I know that there is the addition rule of probability, but I want to understand the intuition behind it. Specifically, why does OR signifies addition in probability theory?

Esther Hoffman
2022-04-30
Answered

I know that there is the addition rule of probability, but I want to understand the intuition behind it. Specifically, why does OR signifies addition in probability theory?

You can still ask an expert for help

Brianna Sims

Answered 2022-05-01
Author has **19** answers

I suppose that you question is this:

If two events $A$ or $B$ cannot occur simultaneously, then why is the probability that $A$ occurs or $B$ occurs equal to the probability that $A$ occurs plus the probability that $B$ occurs?

Suppose that there are, say, 100 possibilities, that $A$ takes place in 50 of them and that $B$ takes place in 20 of them. Then the probability that $A$ occurs is $\frac{1}{2}(=\frac{50}{100})$ and the probability that $B$ occurs is $\frac{1}{5}(=\frac{20}{100})$. What is the probability that $A$ occurs or $B$ occurs? Well, out of those 100 possibilities, $A$ occurs or $B$ occurs exactly in 70 of them (this is where I use the fact that $A$ or $B$ cannot occur simultaneously). So, the probability that $A$ occurs or $B$ occurs is

$\begin{array}{rl}\frac{70}{100}& =\frac{50}{100}+\frac{20}{100}\\ & =\text{probability that}A\text{occurs}+\text{probability that}B\text{occurs.}\end{array}$

If two events $A$ or $B$ cannot occur simultaneously, then why is the probability that $A$ occurs or $B$ occurs equal to the probability that $A$ occurs plus the probability that $B$ occurs?

Suppose that there are, say, 100 possibilities, that $A$ takes place in 50 of them and that $B$ takes place in 20 of them. Then the probability that $A$ occurs is $\frac{1}{2}(=\frac{50}{100})$ and the probability that $B$ occurs is $\frac{1}{5}(=\frac{20}{100})$. What is the probability that $A$ occurs or $B$ occurs? Well, out of those 100 possibilities, $A$ occurs or $B$ occurs exactly in 70 of them (this is where I use the fact that $A$ or $B$ cannot occur simultaneously). So, the probability that $A$ occurs or $B$ occurs is

$\begin{array}{rl}\frac{70}{100}& =\frac{50}{100}+\frac{20}{100}\\ & =\text{probability that}A\text{occurs}+\text{probability that}B\text{occurs.}\end{array}$

Tyler Velasquez

Answered 2022-05-02
Author has **19** answers

Because the probability is the number of favorable draws over the total number of draws.

And the number of favorable draws are additive: the number of [red or green] balls is the number of red plus the number of green.

Caution:

The additive rule is only valid for disjoint categories. For example, if you have black/white balls and dice, it is not necessarily true that

$\text{\#(black or ball)}=\text{\#black}+\text{\#balls}.$

And the number of favorable draws are additive: the number of [red or green] balls is the number of red plus the number of green.

Caution:

The additive rule is only valid for disjoint categories. For example, if you have black/white balls and dice, it is not necessarily true that

$\text{\#(black or ball)}=\text{\#black}+\text{\#balls}.$

asked 2022-05-09

What is $P(A\cup (B\cap C))$?

The question says it all. I know

$P(A\cap (B\cup C))=P(A\cap B)+P(A\cap C).$

Would this mean,

$P(A\cup (B\cap C))=P(A\cup B)+P(A\cup C)?$

Just want to make sure.

The question says it all. I know

$P(A\cap (B\cup C))=P(A\cap B)+P(A\cap C).$

Would this mean,

$P(A\cup (B\cap C))=P(A\cup B)+P(A\cup C)?$

Just want to make sure.

asked 2022-04-07

There's an 80% probability of a certain outcome, we get some new information that means that outcome is 4 times more likely to occur.

What's the new probability as a percentage and how do you work it out?

As I remember it the question was posed like so:

Suppose there's a student, Tom W, if you were asked to estimate the probability that Tom is a student of computer science. Without any other information you would only have the base rate to go by (percentage of total students enrolled on computer science) suppose this base rate is 80%.

Then you are given a description of Tom W's personality, suppose from this description you estimate that Tom W is 4 times more likely to be enrolled on computer science.

What is the new probability that Tom W is enrolled on computer science.

The answer given in the book is 94.1% but I couldn't work out how to calculate it!

Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.

What's the new probability as a percentage and how do you work it out?

