Union/Intersection of fractions or percentages

[tmt1]

In a few exercises, I've needed to get the union or intersection of fractions or percentages.

I'm not confident I fully understand.

If I need the intersection of $A = 0.75$ and $ B = 0.5$, do I just take the larger number and subtract the smaller number? So the $ A \cap B$ would be $0.25$? And the union would just be the larger number i.e. $A$ or $0.75$?

And I suppose the same must be true for fractions.

Is this correct? Seems too easy.

 

[I like Serena]

In a few exercises, I've needed to get the union or intersection of fractions or percentages.

I'm not confident I fully understand.

If I need the intersection of $A = 0.75$ and $ B = 0.5$, do I just take the larger number and subtract the smaller number? So the $ A \cap B$ would be $0.25$? And the union would just be the larger number i.e. $A$ or $0.75$?

And I suppose the same must be true for fractions.

Is this correct? Seems too easy.

Hi tmt!

Since you've posted in probability and statistics, are we talking about the intersection and union of chance events?
And the resulting probabilities? (Wondering)

If so, then the product rule states that if events $A$ and $B$ are independent, that then $P(A\cap B)=P(A)P(B)$.
So it would be the product, not the subtraction. If they're not independent, it becomes more complicated.

And the sum rule states that if events $A$ and $B$ are mutually exclusive, that then $P(A\cup B)=P(A)+P(B)$.
So it would be the sum, and not the larger number.
However, we can already see in this case that the sum would be greater than $1$, which is not possible for probabilities, so $A$ and $B$ are definitely not mutually exclusive. In that case we need the general sum rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

 

[tmt1]

Hi tmt!

Since you've posted in probability and statistics, are we talking about the intersection and union of chance events?
And the resulting probabilities? (Wondering)

If so, then the product rule states that if events $A$ and $B$ are independent, that then $P(A\cap B)=P(A)P(B)$.
So it would be the product, not the subtraction. If they're not independent, it becomes more complicated.

And the sum rule states that if events $A$ and $B$ are mutually exclusive, that then $P(A\cup B)=P(A)+P(B)$.
So it would be the sum, and not the larger number.
However, we can already see in this case that the sum would be greater than $1$, which is not possible for probabilities, so $A$ and $B$ are definitely not mutually exclusive. In that case we need the general sum rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

Thanks that is very helpful.

What if it's something thinking more like if $P(A) = 0.75$ and $P(B) = 0.5 $, what is $P(A | B)$?

I need to do $\frac{P(A\cap B)}{P(B)}$ which is $\frac{P(A\cap B)}{0.5}$, so what is $P(A\cap B)$? Is it simply the larger number minus the smaller number so $0.25$? Or would it be $P(A) + P(B) - 1$ which is also $0.25$?

 

[I like Serena]

What if it's something thinking more like if $P(A) = 0.75$ and $P(B) = 0.5 $, what is $P(A | B)$?

That is unknown.
If A and B are independent, then $P(A \mid B) = P(A)$.
If they are dependent, then $P(A \mid B)$ can be anything between $0$ and $1$.

I need to do $\frac{P(A\cap B)}{P(B)}$ which is $\frac{P(A\cap B)}{0.5}$, so what is $P(A\cap B)$? Is it simply the larger number minus the smaller number so $0.25$? Or would it be $P(A) + P(B) - 1$ which is also $0.25$?

If A and B are dependent we have insufficient information to find $P(A\cap B)$.
If they are independent we have $P(A\cap B)=P(A)P(B)$.

 

[tmt1]

That is unknown.
If A and B are independent, then $P(A \mid B) = P(A)$.
If they are dependent, then $P(A \mid B)$ can be anything between $0$ and $P(A)$.

If A and B are dependent we have insufficient information to find $P(A\cap B)$.
If they are independent we have $P(A\cap B)=P(A)P(B)$.

Ok, I have 2 examples.

I have a fair coin and a two-headed coin. I choose one of them with equal probability and I flip it. Given that I flipped a heads, what is the probability that I chose the two-headed coin?

Let the event of flipping heads be H. So $P(H) = \frac{3}{4} $ OR $P(H) = 0.75$.

Let the event of picking a coin with 2 heads be 2-HC. So $P(2-HC) = 0.5$. To solve this question I need to find

$$\frac{P(H \cap 2-HC)}{P(H)}$$

OR

$$\frac{P(H \cap 2-HC)}{0.75}$$

So I need to find $P(H \cap 2-HC)$. So how would I find the intersection of these 2? Just take the smaller of 2 numbers, so 0.5?

Then I would get $0.5/ 0.75$ which is $2/3$ which seems to be correct.

The other example I have is:

80% of the customers at a pub order drinks and 35% order food. Assuming that there are no customers who order neither food nor drinks, what is the probability that a customer who orders drinks will also order food?

So let the event of the customers ordering drinks be D. Therefore, $P(D) = 0.8$. Let the event of customers ordering food be F. So, $S(F) = 0.35$.

I need to find

$$\frac{P(F \cap D)}{P(D)}$$ Or

$$\frac{P(F \cap D)}{0.8}$$

So what is $P(F \cap D)$?

The smaller of the 2 numbers? That would be $0.35$ which is incorrect.

I suppose it could be $F + D - 1$ or $0.15$ and this would yield the correct answer.

But is I try this formula in my first example, $P(H \cap 2-HC)$ would equal $0.25$ which would yield the wrong answer.

Furthermore, what about 2 probabilities that don't equal more than 1, than this formula ($A + B -1 $) wouldn't work at all.

 

[I like Serena]

So I need to find $P(H \cap 2-HC)$. So how would I find the intersection of these 2? Just take the smaller of 2 numbers, so 0.5?

Then I would get $0.5/ 0.75$ which is $2/3$ which seems to be correct.

Not quite.
We have 4 possible outcomes that are equally likely: (H, fair), (T, fair), (H1, 2-HC), (H2, 2-HC).
Only 2 of them match with $(H \cap 2-HC)$.
So the probability is:
$$P(H \cap 2-HC) = \frac{\text{# favorable}}{\text{# total}}= \frac 24=0.5$$
As a result we get indeed that:
$$P(2-HC \mid H)=\frac{0.5}{0.75}=\frac 23$$

The other example I have is:

80% of the customers at a pub order drinks and 35% order food. Assuming that there are no customers who order neither food nor drinks, what is the probability that a customer who orders drinks will also order food?

So let the event of the customers ordering drinks be D. Therefore, $P(D) = 0.8$. Let the event of customers ordering food be F. So, $S(F) = 0.35$.

I need to find

$$\frac{P(F \cap D)}{P(D)}$$

Let's see, 80% order drinks, and everyone has to order something.
That means that 20% order food, but not a drink.
Consequently 15% orders both drink and food.

It looks like this:
[TIKZ]
\filldraw[gray!20!white]
(-3,2) node[below right] {0\%} rectangle (3,-2);
\begin{scope}[blend group=soft light]
\filldraw[orange!50!white]
(-1,0) circle (1.5);
\filldraw[olive!40!white]
(1,0) circle (1.5);
\end{scope}
\draw
(-3,2) node[below right] {0\%} rectangle (3,-2);
\draw
(-1,0) node

{65\%}
(-1,1.5) node {80\%};
\draw
(1,0) node

{20\%}
(1,1.5) node {35\%};
\draw
node at (0,0) {15\%};
[/TIKZ]​

 

[tmt1]

ok thanks