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John wrote a phone no. on a note that was later lost. John

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John wrote a phone no. on a note that was later lost. John [#permalink]

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08 Apr 2006, 18:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

John wrote a phone no. on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?
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10 Apr 2006, 10:47
Here is my take -

there are 7 digits : - - - - - - -

Since 1 appeared in the last three places, we now have a 4 digit number.

- - - -

What is the probability that 2 of the numbers are prime? These 4 digits can take have numbers 2 to 9. The primes are (2,3,5,7).

=> total combinations (8^4)
=> required combinations (4^2*4^2 + 4^3*4^1 + 4^4*4^0)

=> probability = 3 * 4^4 / 8^4

=> 3/16
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10 Apr 2006, 13:14
Is it 1321/2187?
P(atleast 2 primes) = 1 - {P(no primes) + P(exactly 1 prime)}

Primes = 2,3,5,7
Non-primes = 1,4,6,8,9
P{no primes} = 5*5*5*5/9^4
P{exactly 1 prime} = (4*5*5*5/9^4)*4C1 [ The given prime can occupy any 1 of the 4 places]

P(atleast 2 primes) = 1 - {625+2000}/9^4 = 1321/2187
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10 Apr 2006, 15:55
4436/6561

The numbers where there are less than 2 primes (X = non primes, P = Primes)

XXXX, XXXP, XXPX, XPXX, PXXX

The probabilities for each of the above:

(625/6561) + (500/6561) + (500/6561) + (500/6561) = 2125/6561

So Probablity for all numbers where there are more than 2 prime digits:

1 - 2125/6561 = 4436/6561!
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14 Apr 2006, 17:47

Thanks,
Vipin
14 Apr 2006, 17:47
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