A bag contains 10 marbles that are numbered 0 through 9. Precisely three marbles are drawn at random from the bag without replacement.
Determine the probability that a threedigit prime number (with non leading zero) can be constituted by rearrangement of digits corresponding to the three marbles (including the original order of the digits.)
As a bonus determine the corresponding probability if the three marbles were drawn with replacement at the outset.
In addition to the 53 sets of digits without digit duplication, there are 33 with digit duplication.
digits primes
011 101
112 211
113 113 131 311
115 151
118 181 811
119 191 911
133 313 331
166 661
188 881
199 199 919 991
223 223
227 227
229 229
233 233
277 277 727
299 929
334 433
335 353
337 337 373 733
338 383
344 443
377 773
388 883
449 449
499 499
557 557
577 577 757
599 599
677 677
778 787 877
779 797 977
788 887
799 997 86
The probability that the three digits drawn will contain no duplicates is (9/10)*(8/10) = 72/100; the probability the digits are all the same is 1/100 and the probability that there is just a matching pair is then 1  72/100  1/100 = 1  73/100 = 27/100.
As before, under the condition that all the digits are different, the probability they'd have the digits to make a prime is 53/120. But the condition itself has 72/100 probability to make the overall probability contribution of this case equal to 159/500.
The situation of all the digits the same contributes nothing to the probability of forming a prime, as no primes have all three the same digit, as 111 is not prime.
In the 27/100 case that there is just a pair of matching digits, the conditional probability of being able to form a prime is the 33 cases above divided by the number of possible such combinations. That number of combinations is 10*9 = 90 as there are 10 ways of choosing the individual number and 9 ways remaining of choosing the doubled digit. Or you could look at it the other way around in order of choosing and it's still the same product. So the contribution to the overall probability of the doubles is (27/100)*(33/90) = 99/1000.
So the probability, when drawing is with replacement is
159/500 + 99/1000 = 417/1000.

Posted by Charlie
on 20100119 17:00:55 