Seven sportsmen (named A,B,C,D,E,F and G) buy 7 hats choosing randomly out of 4 available
colors (say G,R,W and Y).
What is the probability that all
the colors were chosen?
Using "&" to represent "and", and "" to represent "or":
The inclusion/exclusion principle gives us:
p(g & r & w & y) = p(g)+p(r)+p(w)+p(y)
 p(gr)  p(gw)  p(gy)
 p(rw)  p(ry)  p(wy)
+ p(grw) + p(gry) + p(gwy) + p(rwy)
 p(grwy)
Now p(g), or any of the single color probabilities (that is the probability that a given color shows up among the sportsmen, not necessarily exclusively), is 1  (3/4)^7.
The probability that a given pair of colors will have at least one representation among the sportsmen, such as p(gr), is 1  (1/2)^7.
And the probability of one of the triples is 1  (1/4)^7.
And of course p(grwy) = 1, as one of these colors must be represented at least once among the sportsmen.
So the probability that all four colors will be represented is:
4 * (1  (3/4)^7)  6 * (1  (1/2)^7) + 4 * (1  (1/4)^7)  1
which is equal to 525/1024, or about 0.5126953125.
Simulation verification:
FOR tr = 1 TO 1000000
REDIM used(3)
FOR spm = 1 TO 7
choice = INT(RND(1) * 4)
used(choice) = 1
NEXT
good = 1
FOR i = 0 TO 3
IF used(i) = 0 THEN good = 0
NEXT
IF good THEN hits = hits + 1
PRINT tr, hits / tr, 1024 * hits / tr
NEXT
finds as its last line:
1000000 .512473 524.7723
meaning that after a million trials, the fraction of successes is .512473, or, relative to a denominator of 1024: 524.7723/1024.

Posted by Charlie
on 20100426 13:50:11 