In a box there are $16$ ice-creams: $6$ lemon flavor,$4$ mint flavor and $6$ strawberry flavor.When we extract two ice-creams,what's the probability of getting two different flavors,given that at least one is strawberry flavor.
That's my solution :
$P(A)$ = "Different flavors" = $\frac{\binom{16}{2}}{16!}$
$P(B^c)$ = "At least one is strawberry flavor" = $1 - \frac{\binom{10}{2}}{16!}$
So...We want $P(A|B^c)$ using conditional probability, where I go wrong?
I dont see exactly what you did (edit: I will provide a complete anwer to clarify things).
There is a way to do it using the law of total probability: assuming that each ice-cream have the same probability to be taken then it is the probability that the first one will be some flavor and that the second will be different, that is:
$$P[\text{ two distinct flavor }]=\\=\sum_{x\in\{\text{ flavors }\}}P[\text{ first extraction is }x\text{ and second extraction is not } x]$$
Using formal notation: let $X_1$ the random variable that represent choose the first ice-cream, and $X_2$ the random variable that choose the second. The sample space is
$$\Omega:=\{\text{ lemon , strawberry, mint }\}$$
Now we assign $X_j(\text{ lemon }):=1$, $X_j(\text{ strawberry }):=2$, $X_j(\text{ mint }):=3$, for $j=1,2$. Thus the above can be rewritten formally as
$$P[X_1\neq X_2]=\sum_{x=1}^3P[X_1=x,X_2\neq x]=\sum_{x=1}^3P[X_2\neq x|X_1=x]P[X_1=x]\\=\frac{4+6}{15}\cdot\frac6{16}+\frac{4+6}{15}\cdot\frac6{16}+\frac{6+6}{15}\cdot\frac4{16}\\=\frac1{15\cdot 16}(2\cdot 6\cdot10+4\cdot 12)=\frac7{10}$$
For the second probability you just need to consider two cases and add up: if the first extracted ice-cream is of strawberry the probability that the second is of different flavor, and the probability that the first extracted ice-cream is not strawberry, that is
$$P[\text{ second is not strawberry considering the first one is }]+\\+P[\text{ first is not strawberry considering that the second will be }]$$
or alternatively using conditional probability notation
$$P[\text{ two distinct flavors }|\text{ first one is strawberry }]+\\+P[\text{ two distinct flavors }|\text{ second one is strawberry }]$$
Using the above formal notation this can be written as
$$P[X_1\neq X_2|X_1=2]+P[X_1\neq X_2|X_2=2]\\=P[X_1=2|X_2\neq 2]+P[X_1\neq 2|X_2=2]\\=\frac{P[X_1=2,X_2\neq 2]}{P[X_2\neq 2]}+\frac{P[X_1\neq 2,X_2=2]}{P[X_2=2]}\\=P[X_2\neq 2|X_1=2]\frac{P[X_1=1]}{P[X_2\neq 2]}+P[X_2= 2|X_1\neq2]\frac{P[X_1\neq1]}{P[X_2= 2]}$$
where
$$P[X_2\neq 2]=P[X_2\neq 2|X_1=2]P[X_1=2]+P[X_2\neq 2|X_1\neq 2]P[X_1\neq 2]\\=\frac{6+4}{15}\cdot\frac6{16}+\frac{6+4-1}{15}\frac{4+6}{16}=\frac1{15\cdot 16}(60+90)=\frac58$$
I leave the other calculations to you.