Cara is playing a number game where she has two tiles for each number 0 to 9. A tile is chosen at random. How many favorable outcomes are there for choosing a multiple of 3?

Step-by-step explanation:

The number of possible numbers on the tile are:

0, 1 , 2 , 3 , 4,  5,  6,  7 ,  8 and 9.

Now, the chosen tile can have any of the ten options mentioned above.

The favorable outcomes are  3 , 6 and 9 as they are the multiple of 3.

So, the number of possible favorable outcomes  = 3 ways

Hence, if a card is chosen at random, there are 3 possible favorable outcomes for choosing a multiple of 3.

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

i don't even know man. edge inuity be messing with us. i had to ask a question too.