Question

A mini license plate for a toy car must consist of a two letters followed by a one digit odd number. Each letter must be a J or a Z. Repetition of letters is permitted.

Use the counting principle to determine the number of points in the sample space.
Construct a tree diagram to represent this situation
List the sample space.
Determine the exact probability of creating a mini license plate with a Z. Give solution exactly in reduced fraction form.

Answer 1

OK, _ _ _ ....you have three spaces: space 1 can be either a "J" or "Z", space 2 can be either a "J" or "Z", space 3 can be either {1,3,5,7,9}

Answer 2

you therefore have 2 choices for space1, 2 choices for space2, and 5 choices for space3...the letters can repeat therefore the choices for letters does not change here

Answer 3

(2) (2) (5)

Answer 4

the number of points is then the product of the number of choices available for each item: 2 letters * 2 letters * 5 odd numbers

Answer 5

so it would be 2*2*5=

Answer 6

yes

Answer 7

so 20 would be the number of points ?

Answer 8

yes, the sample space is basically all the combinations that are possible

Answer 9

can you determine how many license plates have a "Z" in them?

Answer 10

we know there are 20 combinations total

Answer 11

one way to determine the number of plates with a "Z" in them would be: (1) determine the number of plates that have only "J" in them, and subtract that from the total, or (2) outright, determine the number of plates that have either 1 "Z" or 2 "Z's"

Answer 12

so, we can count these situations in the same way that we counted the total

Answer 13

brb phone

Answer 14

I bet you can figure this out now: X combinations have "Z" in them, so X/20 is probability

Answer 15

ok, no problem

Answer 16

honestly i dont know how to do this.. i'm sorry.. :(

Answer 17

use the same principle we used to solve for the total

Answer 18

_ _ _ --> #choices for space 1 * #choices for space 2 * #choices for space 3

Answer 19

so half would have z 10

Answer 20

why half?

Answer 21

well there is 20 z and J total

Answer 22

just a second

Answer 23

OK, here we go

Answer 24

so, let's look at the combinations that only have a "J" in them...that would be: J J then a number

Answer 25

1 choice of letter and 5 choices of number

Answer 26

J J 1, J J 2, J J 3, J J 4, J J 5,

Answer 27

so there exist other combinations of letters: J Z {1,2,3,4,5}, Z J {1,2,3,4,5}, Z Z {1,2,3,4,5}

Answer 28

you'll see that each group has 5 total, add them all up and we get 20, the total we found at first

Answer 29

now, we want to consider those that have a "Z"

Answer 30

well, we know that 5 only have J's, the rest then have 1 Z or 2 Z's

Answer 31

so 20-5 = 15

Answer 32

15 combinations out of the total combinations have a Z in them

Answer 33

so, how do you feel about this problem?