Tree Diagrams Probability

Question

4. 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, $[2]$ b. one of each chocolate is taken, $[2]$ c. both chocolates have hard centres, given that the second chocolate has a hard centre. $[4]$

a. $P(SS)= (15/20)(14/19)= 21/38$

b. $P(\text{sorting})= (SH)+(HS)= (15/20 x5/19)+(5/20 x15/19)= 15/38$

c. P(Both H and second is H) divide by P(second is H)

Using formula: $P(BH | 2H)/(2H)$

How to tackle with it, i try all combination but do not get the answer $(20/83)$

Answer 1

The answer for (c) does not come out to be 20/83 according to my method either, so can you show what answer you get?

I get 4/19 as the required probability for (c) by

P(H ∩ H) / (P(H ∩ H) + P(S ∩ H))

where, S is the event of choosing a soft centered chocolate and H is the event of choosing a hard centered chocolate.

Answer 2

Possibility Space = $\{HH,SS,SH,HS\}$

Sample Space $= 20P2 = 380$

$n(HH) = 5P2 = 20\\ n(SS) = 15P2 = 210\\ n(SH)+N(HS) = 380-210-20 = 150\\ n(HS) = 75\\ N(SH) = 75$

Given the second chocolate is HARD leaves $4$ possibility for the FIRST to be HARD. therefore the number of possible combinations with the SOFT chocolates are $4$ Hard $\cdot 15$ Soft $= 4\cdot15 = 60$ $\begin{align}n(HH)' &= \text{Possibility Space} - n(HH)\\ &= 4-1\\ &= 3\end{align}$

The probability of first and second being hard given the second is hard is

$\begin{align}P(HH) &= n(HH)/[(4\text{Hard}\cdot15\text{Soft}) +n(HH)+ n(HH)']\\ &= 20/(60+20+3)\\ &= 20/83\end{align}$