Conditional probability of dying from eating a poison fruit

[Addez123]

Summary:: There's 11 fruits, 3 of which is poisionous.
A guy eats 4 of them, a girl eats 6 and a dog gets the last one.

What is the conditional probability of both the girl and guy dying IF the dog made it? One fruit is enough to kill you.

$$P(dog lives) = 8/11$$

$$P(allPeopleDie | dog lives) = P(allPeopleDie \cap dog lives)/P(dog lives) $$
$$P(allPeopleDie \cap dog lives) = ??$$

Even if I could calculate guy & girl both getting atleast one posionous fruit each (allPeopleDie), I can't just multiply that with P(dog lives) because they are not independent. So I have no idea how to solve this. Maybe my whole approach is wrong?

 

[member 587159]

Hint:

$$\{\mathrm{all \ people \ die \ and \ dog \ lives}\}= $$$$\{\mathrm{\ boy \ eats \ two \ poisonous \ fruits \ and \ girl \ one }\}\cup \{\mathrm{\ girl \ eats \ two \ poisonous \ fruits \ and \ boy \ one }\}$$

I wonder who came up with such a sadistic problem :P

 

[Addez123]

I just can't seem to get the right answer.
The answer is 4/5.

A. How many ways can the boy and girl eat 3 fruits and each have atleast one?
A = 2 ways, as you showed above.
B. How many ways can 3 people eat 11 fruits given the guy eats 4, the girl eats 6 and the dog eats 1?
That's an extremely complex question and I've tried for hours and come up with no good answer.

Solution would be $$P(all people die \cap dog lives) = A/B$$ but I don't have B.

 

[WWGD]

I just can't seem to get the right answer.
The answer is 4/5.

A. How many ways can the boy and girl eat 3 fruits and each have atleast one?
A = 2 ways, as you showed above.
B. How many ways can 3 people eat 11 fruits given the guy eats 4, the girl eats 6 and the dog eats 1?
That's an extremely complex question and I've tried for hours and come up with no good answer.

Solution would be $$P(all people die \cap dog lives) = A/B$$ but I don't have B.

Notice , as QED wrote, that boy and girl must finish up all poison fruits before dog gets anything. Can you see and count the number of ways in which these 3 fruits can be chosen? Look at QED's hint in #2 again.

 

[haruspex]

Note that the order does not matter, so simplest is to deal with the dog first. Dog eats one ok fruit, leaving 7 ok, 3 bad, and we can forget about the dog.
If Romeo and Juliet are both to die, what combinations might the boy eat?

Edit: strike that last line - always look for the smallest set, in this case the set of 3 poisonous fruits.
Consider how these might be selected from the 6+4 eaten by the kids if both are to die.