Question

Does anyone know how Adding Probablity works?? I need helppp

Answer 1

What's the problem?

Answer 2

@randogirl123

Answer 3

this is my homework from the day I was one and so the teacher explained it but I am confused how it works

Answer 4

To add probabilities, you use the Addition Rule: P(A or B) = P(A) + P(B) for events that can't happen together (mutually exclusive), or P(A or B) = P(A) + P(B) - P(A and B) for events that can overlap, subtracting the overlap to avoid double-counting, which is essential for "or" scenarios like rolling a die and getting an even number or a multiple of 3.

Answer 5

Yep

Answer 6

Google helps for a lot ngl

Answer 7

Google don't help

Answer 8

oh dear?

Answer 9

I've done research and it gets confusing and confusing

Answer 10

well watch a video? that kinda helps me

Answer 11

Well this is the formula $$P(A \text{ or } B) = P(A) + P(B)$$

Answer 12

I can do one of them for you

Answer 13

what a great teacher

Answer 14

Probability of working part-time: P(PT) = 0.5

Probability of belonging to a club: P(C) = 0.4

Probability of both (the overlap): P(PT and C) = 0.05

Now we're gonna use this formula: $$P(PT \text{ or } C) = P(PT) + P(C) - P(PT \text{ and } C)$$

Then we'll plug in the numbers to the equation:

$$P(PT \text{ or } C) = 0.5 + 0.4 - 0.05$$$$P(PT \text{ or } C) = 0.9 - 0.05$$$$P(PT \text{ or } C) = 0.85$$

Then, 0.85 (or 85%) will be your final answer

Answer 15

Sorry this was so long :sob:

Answer 16

oooo hmm

Answer 17

it kinda makes sense

Answer 18

Lemme take a look rq

Answer 19

ok

Answer 20

u still typing ;-;

Answer 21

So we know that there's 6 sides on a die, so there will be a total of 6 possible outcomes

For #1 since we know that there's only one "4" on a standard dice we can tell that the Successful Outcome would be {4} and the result would be 1/6.

For #2 the numbers are either even or odd because every number falls into one of the two categories:

Even: 3/6

Odd: 3/6

Calculation: 3/6 + 3/6 = 6/6 = 1

So for this one the result would be 1 or a 100% chance


For #3 since "3" is larger than one, we can avoid double-counting by listing the unique values:

Rolling a 3: {3}

Numbers larger than one: {2,3,4,5,6}

Unique set: {2,3,4,5,6}

Result: 5/6


For #4 we're going to do the same thing like the others, which is putting everything in a separate category:

Multiples of 3: {3,6}

Odd numbers: {1,3,5}

The overlap: 3, because it's in both lists

Calculation using the addition rule: P(mult of 3) + P(odd) - P(both)

2/6 + 3/6 - 1/6 = 4/6 (simplified to 2/3)


Final answers:

1. 1/6
2. 1
3. 5/6
4. 2/3

Answer 22

more helppp