lasonya39

04-15-2008, 12:32 AM

Please help! I'm so lost on obtaining these answers....

1. If three cards are drawn without replacement from an ordinary deck, find the probability that the third card is a heart, given that the first two cards were hearts.

2. Two fair dice are rolled. The sum of the numbers on the dice is 6 or 9.

3. Assume that 2 marbles are drawn without replacement from a box with 1 blue, 3 white, 2 green, and 2 red marbles. Find the probability of the indicated result:

a. The second marble is white, given that the first marble is blue.

b. The second marble is blue, given that the first marble is white.

If you could explain, then I can answer my remaining 30 questions.

Thanks in advance.

tkhunny

04-15-2008, 09:30 AM

"If you could explain, then I can answer my remaining 30 questions."

I'm not sure I agree with this conclusion. These are very simple counting problems. If you are TRULY struggling with them, I suspect the other 30 will not be your friends.

Deck of Cards: 52 Total 13 Hearts

Pull out 1 Heart: 51 Total 12 Hearts

Pull out another Heart: 50 Total 11 Hearts

Now answer Question #1.

For #2, There are only 36 possible outcomes. List them and count the ones that add up to 6 or 9. Then answer the question.

Let's see what you get on #3.

soroban

04-15-2008, 10:14 AM

Hello, lasonya39!

These don't require any fancy formulas . . . just a little Thinking.

1. If three cards are drawn without replacement from an ordinary deck,

find the probability that the third card is a heart, given that the first two cards were hearts.

If the first two cards were Hearts, there are 11 Hearts among the remaining 50 cards.

\text{The probability is: }\:\frac{11}{50}

2. Two fair dice are rolled. The sum of the numbers on the dice is 6 or 9.

With dice problems, there are no neat formulas.

We must visualize (or write out) the possible outcomes.

\text{There are: }\:6 \times 6 \:=\:36\text{ possible outcomes.}

\text{Sum of 6: }\;(1,5),\:(2,4),\:(3,3),\:(4,2),\:(5,1)\quad\hdot s\text{ 5 ways}

\text{Sum of 9: }\;(3,6),\:(4,5),\:(5,4),\:(6,3)\quad\hdots \text{ 4 ways}

\text{Therefore: }\;P(\text{6 or 9}) \;=\;\frac{9}{36} \;=\;\frac{1}{4}

3. Assume that 2 marbles are drawn without replacement from a box

with 1 blue, 3 white, 2 green, and 2 red marbles.

Find the probability of the indicated result:

a. The second marble is white, given that the first marble is blue.

If the first marble was blue, there are 7 marbles left: 3 white, 2 green, 2 red.

\text{Therefore: }\:P(\text{2nd is white}) \;=\;\frac{3}{7}

b. The second marble is blue, given that the first marble is white.

If the first marble was white, there are 7 marbles left: 1 blue, 2 white, 2 green, 2 red.

\text{Therefore: }\:P(\text{2nd is blue}) \;=\;\frac{1}{7}

tkhunny

04-15-2008, 04:28 PM

. . . just a little Thinking.[/size]

...and we'll never know if you did any thinking unless you show us YOUR work. Let's see whatever is next in your problem set. Show us what you get and how you got it.

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