Question

********What is the probability of drawing an even card, not replacing it, and then drawing a odd card?************************************ (here are the details there are 5 cards. 3 blue 2 red.here are the cards in order our first card we have is red and has a 4 on it,the second card is blue and has a 5 on it,the third card is red and has a 6 on it.the fourth card is blue and has a 7 on it and the last card is blue and has an 8 on it)

Answer 1

How many even cards are there when you pick the first card?

Answer 2

color is irrelevant

you have 5 cards; 4,5,6,7,8 ; 3 are even, and 2 are odd

the probability of: P(even) and P(odd) are then

P(even)*P(odd|no replacement) ... of something like that

Answer 3

what you mean when i pick the 1st card? theres only going to be 1 even left lying down

Answer 4

How many are there before you pick the first card?

Answer 5

5

Answer 6

There are 5 cards, only 3 are even when you go to pick the first card. Right?

Answer 7

Are u using a std. deck of cards?

Answer 8

So the probability to draw an even card on the first pick is \(3 \over 5\).

Then once you pick that one, how many cards are there left? And how many of those are odd?

Answer 9

3/5 is not an option polpak!

Answer 10

oops i meant 2 are left and theyre both odd

Answer 11

Right there are 2 odds left and 4 cards left. So the odds of pulling an odd after an even is the probability of doing the one, times the probability of doing the other.

\[\frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}\]