What is the key word that indictes that the Addition Rule for Probabilities will be used?

Cohen Ritter
2022-10-20
Answered

What is the key word that indictes that the Addition Rule for Probabilities will be used?

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giosgi5

Answered 2022-10-21
Author has **15** answers

The key word for events with probabilities that can be added is mutually exclusive.

If two events, A and B, are mutually exclusive, the probability of either one or another to happen equal to a sum their probabilities (notice different symbols used to describe logical condition or:$P\left(A\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}B\right)=P(A+B)=P(A\cup B)=P\left(A\right)+P\left(B\right)$

An explanation and intuitive justification of this rule is a similarity of a concept of probability and a concept of an area of a flat figure on a plane. Mutually exclusive events are similar to two figures on a plane that do not have any points in common. Then it's obvious that the area of both figures (the probability of one event or another) equals to a sum of areas of these two figures (equals to a sum of probabilities of individual events).

If two events, A and B, are mutually exclusive, the probability of either one or another to happen equal to a sum their probabilities (notice different symbols used to describe logical condition or:$P\left(A\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}B\right)=P(A+B)=P(A\cup B)=P\left(A\right)+P\left(B\right)$

An explanation and intuitive justification of this rule is a similarity of a concept of probability and a concept of an area of a flat figure on a plane. Mutually exclusive events are similar to two figures on a plane that do not have any points in common. Then it's obvious that the area of both figures (the probability of one event or another) equals to a sum of areas of these two figures (equals to a sum of probabilities of individual events).

asked 2022-07-20

What is the addition rule for mutually exclusive events?

asked 2022-09-28

How do you add probabilities?

asked 2022-07-02

I am trying to improve my stat skills from a book that gives an exercise, I cannot get my head around.<br.It goes like this:<br.There are 5 green balls and 1 red ball in a box. We draw four times randomly with replacement. What is the probability that we draw at least two red balls?<br.My guess would be<br.(1/6)^2 + (1/6)^3 + (1/6)^4 because the probability of at least 2 red balls out of 4 draws must be equal to drawing two or three or four times a red ball. Hence I would use the addition rule, which gives 0.03 = 3 percent<br.However, the book says it is 13 percent, and gives the following explanation (150+20+1)/1296.<br.Why is this so?<br.My attempt to trace back the terms:<br.I can possibly see how it got the first and third term in the fraction: The first could be (5/6)^2 = 25/36, which is the probability of drawing two green balls; and the third could be (1/6)^4 = 1/1296, which is the probability of drawing four red balls. I have no idea though where they could possibly get the 20 from.

asked 2022-06-25

The general addition rule for non-mutually exclusive is

P(A OR B) = P(A) + P(B) - P(A AND B);

Now I know how to calculate P(A) and P(B), but how do you calculate P(A AND B)?

Example: What is the probability of rolling a 5 or odd number? To find P(A AND B) in this case is very simple, we know without calculations that it is 1, right? But how to find P(A AND B) when there is a lot of counting to do... like lets say there are 9999 slots which are numbered, some are black some are white, some are even some are odd. Now to find the probability of getting a black and even when a die is rolled upon the slots, how to find P(A AND B)? or we just have to count it just like that?

P(A OR B) = P(A) + P(B) - P(A AND B);

Now I know how to calculate P(A) and P(B), but how do you calculate P(A AND B)?

Example: What is the probability of rolling a 5 or odd number? To find P(A AND B) in this case is very simple, we know without calculations that it is 1, right? But how to find P(A AND B) when there is a lot of counting to do... like lets say there are 9999 slots which are numbered, some are black some are white, some are even some are odd. Now to find the probability of getting a black and even when a die is rolled upon the slots, how to find P(A AND B)? or we just have to count it just like that?

asked 2022-06-23

This is one of our probability exercise:

We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color?

(a) Assume that the balls are sampled without replacement.

(b) Assume that the balls are sampled with replacement.

(c) When is the answer to part (a) larger than the answer to part (b)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

For (a), I think there are only two possible outcomes of this experiment, which are either two green balls or two yellow balls. Since they are independent on each other, so we can use the addition rule. Since it is without replacement, so we can directly use combinations.

