to successfully obtain the first two answers correct, but question (c) don't corresponds to any solutions that I could ever think of.

    My Solution:


    Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawshaw-tree-diagram.jpg


    (a)
    =P(S\cap S)

    =\frac{15}{20}\times \frac{14}{19}

    =\frac{21}{38} MATCHED


    (b)
    =P(S\cap H)+P(H\cap S)

    =\left (\frac{15}{20}\times \frac{5}{19}  \right )+\left (\frac{5}{20}\times \frac{15}{19}  \right )

    =\frac{15}{38} MATCHED


    (c) I need Help here, this conditional probability is ok I hope:


    =\frac{P(H\cap H)}{P(S\cap H)+P(H\cap H)}

    =\frac{\left (\frac{5}{20}\times \frac{4}{19}  \right )}{\left (\frac{15}{20}\times \frac{15}{19}  \right )+\left (\frac{15}{20}\times \frac{4}{19}  \right )}

    =\frac{4}{19}\neq \frac{20}{83} MISMATCHED !!!


    Is it a mistake of interpretation from my part or from the Book? Your Help is welcomed, Thank you
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    Re: Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawsha

    According to my calculations you are correct and not the textbook.
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    Re: Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawsha

    I think you are right too.
    I just said one dark choc doesn't count-it's already accounted for. Therefore the probability of the first one being a dark chocolate is 4/19
    Mind you, I'd believe Plato way before I'd believe me. I am here to ask another probability question.
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