What is another name for mutually exclusive events?

Kassandra Mccall
2022-10-03
Answered

What is another name for mutually exclusive events?

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omeopata25

Answered 2022-10-04
Author has **5** answers

Another name for mutually exclusive events would be disjoint events. Note that collectively exhaustive is not equivalent to mutual exclusivity, as if a set of events are collectively exhaustive, at least one of the events must occur.

As an example, suppose that you roll a fair six sided die once. The event that you roll a 1 is mutually exclusive with the event that you roll a 6; that is, you cannot roll both a 1 and a 6 by rolling the die a single time. However, there exists a possibility that you will roll neither a 1 nor a 6; ergo, the possibilities are not collectively exhaustive.

(As a note, if you consider six events A-F, where A is the event that you roll a 1 on a single roll of the six sided die, B is the event that you roll a 2, C the event that you roll a 3, D the event you roll a 4, E the event you roll a 5, and F the event you roll a 6, then the set of events A-F are both mutually exclusive and collectively exhaustive; you must roll one of the numbers on the die when you roll the die.)

As an example, suppose that you roll a fair six sided die once. The event that you roll a 1 is mutually exclusive with the event that you roll a 6; that is, you cannot roll both a 1 and a 6 by rolling the die a single time. However, there exists a possibility that you will roll neither a 1 nor a 6; ergo, the possibilities are not collectively exhaustive.

(As a note, if you consider six events A-F, where A is the event that you roll a 1 on a single roll of the six sided die, B is the event that you roll a 2, C the event that you roll a 3, D the event you roll a 4, E the event you roll a 5, and F the event you roll a 6, then the set of events A-F are both mutually exclusive and collectively exhaustive; you must roll one of the numbers on the die when you roll the die.)

asked 2022-05-28

In Texas holdem, one is dealt a Decent Hand (Any pocket pair or any two broadway cards) $\sim $15 percent of the time. If there are three people left in the hand, I can use the probability addition rule to say at least one of those three people left will show up with a Decent Hand $\sim $45 percent of the time, correct? What about when there are 10 people left? Does someone show up with a Decent Hand $\sim $150% of the time?

Edit: Why does the addition rule not suited for this case?

Edit: Why does the addition rule not suited for this case?

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)]

asked 2022-07-09

A game is played by rolling a six sided die which has four red faces and two blue faces. One turn consists of throwing the die repeatedly until a blue face is on top or the die has been thrown 4 times

Adnan and Beryl each have one turn. Find the probability that Adnan throws the die more turns than Beryl

I tried : Adnan throws two times and Beryl throws once = $\frac{2}{3}$ x $\frac{1}{3}$

Adnan throws three times and Beryl throws once = $\frac{4}{9}$ x $\frac{1}{2}$

Adnan throws three times and Beryl throws twice = $\frac{4}{9}$ x $\frac{2}{3}$

Adnan throws four times and Beryl throws once = $\frac{8}{27}$ x $\frac{1}{2}$

Adnan throws four times and Beryl throws twice = $\frac{8}{27}$ x $\frac{2}{3}$

Adnan throws four times and Beryl throws three times = $\frac{8}{27}$ x $\frac{4}{9}$

The answer says 0.365

Adnan and Beryl each have one turn. Find the probability that Adnan throws the die more turns than Beryl

I tried : Adnan throws two times and Beryl throws once = $\frac{2}{3}$ x $\frac{1}{3}$

Adnan throws three times and Beryl throws once = $\frac{4}{9}$ x $\frac{1}{2}$

Adnan throws three times and Beryl throws twice = $\frac{4}{9}$ x $\frac{2}{3}$

Adnan throws four times and Beryl throws once = $\frac{8}{27}$ x $\frac{1}{2}$

Adnan throws four times and Beryl throws twice = $\frac{8}{27}$ x $\frac{2}{3}$

Adnan throws four times and Beryl throws three times = $\frac{8}{27}$ x $\frac{4}{9}$

The answer says 0.365

asked 2022-07-15

What is the difference between events that are mutually exclusive and those that are not mutually exclusive?

asked 2022-06-26

If I toss a coin 3 times and want to know the probability of at least one head, I have understood that the answer is $1-{0.5}^{3}=99\mathrm{\%}$. However, why cannot I not use the additon rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, i.e. $0.5+0.5+0.5-{0.5}^{3}$?

asked 2022-06-09

Came across this,

$\varphi =$ probability of event

$p(y,\varphi )={\varphi}^{y}(1-\varphi {)}^{(1-y)}=Exp(\mathrm{log}({\varphi}^{y}(1-\varphi {)}^{(1-y)}))$ and now, somehow $Exp(\mathrm{log}({\varphi}^{y}(1-\varphi {)}^{(1-y)}))=Exp[\mathrm{log}(\frac{\varphi}{1-\varphi})y+\mathrm{log}(1-\varphi )]$. I looked to use the power rule for logs to bring down the $y(1-y)$ in front, the product rule for logs to get an addition to the equation, and the subtraction rule for logs to get the division part of the expression. Upon simplifying, I don't yield the desired expression

My attempt is shown below:

$\begin{array}{rl}Exp(\mathrm{log}({\varphi}^{y}(1-\varphi {)}^{(1-y)}))& =Exp[(1-y)\mathrm{log}({\varphi}^{y}(1-\varphi ))]\\ & =Exp[(1-y)\mathrm{log}({\varphi}^{y})+\mathrm{log}(1-\varphi )]\\ & =Exp[(1-y)y\mathrm{log}(\varphi )+\mathrm{log}(1-\varphi )]\end{array}$

So, now I'm not sure if this can arrive at the right answer

$\varphi =$ probability of event

$p(y,\varphi )={\varphi}^{y}(1-\varphi {)}^{(1-y)}=Exp(\mathrm{log}({\varphi}^{y}(1-\varphi {)}^{(1-y)}))$ and now, somehow $Exp(\mathrm{log}({\varphi}^{y}(1-\varphi {)}^{(1-y)}))=Exp[\mathrm{log}(\frac{\varphi}{1-\varphi})y+\mathrm{log}(1-\varphi )]$. I looked to use the power rule for logs to bring down the $y(1-y)$ in front, the product rule for logs to get an addition to the equation, and the subtraction rule for logs to get the division part of the expression. Upon simplifying, I don't yield the desired expression

My attempt is shown below:

$\begin{array}{rl}Exp(\mathrm{log}({\varphi}^{y}(1-\varphi {)}^{(1-y)}))& =Exp[(1-y)\mathrm{log}({\varphi}^{y}(1-\varphi ))]\\ & =Exp[(1-y)\mathrm{log}({\varphi}^{y})+\mathrm{log}(1-\varphi )]\\ & =Exp[(1-y)y\mathrm{log}(\varphi )+\mathrm{log}(1-\varphi )]\end{array}$

So, now I'm not sure if this can arrive at the right answer

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.