As I remember it the question was posed like so:

Suppose there's a student, Tom W, if you were asked to estimate the probability that Tom is a student of computer science. Without any other information you would only have the base rate to go by (percentage of total students enrolled on computer science) suppose this base rate is 80%.

Then you are given a description of Tom W's personality, suppose from this description you estimate that Tom W is 4 times more likely to be enrolled on computer science.

What is the new probability that Tom W is enrolled on computer science.

The answer given in the book is 94.1% but I couldn't work out how to calculate it!

Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.

asked 2022-04-12

I came across the following question in a textbook (bear in mind that this is the only information given)-

There is a 50 percent chance of rain today. There is a 60 percent chance of rain tomorrow. There is a 30 percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?

My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of 30 percent. However, the answer given is 40 percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.

There is a 50 percent chance of rain today. There is a 60 percent chance of rain tomorrow. There is a 30 percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?

My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of 30 percent. However, the answer given is 40 percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.

asked 2022-05-09

Getting 2 or 5 in two throws should be $P(2)+P(5)$. $P(2)=1/6,P(5)=1/6$ so the combined so it should be 1/3.

I tried to visualize but not able to do so correctly.

11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6

total of 36 possibilities.

12,15,21,22,23,24,25,26,31,35,42,45,51,52,53,54,55,56,61,65

out of which 20 possibilities, so the probability should be 20/36 which is not 1/3.

Where am I going wrong?

I tried to visualize but not able to do so correctly.

11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6

total of 36 possibilities.

12,15,21,22,23,24,25,26,31,35,42,45,51,52,53,54,55,56,61,65

out of which 20 possibilities, so the probability should be 20/36 which is not 1/3.

Where am I going wrong?

asked 2022-05-02

I'm studying the Cartesian product, which is bound to the idea of a binary relation. Even with Cartesian products of several sets, n-ary Cartesian products, we have to think combinatorically as two sets at a time, recursively. Is there any sort of operation where there is, say, a triary operation happening. Obviously, when we have $3-2-1$ we have to think of associative rules, forcing subtraction to be a binary operator "step-through" affair. There is no $3\oplus 2\oplus 1$ operator that does something in one fell swoop to all three numbers is there? Lisp has (+ 1 2 3) and grade-school math has vertical addition with the plus-sign and a line under the lowest number, but these are not "internally" non-binary. The only thing that takes "three at once" is, yes, a product-based operator, again, an n-ary Cartesian product. Correct? Another example would be playing poker and being dealt five cards. The cards were shuffled, which is a combinatoric permutation of the stack of cards, which are then dealt into hands. Is there anything from probability (or anywhere) that doesn't start with a shuffle permutation therefore binary, rather, just looks at the five cards coming together non-permutation-, non-binary-wise?

asked 2022-05-09

Okay, so, my teacher gave us this worksheet of "harder/unusual probability questions", and Q.5 is real tough. I'm studying at GCSE level, so it'd be appreciated if all you stellar mathematicians explained it in a way that a 15 year old would understand. Thanks!

So, John has an empty box. He puts some red counters and some blue counters into the box.

The ratio of the number of red counters to blue counters is 1:4

Linda takes out, at random, 2 counters from the box.

The probability that she takes out 2 red counters is 6/155

How many red counters did John put into the box?

So, John has an empty box. He puts some red counters and some blue counters into the box.

The ratio of the number of red counters to blue counters is 1:4

Linda takes out, at random, 2 counters from the box.

The probability that she takes out 2 red counters is 6/155

How many red counters did John put into the box?

asked 2022-05-09

20 blue balls and 11 yellow, drawing 6 times with no replacement, what is the chance that at least one is yellow OR the first two draws are the same?

I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:

1) the first two are yellow

2) the first two are blue

I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?

Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?

I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:

1- [P(no yellow) or P(first two are different)]

I'm not sure my intuition for the solution of the problem is correct, in particular, I have some issue with the addition rule part.I started out this problem by thinking about the two scenarios:

1) the first two are yellow

2) the first two are blue

I'm a bit confused about the first scenario: P(at least 1 yellow) OR P(first two are the same) if the first two are yellow, that also satisfy the P(at least 1 yellow) so in this case does the probability equal to 1?

Is this similar to flipping a coin P(head) or P(tail)? Or perhaps it's only a "double counted intersection" that we have to subtract out?

I also thought about using compliment to solve this problem but I'm not quite sure if it works this way:

1- [P(no yellow) or P(first two are different)]