Damn probability! Answer this question please?

reznov

The new project Maths book offers little information in solving probability problems. My teacher has one answer - book, another.
Solve this question please? It's not for homework.

A bag contains eight black, three white and five red beads. Three beads are picked at random without replacement. Find probability that one is white, the and the others are not.

I assume it is: 1 - P(none white)?
Teacher's answer: 557/560
Book's answer: 117/280

>_<

BeanbagBallbag

16 balls

If you want one white only it can be 1st OR 2nd OR 3rd;

If it's first that's 3/16..you need the second and third ball to be not white;
3/16 X 13(non white)/15(remaining balls) X 12(non whites)/14(remaining balls) = 39/280

Similarly, if the white ball comes 2nd it will be 39/280, and if it comes third it will be 39/280.

So that's 3 X 39/280 = 117/280

Hope that helps .

reznov

Thanks a million! Life saver. It's interesting how my teacher was incorrect also. Probability of her being effective, 0.0005

**Timbuk2**

Often with probability, you can get an idea if your answer is right based on how plausible it looks.

Even if you knew nothing about how to do the question, your teachers answer of 557/560 doesn't look realistic - that is saying there is almost a 99.5% chance that there is only one white, which doesn't seem right given that there are 3 whites, 8 black and 5 reds in the bag!

crayon1

reznov wrote: »

The new project Maths book offers little information in solving probability problems. My teacher has one answer - book, another.
Solve this question please? It's not for homework.

A bag contains eight black, three white and five red beads. Three beads are picked at random without replacement. Find probability that one is white, the and the others are not.

I assume it is: 1 - P(none white)?
Teacher's answer: 557/560
Book's answer: 117/280

>_<

Total = 16
P(1st one white) = (3 Choose 1) x (1/16)
P(2nd not white) = (13 Choose 1) x (1/15)
P(3rd not white) = (12 Choose 1) x (1/14)

3/16 x 13/15 x 12/14 x 3!/2! = 117/280

Theres your answer from the book.....I'm not a teacher so sorry if I'm wrong....I'm open to correction! But I tried!

reznov

Oh I know her answer did not resemble something plausible, and I do trust the book more in relation to answers. It just puts you off!

Sorry, I don't know where the thank you button is :P

**Timbuk2**

You've already got two answers, and this one is similar, but this is how I'd do it.

Instead of considering three colours, consider 'white' and non-white. Now the problem becomes what's the probability of picking exactly one white?

Effectively this is saying - what is the probability of picking one white and two non-whites.

How many ways can you pick one white AND two non-whites?
3C1 * 13C2 (here 8 black + 5 red = 13 non-white) = 234 ways

How many ways can you pick 3 beads? 16C3 = 560

So the probability of picking exactly one white is 234/560 = 117/280.