How do you add probabilities?

pramrok62
2022-09-28
Answered

How do you add probabilities?

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oldgaffer1b

Answered 2022-09-29
Author has **9** answers

We can add two probabilities as long as the events are disjoint.

Explanation:

Suppose A and B are two events.

$P\left(A\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}B\phantom{\rule{1ex}{0ex}}\text{occurs}\right)$=$P(A\cup B)=P\left(A\right)+P\left(B\right)+P(A\cap B)$

Here $P(A\cap B)$ is the probability that the two events occur simultaneously. If these events are disjoint, $P(A\cap B)=0$ and in that case we can add the probabilities.

For example: Suppose events A and B are disjoint, $P\left(A\right)=0.1,P\left(B\right)=0.2$, then $P(A\cup B)=0.1+0.2=0.3$

Explanation:

Suppose A and B are two events.

$P\left(A\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}B\phantom{\rule{1ex}{0ex}}\text{occurs}\right)$=$P(A\cup B)=P\left(A\right)+P\left(B\right)+P(A\cap B)$

Here $P(A\cap B)$ is the probability that the two events occur simultaneously. If these events are disjoint, $P(A\cap B)=0$ and in that case we can add the probabilities.

For example: Suppose events A and B are disjoint, $P\left(A\right)=0.1,P\left(B\right)=0.2$, then $P(A\cup B)=0.1+0.2=0.3$

asked 2022-07-15

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

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?

asked 2022-08-16

What is the probability that, in a single draw from a standard deck of cards, we will get either a Jack or a Diamond?

asked 2022-07-09

If the probability that Joe will buy a pizza is 0.5 and the probability that Elaine will buy a pizza is 0.35, then what is the probability that at least one of the two will buy a pizza on their next visit to the pizza place?

I know that this is a very basic problem. However, I am a little confused about the "at least" statement in the question. I would really like to understand why the correct answer is correct, so please go in depth for how you got your answer. I am thinking that you might use the addition rule for probability, but again, the "at least" confuses me.

I know that this is a very basic problem. However, I am a little confused about the "at least" statement in the question. I would really like to understand why the correct answer is correct, so please go in depth for how you got your answer. I am thinking that you might use the addition rule for probability, but again, the "at least" confuses me.

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 2022-07-09

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