Anyone seen this before

Tim Robbins

You have a draw with Red (r) and Blue socks (b). When two socks are drawn at random the probability that both of the socks are is 1 /2. How small can the number of socks be?


Ok.

so the probabitly of the first sock being red is: r / r + b
the probability of the second sock being red is: r - 1 / r - 1 + b

where r = number of reds socks in drawer. b = number of blue socks in drawer.

So we want:

(r / r + b) * (r - 1 / r + b - 1) = 1/ 2


The he says...

Since
(r / r + b) > (r - 1 / r + b - 1) for b > 0 (Not sure how he derives this)

Then he says that since the above is true we can say:

(r / r + b) pow 2 > 1 / 2 > ( r - 1 / r + b - 1) pow 2 (really don't see where he gets this).

This is taken from Fifty Challenging Problems in Probability - page 15.

Can you explain how this can be concluded?

mikhail

Tim, the question is incomplete: "When two socks are drawn at random the probability of the number of red socks is 1 /2" is not a sentence.

Tim Robbins

mikhail wrote: »

Tim, the question is incomplete: "When two socks are drawn at random the probability of the number of red socks is 1 /2" is not a sentence.

Well spotted. Fixed.

Tim Robbins

OK I got this part:

(r / r + b) > (r - 1 / r + b - 1) for b > 0 (Not sure how he derives this)

If you go backwards the assumption that it's true you get

(r)(r + b - 1) > (r - 1)(r + b)

r pow 2 + rb -r > r pow 2 - r + br - b
you get...

0 > -b

or b > 0.

Tim Robbins

Ok got the second part now.

Again it's just algerba.

Think about this way...

Let x = (r / r + b)
Let y = (r - 1 / r + b -1)

We know
x > y ... eq 1
and we know
x * y = 1 / 2 .. eq2

substituiting for x we get
1 / 2 > y pow 2

and substituiting for y we get

x > 1 / 2

That's it. Just some algerbra.