Basic- Discrete probability distribution - Probability with conditional event?

[Njords]

Hi, I'm having difficulty in what I think should be an easy calculation..

I'm thinking that the probability 7 are sold is just it's probability divided by the sum of others excluding Pr X=0?

Or is there are more generalized formula or method?

Thanks.

 

[JeffM]

Hi, I'm having difficulty in what I think should be an easy calculation..

I'm thinking that the probability 7 are sold is just it's probability divided by the sum of others excluding Pr X=0?

Or is there are more generalized formula or method?

Thanks.

View attachment 10439

You are correct. Well done.

There is no more generalized formula for a conditional probability because that is the definition of conditional probability.

\(\displaystyle \text {P(A)} \ne 0 \implies \text { P(B given A) } \equiv \dfrac{\text {P(A and B)}}{\text {P(A)}}.\)

By the way, \(\displaystyle \text {P(B given A)}\) is usually written as \(\displaystyle \text {P(B | A)}\).

EDIT: It occurs to me that P(A and B) may not be obvious to a beginning student. Let A be the event of selling at least one car in a day, and B be the event of selling exactly seven cars in a day.

From our definition

\(\displaystyle \text {P(A | B) } \equiv \dfrac{\text {P(A and B)}}{\text {P(B)}} \implies \text {P(A and B) } \equiv \text { P(A | B) } * \text { P(B) if P(B)} \ne 0.\)

If 7 cars are sold, then clearly at least 1 car is sold so

\(\displaystyle \text {P(A | B) } = 1 \implies \text { P(A and B) } = 1 * \text { P(B) } = \text { P(B) } \implies \)

\(\displaystyle \text {P(B | A) } = \dfrac{\text {P(A and B)}}{\text {P(A)}} = \dfrac{\text {P(B)}}{\text {P(A)}}.\)

 

[Njords]

You are correct. Well done.

There is no more generalized formula for a conditional probability because that is the definition of conditional probability.

\(\displaystyle \text {P(A)} \ne 0 \implies \text { P(B given A) } \equiv \dfrac{\text {P(A and B)}}{\text {P(A)}}.\)

By the way, \(\displaystyle \text {P(B given A)}\) is usually written as \(\displaystyle \text {P(B | A)}\).

EDIT: It occurs to me that P(A and B) may not be obvious to a beginning student. Let A be the event of selling at least one car in a day, and B be the event of selling exactly seven cars in a day.

From our definition

\(\displaystyle \text {P(A | B) } \equiv \dfrac{\text {P(A and B)}}{\text {P(B)}} \implies \text {P(A and B) } \equiv \text { P(A | B) } * \text { P(B) if P(B)} \ne 0.\)

If 7 cars are sold, then clearly at least 1 car is sold so

\(\displaystyle \text {P(A | B) } = 1 \implies \text { P(A and B) } = 1 * \text { P(B) } = \text { P(B) } \implies \)

\(\displaystyle \text {P(B | A) } = \dfrac{\text {P(A and B)}}{\text {P(A)}} = \dfrac{\text {P(B)}}{\text {P(A)}}.\)

Ohhhhhhhhhhh! Thanks for the reassurance. It's clicked in!

What was causing brain.exe to crash was how I was defining probability A. Merely reading directly off the table which I understand why it is wrong now, Intuitively I knew it to be the sum
of the rest excluding zero, I dunno maybe I read too fast and miss vital bits of the language, this topic shouldn't be difficult but I do get caught out a lot with simple but wrong deductions at times.
.... At least 1 sold = \(\displaystyle \text{P(A)}\) or
1- Probability none are sold.
Thank you once again!

 

[JeffM]

A lot of basic math, like algebra and probability theory, involves learning to trust the mechanical rules over uninformed intuition. Particularly in probability theory, our uninformed tuition is too often wrong. Set things up according to the basic definitions and reason from there. In this case, your intuition was correct, but when you cannot justify your intuition from the basics, you are absolutely right to distrust your results.

I am sorry that I did not initially see why you were unsure of your answer. I view my role as helping people see around what seemed obscure to them, and I cannot do that when I myself am too dim to see what that obscurity is.

 

[Njords]

A lot of basic math, like algebra and probability theory, involves learning to trust the mechanical rules over uninformed intuition. Particularly in probability theory, our uninformed tuition is too often wrong. Set things up according to the basic definitions and reason from there. In this case, your intuition was correct, but when you cannot justify your intuition from the basics, you are absolutely right to distrust your results.

I am sorry that I did not initially see why you were unsure of your answer. I view my role as helping people see around what seemed obscure to them, and I cannot do that when I myself am too dim to see what that obscurity is.

No need to apologize as my original post was the cause of any obscurity, I could have been a bit more clear in that regard. However, I did need help with seeing it in terms of conditional probability and your reply absolutely helped in that regard. Perhaps it was exactly as needed in order for me to see it and relate the rules for conditional probability to the context of the problem. So there was a guided push from your part in that answer, that led to a bit of self learning or readjustment in the way I was thinking about it, which I find is most ideal for learning hence I'd say you were successful in your stated role. I really appreciate this forum and especially the willingness for the people that help.