Probability

[ithamarsorek]

Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?

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Thanks for your help

[721tjm]

What's the probability she doesn't draw out a quarter after X trials?
X=1: (10+5)/21 = 15/21 chance she doesn't draw a quarter first
X=2: (10+5-1)/(21-1)=14/20 chance she doesn't draw a quarter on the second try after not drawing one the first time

(15/21)(14/20) = 210/420 = .5 chance she won't have drawn a quarter = .5 chance she will have drawn a quarter

So she needs to draw at least two

IMO B

[aleph777]

Does this mean she's had a 50% chance of pulling a quarter on the third pull or the second?

Thanks!

[721tjm]

Apologies for making that jump without explaining it.

Mechanically, drawing N coins at once is the same as drawing one coin, then another coin, and additional coins until you've drawn N coins. To frame the problem in terms we can better describe, we can consider the latter case so that we can refer to a first coin, a second coin, etc.

And the probability she has pulled a quarter after N draws is 1 minus the probability that she hasn't drawn a quarter after N draws. Rephrasing in terms of this opposite event allows us to ignore the separate possibilities of drawing 1 or 2 quarters in 2 tries and instead lumps them both into the easier-to-handle case of whether or not she's drawn any quarters.

The probability she pulls a quarter the first time is the number of non-quarters over the total number of coins: 15/21
The probability she does it again after drawing a non-quarter the first time is the same calculation except accounting for the coin already withdrawn: (15-1)/(21-1)=14/20

P(Drawing at least one quarter)
= 1 - P(Not drawing any quarters)
= 1- P(Not drawing a quarter the first time)*P(Not drawing a quarter the second time after not drawing one the first time)*...*P(Not drawing a quarter the Nth time after not drawing one the N-1 times)

You could keep doing this for up to 15 trials (after which she'd definitely pull a quarter), but in this case we just need to know how many draws it would take for it to be no more than 50% likely that she has continuously drawn non-quarters up to that point.

In this case, we can stop after two tries:
1 - (15/21)*(14/20) = 1 - .5 = .5

Hope this helps

[prachich1987]

a tough question
what's the source?

[fskilnik@GMATH]

Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?

1
2
5
6
7

Hi there!

Beautiful problem... well done (and explained) by 721tjm !

Let me give a second look at the problem, in a way that perhaps will help to see things in a clearer perspective!

For any integer 1<= n <= 15, let An be the event "no quarters obtained in the simultaneous extraction of n coins". It is really easy to find a formula for the probability of An ... have a look:

(*) Prob(An) = x/y where

x = # ways to choose n coins between the (5+10) dimes and nickes ;
y = # ways to choose n coins between all of them (6+5+10 = 21 coins) .

(The fact that all y possibilities are equiprobable allows us to use the formula (*) given above.)

Well, x = C(15, n) and y = C(21,n) ... (explain!)

From the question stem (and 721tjm´s good explanation), we are looking for n such that Prob(An) <= 0.5 ...

From the fact that the expression C(21,n)/C(15,n) doesn´t seem nice, I guess it is smarter to put values for n, from the alternatives (of course):

If n = 1 we get C(21,n)/C(15,n) = 21/15 > 0.5 so n=1 is not the right choice ;
If n = 2 we get C(21,2)/C(15,2) = 0.5 (check that), therefore we are done.

Regards,
Fabio.