A grab bag contains 7 football cards and 3 basketball cards. An experiment consists of taking one card out of the bag, then selecting another card.What is the probability of selecting a football card and then a basketball card if: (a) the first card is not replaced?(b) the first card is replaced?Express your answers in decimal form. If necessary, round answers to the nearest hundredth.

a. (a) 0.23(b) 0.21
b. (a) 0.09(b) 0.49
c. (a) 0.21(b) 0.23
d. (a) 0.49(b) 0.09

Question

A grab bag contains 7 football cards and 3 basketball cards. An experiment consists of taking one card out of the bag, then selecting another card.What is the probability of selecting a football card and then a basketball card if: (a) the first card is not replaced?(b) the first card is replaced?Express your answers in decimal form. If necessary, round answers to the nearest hundredth.

a. (a) 0.23(b) 0.21
b. (a) 0.09(b) 0.49
c. (a) 0.21(b) 0.23
d. (a) 0.49(b) 0.09

terenzreignz · Answer

Let's see.. what's the probability of just selecting a football card on the first go?

anonymous · Answer

I think it's 7/10.

terenzreignz · Answer

That's right.  :)
And once a football card is picked, what's the probability of selecting a basketball card on the second go?

anonymous · Answer

I think that once a football card is picked the probability is 3/9. =)

terenzreignz · Answer

Yes, or 1/3
Now, just multiply them together, and you'd get the answer to (a). :)

anonymous · Answer

Thank you, that made it soo much easier.

terenzreignz · Answer

Now, as for (b) (which may not be necessary, as only one of the choices has the correct answer to [a])

Picking a football card on the first go is still 7/10, but picking a basketball card on the second go is now 3/10, since the first card will be returned to the pool...

so...7/10 * 3/10 = 21 / 100 = 0.21

:)

anonymous · Answer

Thank you very much!! I'm always grateful for the help that I receive, you're always so helpful!

terenzreignz · Answer

^_^