Question

Conditional Probability.. I think

Suppose 20 numbered balls are in an urn, each with a unique number, 1 through 20 on it. If you draw 4 (and order doesn't matter), approximate the probability that the set of 4 will contain at least 2 consecutive numbers. (for example, k and k+1 for 9 and 10.)

Any suggestions on how to set a problem like this up?

Answer 1

i wonder why it says "approximate"?

Answer 2

i can think of a (not necessarily right) procedure that might be an approximation, but it would certainly not be exact

Answer 3

no, i see it is wrong. damn

Answer 4

lets start anyway, maybe @Zarkon will correct it
pick a number
assuming it is not one or 20 when you pick the next number there are 2 out of the remaining 19 that could be adjacent
if so you are done and the probability is \(\frac{2}{19}\)
if not, with probabilty \(\frac{17}{19}\) you pick one that is not adjacent. then the probability that the next one is adjacent to one of the first two is either \(\frac{3}{18}\) or \(\frac{4}{18}\) depending on the first two
damn this won't work

Answer 5

maybe we can do it by passing to the compliment
what is the probability that none are consecutive

Answer 6

\[1-\frac{{20-4+1\choose 4}}{{20\choose 4}}\]

Answer 7

@satellite73

That is the way to do it

number the 4 balls to be chosen in order

\[N_1,N_2,N_3,N_4\]
now they need to be chosen so that there is at least one ball between them...mark that ball with a star

\[N_1\star,N_2\star,N_3\star,N_4\]

so we still have 4 choices but treat \(N_1\star,N_2\star,N_3\star\) as 3 balls (not 6). that gives us a total of 17 balls to choose from (when picking 4)...hence


\[{ 17\choose 4}={20-4+1\choose 4}\]

the denominator is then obviously \(\displaystyle {20\choose 4}\)

Answer 8

this is why i should stay up later. missed this answer last night