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heidi18
05-30-2008, 07:46 PM
I am auditing a college class and I did not understand these problem by the teacher. Go easy this is my first go around with this stuff.

A bag contains five white, three black, and four red balls. A single drawing of three balls is made. Find the probability that the three balls will all be of a different color?

Is it (5/12) times (4/11) times (3/10)??????

The other one that did not make sense was...

From a deck of 52 cards, three aces have been drawn one at a time in three consecutive draws. What is the probability of the fourth ace being drawn on the next draw, and what is the probability of these four aces being drawn as described?

Is the second part of that question (4/52)(3/51)(2/50)(1/49)? I am not sure of the first part of the question, where do I start?

I do not want to be a bother, ill be greatful to anyone who can make me better and smarter and help me with this!

galactus
05-30-2008, 08:24 PM
A bag contains five white, three black, and four red balls. A single drawing of three balls is made. Find the probability that the three balls will all be of a different color?

Is it (5/12) times (4/11) times (3/10)??????

You are on the right track, except you need to multiply by 6. That's because they can come out in 3!=6 different ways. For instance, say they came out BWR, RWB, WRB, etc. 6(5/12)(4/11)(3/10).

Another way to do this is using combinations. \frac{5\cdot{4}\cdot{3}}{C(12,3)}. Because we are choosing 3 from 12 and and there are 5 whites to choose, 4 reds, and 3 blacks.

The other one that did not make sense was...


From a deck of 52 cards, three aces have been drawn one at a time in three consecutive draws. What is the probability of the fourth ace being drawn on the next draw, and what is the probability of these four aces being drawn as described?

Is the second part of that question (4/52)(3/51)(2/50)(1/49)? I am not sure of the first part of the question, where do I start?

I do not want to be a bother, ill be grateful to anyone who can make me better and smarter and help me with this!

You are correct. The last Ace will have a probability of 1/49 of being drawn. But remember, they can be drawn in 4!=24 different ways because there are 4 different Aces. So multiply by 24. 24(4/52)(3/51)(2/50)(1/49)=

heidi18
05-30-2008, 09:05 PM
I think I am starting to understand...

A box holds 4 apples, 5 oranges, and 8 pears. How many ways can 2 apples, 1 orange, and 2 pears be chosen?

so I would have multipication because of the word AND, (4/17)(3/16)(5/15)(2/14)(1/13)=

...except I probably need to multiply this by all the ways they can come out of the box, right???

that seems like a lot of ways, do I have to list all the of them to find out? Like AAOPP AOPPA...???

galactus
05-30-2008, 09:17 PM
You are on the right track. Seems you're catching on and you are correct in that you have to multiply by the number of arrangements. 5!/(2!1!2!)=30

There are 5 items you are choosing, but there are 2 the same, 1 , and 2 the same. That is called distinguishable permutations.

30/6188=30/221.

Here is another way. You get the same answer.

C(4,2)C(5,1)(C(8,2)=840

Because we are choosing 2 from 4, 1 from 5, 2 from 8.

Now, if you wanted the probability of choosing what you stated, then divide by C(17,5)

Because we are choosing 5 from 17 total.

\frac{C(4,2)C(5,1)(C(8,2)}{C(17,5)}=\frac{30}{221}

heidi18
05-30-2008, 09:39 PM
I dont mean to bother you, but where did you get the 30/6188=30/221? Particularly the 30/6188? sorry

galactus
05-30-2008, 10:08 PM
That is the calculation. You got the 1/6188 yourself. I just multiplied by 30.