Probability

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Tadeusz Jordan - 29 Apr 2007 23:53 GMT

Hello. I was solving a problem but I got a different answer than my friend.
Could anyone tell me which answer is wrong and where is a mistake?
The problem is:
Three balls are to be randomly selected with replacement from an urn
containing 20 balls numbered 1 through 20. If we bet at least one of the
drawn balls has a number as large as 17 or larger than 17, what is the
probability that we win the bet?
I got:
P{X=3} (3 balls are >=17) = 1/5 *1/5 * 1/5 = 1/125
P{X=2} (2 balls are >=17) = 3 * 1/5 * 1/5 * 4/5 = 12/125
P{X=1} (1 ball is >=17) = 3 * 1/5 * 4/5 * 4/5 = 48/125
P{X=0} (0 balls are >=17) = 4/5 * 4/5 * 4/5 = 64/125

So the probability of winning is P{X=3} + P{X=2} + P{X=1} = 61/125

Now there is another approach:

Let X denote the largest number selected.
P{X=20} = P{X=19} = P{X=18} = P{X=17}= [20 choose 2]/[20 choose 3] = 1/6
So the probability of winning is 4/6 = 2/3
Could anyone tell me which approach is correct and why the second one is not
correct?

Tad

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Mike Terry - 30 Apr 2007 02:01 GMT

> Hello. I was solving a problem but I got a different answer than my friend.
> Could anyone tell me which answer is wrong and where is a mistake?
[quoted text clipped - 10 lines]
>
> So the probability of winning is P{X=3} + P{X=2} + P{X=1} = 61/125

This is correct.

> Now there is another approach:
>
> Let X denote the largest number selected.
> P{X=20} = P{X=19} = P{X=18} = P{X=17}= [20 choose 2]/[20 choose 3] = 1/6
> So the probability of winning is 4/6 = 2/3

I don't understand why you think this would be correct - perhaps you could
explain the thinking behind it?

As an example, P(X=20) = 1 - P(X<20)
= 1 - (19/20)^3
= 1141/8000 (which is not 1/6)

It looks like whatever reasoning you are using could be extended all the way
to P(X=1) = 1/6? This is even more clearly wrong, as P(X=1) = (1/20)^3 =
1/8000....

> Could anyone tell me which approach is correct and why the second one is not
> correct?

The second approach is not correct because it is wrong - nobody has proposed
why it ought to be right, and in maths proofs are wrong by default! :-)

Regards,
Mike.

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Brian M. Scott - 30 Apr 2007 02:05 GMT

> Hello. I was solving a problem but I got a different answer than my friend.
> Could anyone tell me which answer is wrong and where is a mistake?
> The problem is:

> Three balls are to be randomly selected with replacement from an urn
> containing 20 balls numbered 1 through 20. If we bet at least one of the
> drawn balls has a number as large as 17 or larger than 17, what is the
> probability that we win the bet?

The simplest approach is to calculate the probability of
losing: that's just the probability that all three balls are
numbered below 17, so it's (16/20)^3 = (4/5)^3 = 64/125, and
the probability of winning is 1 - 64/125 = 61/125.

> I got:
> P{X=3} (3 balls are >=17) = 1/5 *1/5 * 1/5 = 1/125
> P{X=2} (2 balls are >=17) = 3 * 1/5 * 1/5 * 4/5 = 12/125
> P{X=1} (1 ball is >=17) = 3 * 1/5 * 4/5 * 4/5 = 48/125
> P{X=0} (0 balls are >=17) = 4/5 * 4/5 * 4/5 = 64/125

> So the probability of winning is P{X=3} + P{X=2} + P{X=1} = 61/125

This is correct, though you're working harder than
necessary.

> Now there is another approach:

> Let X denote the largest number selected.
> P{X=20} = P{X=19} =

Not true: P{X=20} = 1 - P{X<20} = 1 - (19/20)^3 = 0.142625,
while P{X=19} = 1 - P{X<19 or X=20} = 1 - [(18/20)^3 +
0.142625] = 0.1283575.

[...]

Brian

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Math Forum / Mathematics / Undergraduate Math / April 2007

Probability