$P(\text{same color})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{m-1}{1}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{n-1}{1}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{m+n}{2}{\textstyle )}}$

The difference between replacement and no replacement is how the candidate pool changes after each action. With replacement, we don't need to modify the number of candidates anymore.

$P(\text{same color})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{m}{1}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}}{(m+n{)}^{2}}$

However, I have no clues how to solve (c). I tried to use algebra to simplify my answer for (a) and (b), but it doesn't help.

Any hints or suggestions would be appreciated!

We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color?

(a) Assume that the balls are sampled without replacement.

(b) Assume that the balls are sampled with replacement.

(c) When is the answer to part (a) larger than the answer to part (b)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

For (a), I think there are only two possible outcomes of this experiment, which are either two green balls or two yellow balls. Since they are independent on each other, so we can use the addition rule. Since it is without replacement, so we can directly use combinations.

$P(\text{same color})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{m-1}{1}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{n-1}{1}{\textstyle )}}{{\textstyle (}\genfrac{}{}{0ex}{}{m+n}{2}{\textstyle )}}$

The difference between replacement and no replacement is how the candidate pool changes after each action. With replacement, we don't need to modify the number of candidates anymore.

$P(\text{same color})=\frac{{\textstyle (}\genfrac{}{}{0ex}{}{m}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{m}{1}{\textstyle )}+{\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}\times {\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}}{(m+n{)}^{2}}$

However, I have no clues how to solve (c). I tried to use algebra to simplify my answer for (a) and (b), but it doesn't help.

Any hints or suggestions would be appreciated!

asked 2022-05-09

Alfonso and Colin each bought one raffle ticket at the state fair. If 50 tickets were randomly sold, what is the probability that Alfonso got ticket 14 and Colin got ticket 23?

The answer should be $\frac{1}{2450}$ which presumably comes from $\frac{1}{50}\times \frac{1}{49}$. But it seems that the order does not count. I did not assume that Alfonso got ticket 14 first then Colin got ticket 23 second.

Update: What is wrong with this reasoning. When I said that I did not assume order, I meant that it's possible

1. Alfonso got ticket 14 first, then Colin got ticket 23,

2. Colin got ticket 23 first, then Alfonso got ticket 14.

Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement. Label the event Alfonso got ticket 14 by ${A}_{14}$ and Colin got ticket 23 by ${A}_{23}$. Then by the addition rule

$Pr(\text{(}{A}_{14}\text{first and}{C}_{23}\text{second) or (}{C}_{23}\text{first and}{A}_{14}\text{second}))\phantom{\rule{0ex}{0ex}}=Pr({A}_{14})\times Pr({C}_{23}\mid {A}_{14})+Pr({C}_{23})\times Pr({A}_{14}\mid {C}_{23})=\frac{1}{50}\times \frac{1}{49}\times 2.$

I realize that once the tickets are sold, then only one of $\{{A}_{14}{C}_{23}\text{},\text{}{C}_{23}{A}_{14}\}$ must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold.

The answer should be $\frac{1}{2450}$ which presumably comes from $\frac{1}{50}\times \frac{1}{49}$. But it seems that the order does not count. I did not assume that Alfonso got ticket 14 first then Colin got ticket 23 second.

Update: What is wrong with this reasoning. When I said that I did not assume order, I meant that it's possible

1. Alfonso got ticket 14 first, then Colin got ticket 23,

2. Colin got ticket 23 first, then Alfonso got ticket 14.

Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement. Label the event Alfonso got ticket 14 by ${A}_{14}$ and Colin got ticket 23 by ${A}_{23}$. Then by the addition rule

$Pr(\text{(}{A}_{14}\text{first and}{C}_{23}\text{second) or (}{C}_{23}\text{first and}{A}_{14}\text{second}))\phantom{\rule{0ex}{0ex}}=Pr({A}_{14})\times Pr({C}_{23}\mid {A}_{14})+Pr({C}_{23})\times Pr({A}_{14}\mid {C}_{23})=\frac{1}{50}\times \frac{1}{49}\times 2.$

I realize that once the tickets are sold, then only one of $\{{A}_{14}{C}_{23}\text{},\text{}{C}_{23}{A}_{14}\}$ must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold.

asked 2022-10-03

What is another name for mutually exclusive events?