- 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 $(20/83)$
Best Answer
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.