Math Is Fun Forum

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Box of Sports Cards

A box has 4 baseball cards and 5 football cards.
Two cards are randomly selected without replacement.
What is the probability of selecting one baseball card and one football card?

Let me see.

Let A = probability of selecting baseball card.

Let B = probability of selecting football card.

This is a without replacement event.

I think P(A and B) = P(A) • P(B|A) is needed.

P(A) = 4/9.

P(B|A) = 4/8 = 1/2.

P(A) • P(B|A) = (4/9)(1/2) = 4/18 = 2/9.

Is this right?

Wait! I don't think the problem ends here.

What if the first card selected is a football card and then a baseball card?

You say?

Last edited by mathxyz (2024-05-11 17:34:56)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Box of Sports Cards

Yes, your second thoughts are correct.  Work out the probability of football then baseball and add the two answers together.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Box of Sports Cards

Yes, your second thoughts are correct.  Work out the probability of football then baseball and add the two answers together.

Bob

Ok. Sounds good. I will continue with the rest of this problem as you suggested later today.