MHB Find the probability that E' occurs at least once

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To find the probability that event E' occurs at least once, the complement rule is applied, which states that P(B) = 1 - P(A), where P(A) is the probability that E' never occurs. The calculated probability of E' never occurring is 25/81, leading to P(B) being 56/81. This method is more efficient than directly calculating the probability by analyzing multiple paths in the tree diagram. The discussion emphasizes the importance of understanding the symbols used in probability equations.
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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 :)
 

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mathlearn said:
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)
 
I like Serena said:
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?
 
mathlearn said:
$\frac{5}{9}*\frac{5}{9}=\frac{25}{81}$

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:

$$P(A)+P(B)=1$$

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

$$P(B)=1-P(A)$$

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
 
MarkFL said:
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:

$$P(A)+P(B)=1$$

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

$$P(B)=1-P(A)$$

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 ? :rolleyes:
 
mathlearn said:
$\displaystyle P(B)=1-P(A)$
$\displaystyle P(B)=1- \frac{25}{81}$
$\displaystyle P(B)=1- \frac{25}{81}$
$\displaystyle P(B)=\frac{81}{81}- \frac{25}{81}$
$\displaystyle P(B)=\frac{81}{81}- \frac{25}{81}$
$\displaystyle P(B)=\frac{56}{81}$

Correct ? :rolleyes:

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

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

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.
 
(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?
 
mathlearn said:
(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)
 
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