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

A box contains20chocolates, of which15have Soft centres and5have Hard centres. Two chocolates are taken at random,one after the other.

Calculate the probability that

(a)both chocolates have Soft centres,[BOOK ANSWER:$\displaystyle \frac{21}{38}$]

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

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

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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)

$\displaystyle =P(S\cap S)$

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

$\displaystyle =\frac{21}{38}$MATCHED

(b)

$\displaystyle =P(S\cap H)+P(H\cap S)$

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

$\displaystyle =\frac{15}{38}$MATCHED

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

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

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

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

Is it a mistake of interpretation from my part or from the Book? Your Help is welcomed, Thank you