help plz math probability, mutually exclusive

[Guest]

I am taking correspondance and I am stuck on this question. I understand simple probability questions, but this is confusing to me. Some one plz help ...

A single card is drawn from a standard deck. Let A and B represent the following events:
A:"A red card is drawn"
B:"A jack is drawn"

a. Are these events mutually exclusive? Why or Why not?

b. Determine P(A and B).

c. Determine P(And or B).

 

[soroban]

Hello, Amy-Marie!

A single card is drawn from a standard deck.
Let A and B represent the following events:
\(\displaystyle \;\;\)A: "a red card is drawn"
\(\displaystyle \;\;\)B: "a Jack is drawn"

a. Are these events mutually exclusive? Why or Why not?

"Mutually exclusive" means the events cannot happen at the same time.

We <u>can</u> draw a card which is both a red card and a Jack (J♥ or J◊)
\(\displaystyle \;\;\)They are not mutually exclusive.

b. Determine P(A and B).

There are 52 possible outcomes.
There are 2 cards which are both red and a Jack.

\(\displaystyle \;\;P(\text{red }\cap\text{ Jack})\;=\;\frac{2}{52}\;=\;\frac{1}{26}\)

c. Determine P(A or B).

There are 26 red cards (including two red Jacks) . . . and 2 black Jacks.
\(\displaystyle \;\;\)Hence, there are 30 ways get a red card or a Jack.

\(\displaystyle \;\;{(\text{red }\cup\text{ Jac})\;=\;\frac{30}{52}\;=\;\frac{15}{26}\)

 

[pka]

There are 26 red cards (including two red Jacks) . . . and 2 black Jacks.
Hence, there are 30 ways get a red card or a Jack.

There is an overcounting error above.
Actually, there are only 28 ways get a red card or a Jack.

The correct answer to part c is;
\(\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{26}}{{52}} + \frac{4}{{52}} - \frac{2}{{52}} = \frac{7}{{13}}\)

 

[Guest]

There are 26 red cards (including two red Jacks) . . . and 2 black Jacks.
Hence, there are 30 ways get a red card or a Jack.

There is an overcounting error above.
Actually, there are only 28 ways get a red card or a Jack.

The correct answer to part c is;
\(\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{26}}{{52}} + \frac{4}{{52}} - \frac{2}{{52}} = \frac{7}{{13}}\)

 

[Guest]

P(A and B)= P(A)*P(B) P(A or B)= P(A)+P(B)

Thank you Pka and soroban, but the formula for P(A and B) = P(A)*P(B).
=26/52 * 4/52
=15/1352????

How did you get the fraction 2/52? I'm confused

 

[pka]

You gave two events:
A:"A red card is drawn"
B:"A jack is drawn"

P(A)=26/52 , P(B)=4/52 & P(AB)=2/52: there are 26 red cards and there are 4 Jacks.
But 2 of the red cards are also Jacks. There are 28 cards that are red or Jacks.
It is every common error as you can see to overcount.

If each of X & Y is a finite set the number of elements in their union is:
\(\displaystyle |X \cup Y| = |X| + |Y| - |X \cap Y|\), the number in X plus the number in Y minus the number in both.
We subtract the number to eliminate the problem of overcounting.

Thus we get the well formula for the probability for OR:
\(\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

 

[pka]

Re: P(A and B)= P(A)*P(B) P(A or B)= P(A)+P(B)

Thank you Pka and soroban, but the formula for P(A and B) = P(A)*P(B).
THIS IS INCORRET!!

That is for independent events.
Events A and B are NOT INDEPENDENT.
P(A and B)=2/52.

 

[Guest]

independent and dependent events

Got it, thanx a bunch!

 

[Guest]

dependent probability

Can you give me the formula for dependent P(A and B )

 

[Molly]

Alright, Im going to take a stab at this.. first, these events are mutually inclusive beacuse they can happen at the same time...for example pulling out a jack of hearts. The P( A and B) would be how likley is it to pull out both a red card and a jack ( red-13/52, and jacks 12/52) but there are only 2 ways for them to happen at the same time..." and". As for P ( A or B), you add the total outcome of A and B toeather, which I think comes out to 23/52, becuse you subtract the probability of getting them both at the same time, which is two, as I mentioned up above!!!! Good Luck.....smiles :lol: