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 Probable Prime Poser (Posted on 2010-01-19)
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.

 No Solution Yet Submitted by K Sengupta Rating: 1.0000 (1 votes)

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 bonus solution | Comment 2 of 4 |

In addition to the 53 sets of digits without digit duplication, there are 33 with digit duplication.

`digits    primes011       101112       211113       113 131 311115       151118       181 811119       191 911133       313 331166       661188       881199       199 919 991223       223227       227229       229233       233277       277 727299       929334       433335       353337       337 373 733338       383344       443377       773388       883449       449499       499557       557577       577 757599       599677       677778       787 877779       797 977788       887799       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 2010-01-19 17:00:55
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