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 three-digit 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.
There are 53 unique sets of 3 different digits that can form 3-digit primes
digits primes
013 103
014 401
016 601
017 107 701
019 109
035 503
037 307
049 409
059 509
067 607
079 709 907
089 809
124 241 421
125 251 521
127 127 271
128 281 821
134 431
136 163 613 631
137 137 173 317
139 139 193
145 541
146 461 641
149 149 419 491 941
157 157 571 751
167 167 617 761
169 619 691
179 179 197 719 971
235 523
236 263
238 283 823
239 239 293
257 257
269 269
278 827
289 829
346 463 643
347 347 743
349 349 439
356 563 653
358 853
359 359 593 953
367 367 673
368 683 863
379 379 397 739 937
389 389 839 983
457 457 547
467 467 647
478 487
479 479 947
569 569 659
578 587 857
589 859
679 769 967
so that's 53 digit sets out of the C(10,3) = 120 equally likely possible sets of 3 digits chosen.
The probability is 53/120.
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Posted by Charlie
on 2010-01-19 13:08:32 |
computer solution to part 1
