3u (maybe even 2u?) probability q. (1 Viewer)

[ND]

I'm posting this in 4u cos it'll get more views here. My answer for this question is different than that in the back of the book. The question is something like this:

"There are 8 10c coins and 8 20c coins. If 2 10c coins and 1 20c coin is drawn from these, find the probability that a particular 10c coin will be drawn."

Surely the answer is 1/4? In the back of the book they had 1/10, and my substitute teacher couldn't understand my logic. Can anyone explain how they got 1/10?

 

[McLake]

Isn't it a 15/56 chance? (slightly better than 1/4)

Choose 1/8 then choose 1/7 so (7+8)/(7*8) = 15/56

Try something else:

7/8 of not getting in the first, then a 6/7 chance of not getting it in the second.

13/56 chance of not getting it
43/56 chance of not getting it in the second

Hmm, I hate probability ...

 

[ND]

But if both coins are taken simultaneously, doesn't that make it 2/8? I suck at probability...

 

[McLake]

Originally posted by ND
But if both coins are taken simultaneously, doesn't that make it 2/8? I suck at probability...

So do I. But my reasoning is that who cares if you take them both at once, it should work the otehr way too ...

 

[Affinity]

McLake:
no it's not 15/56

Choose 1/8 then choose 1/7 so (7+8)/(7*8) = 15/56

it's not 1/8 + 1/7
it's 1/8 (for getting the coin the first go)
and 7/8 * (1/7) for getting it the second go, the factor of 7/8 is there because the coin should not be picked on the first go.

7/8 of not getting in the first, then a 6/7 chance of not getting it in the second.
13/56 chance of not getting it
43/56 chance of not getting it in the second.

13/56 chance of not getting it? <-
"43/56 not getting it in the second" 6*7=42
so the chance of getting it is still 14/56=1/4

ND:
hmm must be 1/4 since:

we only need to consider the 2 10c coin.

there are 7 sets of two coins that contain the particular coin(that coin plus any of the other 7).
and there are in total 8C2 = 28 sets of 2.

therefore, the chance of picking that coin is 1/4

which book/chapter/exercise/question is it by the way

 

[wogboy]

Here's my reasoning:

The 20c coins don't matter because we don't care about them.

For the 10c coins, the probability of picking the first one to be the "special" coin is 1/8 (so we start off with 1/8). If this fails (i.e. the coin picked is not the "special" one), then the probability of picking the "special" one for the second coin is 1/7 (so we add 1/7). But of course the first coin and the second coin can't both be the special one (so we subtract 1/7*1/8).

p = 1/8 + 1/7 - 1/56 = 1/4

(of course you can take the simple minded view that choosing 2 items out of 8 at random will give a probability of 2/8=1/4 of choosing the "special" one).

 

[Lazarus]

I would agree with Affinity's answer (1/8 + 7/8 * 1/7).

However, in an attempt to come up with 1/10 as an answer...

Originally posted by ND
"There are 8 10c coins and 8 20c coins. If 2 10c coins and 1 20c coin is drawn from these, find the probability that a particular 10c coin will be drawn."

Perhaps the question is actually saying that there is a pool of 16 coins, eight of which are 10c coins and eight of which are 20c coins. You randomly draw three coins from the pool, and you happen to come up with two 10c coins and one 20c coin. What's the probability that a particular 10c coin was included in that combination?

(ways of choosing the particular 10c along with any other 10c * ways of choosing any 20c)
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;ways of choosing any three coins from the pool of 16

= (7 * 8) / 16C3

= 1/10

 

[Lazarus]

Alternatively, you could enumerate the possibilities and combine the resulting probabilities, as a friend of mine did -

P(P, 10, 20)
particular: 1/16
10c: 7/15
20c: 8/14
= 56/3360
= 1/60

P(10, P, 20)
10c: 7/16
particular: 1/15
20c: 8/14
= 56/3360
= 1/60

P(20, 10, P)
20c: 1/2
10c: 7/15
particular: 1/14
= 7/420
= 1/60

P(20, P, 10)
20c: 1/2
particular: 1/15
10c: 1/2
= 1/60

P(10, 20, P)
10c: 7/16
20c: 8/15
particular: 1/14
= 56/3360
= 1/60

P(P, 20, 10)
particular: 1/16
20c: 8/15
10c: 1/2
= 8/480
= 1/60


P(P, 10, 20) + P(10, P, 20) + P(20, 10, P) + P(20, P, 10) + P(10, 20, P) + P(P, 20, 10) = 1/10.

 

[ND]

Thanks for the replies.

(of course you can take the simple minded view that choosing 2 items out of 8 at random will give a probability of 2/8=1/4 of choosing the "special" one)

That's how i did it, it may seem obvious, but it works.

Laz: Yeh i guess that's what they meant. Poorly phrased question... in fact that book makes a habit of it.

Affinity: I don't remember the name of the author (i think her 1st name was margaret?), it's not a very popular book from what i can tell.