[K Sengupta]

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 from the digits corresponding to the three marbles.

[mmiguel1]

Spoiler for Let's see

First find all 3 digit primes:
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997

Then of these, find all numbers with 3 distinct digits:
103, 107, 109, 127, 137, 139, 149, 157, 163, 167, 173, 179, 193, 197, 239, 241, 251, 257, 263, 269, 271, 281, 283, 293, 307, 317, 347, 349, 359, 367, 379, 389, 397, 401, 409, 419, 421, 431, 439, 457, 461, 463, 467, 479, 487, 491, 503, 509, 521, 523, 541, 547, 563, 569, 571, 587, 593, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 673, 683, 691, 701, 709, 719, 739, 743, 751, 761, 769, 809, 821, 823, 827, 829, 839, 853, 857, 859, 863, 907, 937, 941, 947, 953, 967, 971, 983,

If I made no mistakes there are 97 of these.
Each of these represents one permutation. Assuming that the probability of obtaining a given permutation from the marble bag is the same for all possible permutations, then the probability of getting any of these is equal to their number divided by the total number of possible permutations.
The total number of possible permutations is 10*9*8=720
The answer I think is 97/720 or 0.134722222...

This post has been edited by mmiguel1: 06 November 2009 - 11:31 PM

[Simon Legree]

mmiguel1, on 06 November 2009 - 05:29 PM, said:

Spoiler for Let's see

First find all 3 digit primes:
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997

Then of these, find all numbers with 3 distinct digits:
103, 107, 109, 127, 137, 139, 149, 157, 163, 167, 173, 179, 193, 197, 239, 241, 251, 257, 263, 269, 271, 281, 283, 293, 307, 317, 347, 349, 359, 367, 379, 389, 397, 401, 409, 419, 421, 431, 439, 457, 461, 463, 467, 479, 487, 491, 503, 509, 521, 523, 541, 547, 563, 569, 571, 587, 593, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 673, 683, 691, 701, 709, 719, 739, 743, 751, 761, 769, 809, 821, 823, 827, 829, 839, 853, 857, 859, 863, 907, 937, 941, 947, 953, 967, 971, 983,

If I made no mistakes there are 97 of these.
Each of these represents one permutation. Assuming that the probability of obtaining a given permutation from the marble bag is the same for all possible permutations, then the probability of getting any of these is equal to their number divided by the total number of possible permutations.
The total number of possible permutations is 10*9*8=720
The answer I think is 97/720 or 0.134722222...

[Simon Legree]

Spoiler for Very good but....

The permutations are 9*9*8 = 648 since the first digit can't be zero but the second can. 97/648 = 0.1497

[mmiguel1]

In what sense are you using the word "constituted"? Can you draw out 3 marbles and rearrange the digits? Or must you construct the number by placing the digits in the same order as they were received?

Spoiler for If it is the first case,

Then referencing my earlier post, we have for 3 digit primes with distinct digits:
103, 107, 109, 127, 137, 139, 149, 157, 163, 167, 173, 179, 193, 197, 239, 241, 251, 257, 263, 269, 271, 281, 283, 293, 307, 317, 347, 349, 359, 367, 379, 389, 397, 401, 409, 419, 421, 431, 439, 457, 461, 463, 467, 479, 487, 491, 503, 509, 521, 523, 541, 547, 563, 569, 571, 587, 593, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 673, 683, 691, 701, 709, 719, 739, 743, 751, 761, 769, 809, 821, 823, 827, 829, 839, 853, 857, 859, 863, 907, 937, 941, 947, 953, 967, 971, 983,

If we can rearrange the marbles, we are interested in combinations, not permutations.
To make a list of combinations of 3 digits that can form 3 digit primes with distinct digits, we must remove duplicate combinations from the above list.
This gives us (as groups of 3 digits, not interpreted as numbers):
013, 014, 016, 017, 019, 035, 037, 049, 059, 067, 079, 089, 124, 125, 127, 128, 134, 136, 137, 139, 145, 146, 149, 157, 167, 169, 179, 235, 236, 238, 239, 257, 269, 278, 289, 346, 347, 349, 356, 358, 359, 367, 368, 379, 389, 457, 467, 478, 479, 569, 578, 589, 679,

There are 53 combinations of 3 digits that can be rearranged to form a 3 digit prime with distinct digits.
There are 10 choose 3 = 720/6 = 120 total combinations
In that case, the probability is
53/120 = 0.44166666...

[mmiguel1]

Hi Simon,
the total permutations reflects just what you might be able to draw out of the bag.
It is very possible that the first marble you draw is a zero.
In that case, you still can draw two marbles to get a valid permutation, you just won't be getting a 3 digit prime.
Because it is just as possible to draw 0 as any other number for the first marble, it should be included in counting the total number of possibilities.

[Simon Legree]

Point taken