Question

Probability help?

Answer 1

(/.\) I want to help, but I can't.

Answer 2

It's fine sammi :) thanks for trying though

Answer 3

@whpalmer4 @ganeshie8 @hartnn

Answer 4

start by counting number of elements in S

Answer 5

S = {1,2,3,4,5,6,7,8,9,10}

E = {1,2,3,4,5,6}

F = {2,4,6,8}

Answer 6

number of elements in S = ?
number of elements in E = ?
number of elements in F = ?

Answer 7

By elements, do you mean like how many numbers in each one?

Answer 8

yes !

Answer 9

once u finish counting,
to get the probabilities, simply take ratio :

P(E) = (number of elements in E) / (number of elements in S)

P(F) = (number of elements in F) / (number of elements in S)

Answer 10

Ohhh ok that makes sense. And then what does it mean with the complements of F and E?

Answer 11

complement of E is all numbers except the numbers in E

Answer 12

since you have E = {1,2,3,4,5,6}

complement of E would be \(E^c\) = {7,8,9,10}

Answer 13

Oh ok and then for complement of F it would be {1,3,5,7,9,10}?

Answer 14

you got it !

Answer 15

Could you just explain how to figure out P(E and complement of F)? That is the one that is tricky now to me.

Answer 16

E = {1,2,3,4,5,6}
\(F^c\) = {1,3,5,7,9,10}

\(E~ and ~F^c\) = {1, 3, 5}

Answer 17

\(P(E~ and ~F^c)\) = 3/10

Answer 18

So for the ones with E and F, it will just be the similar numbers they share?

Answer 19

yes "E and F" is simply the common elements in both E and F

Answer 20

Ok thanks so much!

Answer 21

np :)