Find the probability that E' occurs at least once

[mathlearn]

Data

Below are the possible observations of a random experiment in a tree diagram.Each time it is observed whether the event E defined in part i occurs or not.

View attachment 5990Where do I need help

Find the probability that E' occurs at least once

Many Thanks :)

 

[I like Serena]

Data

Below are the possible observations of a random experiment in a tree diagram.Each time it is observed whether the event E defined in part i occurs or not.

Where do I need help

Find the probability that E' occurs at least once

Many Thanks :)

Hey mathlearn!

The probability that E' occurs at least once is 1 minus the probability that E' never occurs.
What is the probability hat E' never occurs? (Wondering)

 

[mathlearn]

Hey mathlearn!

The probability that E' occurs at least once is 1 minus the probability that E' never occurs.
What is the probability hat E' never occurs? (Wondering)

$\frac{5}{9}*\frac{5}{9}=\frac{25}{81}$

Now to proceed? Subtract 1 from both sides?

 

[MarkFL]

$\frac{5}{9}*\frac{5}{9}=\frac{25}{81}$

Now to proceed? Subtract 1 from both sides?

Here's a way to look at why ILS pointed you to find the probability that E' never occurs, that is, to use the complement:

We know it is certain that either E' never occurs (we'll call that event A) OR E' occurs at least once (we'll call that event B and this is what we are trying to find). We may state this mathematically as:

\(\displaystyle P(A)+P(B)=1\)

Since we are trying to find $P(B)$, let's arrange this as:

\(\displaystyle P(B)=1-P(A)\)

You've already found $P(A)$, so plug that in and then simplify to get $P(B)$.

Sometimes it's simpler and more efficient to find the complement, and use the complement rule, instead of computing the required probability directly. :D

 

[mathlearn]

Here's a way to look at why ILS pointed you to find the probability that E' never occurs, that is, to use the complement:

We know it is certain that either E' never occurs (we'll call that event A) OR E' occurs at least once (we'll call that event B and this is what we are trying to find). We may state this mathematically as:

\(\displaystyle P(A)+P(B)=1\)

Since we are trying to find $P(B)$, let's arrange this as:

\(\displaystyle P(B)=1-P(A)\)

You've already found $P(A)$, so plug that in and then simplify to get $P(B)$.

Sometimes it's simpler and more efficient to find the complement, and use the complement rule, instead of computing the required probability directly. :D

$\displaystyle P(B)=1-P(A)$
$\displaystyle P(B)=1- \frac{25}{81}$
$\displaystyle P(B)=1- \frac{25}{81}$
$\displaystyle P(B)=\frac{81}{81}- \frac{25}{81}$
$\displaystyle P(B)=\frac{81}{81}- \frac{25}{81}$
$\displaystyle P(B)=\frac{56}{81}$

Correct ?

 

[MarkFL]

$\displaystyle P(B)=1-P(A)$
$\displaystyle P(B)=1- \frac{25}{81}$
$\displaystyle P(B)=1- \frac{25}{81}$
$\displaystyle P(B)=\frac{81}{81}- \frac{25}{81}$
$\displaystyle P(B)=\frac{81}{81}- \frac{25}{81}$
$\displaystyle P(B)=\frac{56}{81}$

Correct ?

Well, let's check your answer by computing the probability directly...

\(\displaystyle P(B)=\frac{5}{9}\cdot\frac{4}{9}+\frac{4}{9}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{4}{9}=\frac{20+20+16}{81}=\frac{56}{81}\quad\checkmark\)

You see, by using the complements rule, you only had to analyze one path in the tree diagram, whereas when I computed the probability directly, I had to analyze three paths.

 

[mathlearn]

(Party)(Party)(Happy) Perfectly explained! Thank you very much MarkFL & I Like Serena.

Can $\displaystyle P(A)+P(B)=1$ be replaced with $\displaystyle P(E)+P(E')=1$ if necessary?

 

[MarkFL]

(Party)(Party)(Happy) Perfectly explained! Thank you very much MarkFL & I Like Serena.

Can $\displaystyle P(A)+P(B)=1$ be replaced with $\displaystyle P(E)+P(E')=1$ if necessary?

As long as you know what your symbols stand for, then what you choose is fine. (Yes)