# Thread: Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawshaw

1. ## 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: $\frac{20}{83}$]

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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:

(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

2. ## Re: Mismatching Book Answer - Tree Diagrams - A Level Statistics - Chambers & Crawsha

According to my calculations you are correct and not the textbook.

3. ## 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.