A bag contains five red candies, six white candies, and seven blue candies. Suppose one piece of candy is drawn at random. Find the probability of each of the following.
- A white candy is drawn. _____________
- A red or blue candy is draw. _____________
- Neither a white nor a blue candy is drawn. _____________
You have a total of 18 candies.
We'll assume that they're all equally likely to be chosen.
Now, suppose I'm feeling mean, and I poison 1 of the 18 pieces of candy.
So there's 18 candies, and 1 of them is poisonous.
So if you pick 1, there's a 1 in 18 chance it will be the poisonous one - fair enough?
So Probability( Candy = poisonous) = 1/18.
Now, what if, instead, I poison two candies?
Now there are 2 of the 18 candies which are poisonous.
So if you pick one, you're twice as likely to get a poisonous one, so
P( Candy = poisonous) = 2/18 = 1/9. Ok?
Similarly if I poison 3 candies, the P(Candy = poisonous) = 3/18 = 1/6.
Now, what if I poisoned all the white candies and not the other ones? (And didn't tell you that was what I was doing - because then you wouldn't eat the white ones, so my dastardly plan wouldn't work.)
Well, there are 6 white candies. So what would be the probability of you picking out a poisonous/white one?
............
For the second one, suppose I poisoned all the blue and red candies...
( I've been in the office too long... I think my psychotic tendencies are bubbling to the surface...)
What would be the chances of you picking a poisonous/red or blue candy?
.........
For the last bit - suppose I don't poison the blue candies, and I don't poison the white candies. But I poison all the others, ie. the reds.
What are the chances of you picking a poisonous one now? ie. NOT a white or blue one?