PROBABILITY--Math?

Question

There are 12 doors into the building and 6 stairways from the first floor to the second floor. How many possible ways are there for Laura to go from the outside of the school to a room on the second floor?

I basically just need to understand how to do this?

Answer 1

simple, 72

try something easy.

you have 3 crayons. if you pair each, how may possible pairs can you have if the colors are red yellow and blue. (blue= b. r= red, y= yellow.)

multiply choices by other choices (3x2)

Answer 2

12 doors into the building gives her twelve different options on her first entry. Therefore, she has twelve different ways to enter the building. Once she enters, she can choose one of six ways to go upstairs to the second floor.

In essance: 12 entries (possible options) * 6 (ways upstairs) = 72 ways of getting to the second floor.

Illustrated Choices:
Door1: Stair 1, Stair 2, Stair 3, Stair 4, Stair 5, Stair 6 = 6 choices
Door 2: Stair1, Stair 2, Stair 3, Stair4, Stair 5, Stair 6= 6 choices

continue the pattern and you get six choices for each of the twelve entry ways

Answer 3

its 72. you can multiply 6 by 12 because 1 door can have 6 stairways so 12 is 72 also written 1:6 =12:72

Answer 4

This is classic. There are two choices to make and they are independent

First choose door: 12 ways to do this.

Now, for EACH of those choices, she has 6 choices of stairways:
6 ways to choose stairway.

Since the choices are independent you multiply to get total choices:

12*6 =72 total ways.

This is called the multiplication rule. It applys when choices are independent. It will serve you well.

Answer 5

This is multiplication rule ; m ways x n ways
so 12 x 6 = 72

Answer 6

1 door => 6 sairways,
12doors = 72 different ways

6*12

Answer 7

Q1 - A: experience C is an same as: one dice sows a good huge style, the different an unusual (in trouble-free words then the sum is unusual) If both dice teach the large style 3 & 4, experience A and C are chuffed -> no longer unique If both dice teach the large style 2 & 4, experience A and B are chuffed -> no longer unique yet when both dice teach a good huge style, one in all them cann't be unusual jointly (and vice versa) -> (B,C) is jointly exculisve Q1 - B: If 2 activities are self reliant the equation P(A)*P(B) = P(A and B) valid. this isn't the case if the intersection isn't empty (interest the option isn't genuine!). when you consider that 4&4 is both portion of A and C, the intersection isn't empty and for this reason both activities no longer self reliant. Q2 - A: Assuming that the ten% signifies that they in trouble-free words learn Chemestry, and the 8% in trouble-free words Maths chance = type of fulfillment / type of things (A) = 2%/(10%+2%) = 2/12 = a million/6 (B) = 2%/(8%+2%) = 2/10 = a million/5

Answer 8

xxxxx
there are two independent events required to get from "outside"
to "second floor". You muiltiply the number of ways to perform each event to yield the total number of "way-combinations" for these two events.

there are 12 ways to get from outside to first-floor.

6 ways to get from first-floor to second floor>


12 x 6 = 72
possible paths from "outside" to "first floor"
and then "first floor" to "second-floor"

(number "outside" to "first floor") times
(number "first floor" to "second-floor")

Answer 9

there are 72 ways to go where she wants to go