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bhushan252
my approach...
after three zeros...we left with 4 places which can be filled with 2.3,4,5,6,7,8,9
this means 8 digits with 4 prime & 4 non prime nos to choose from.....

P = [(1P+3NP) OR (2P+2NP) OR (3P+1NP) OR (4P+0NP)]/[8C4]

[4C1*4C3+4C2*4C2+4C3*4C1+4C4*4C0] / [8C4]
[4*4+6*6+4*4+1]/70 = 69/70...BUT I AM WRONG SOMEWHERE :(

First step of getting rid of 3 digits of containing zeros is right: we are left with 4 digit number each of which can be one of 8 digit (2,3,4,5,6,7,8,9 as 0 is already used in first three and 1 is not used).

Next step:
total number of combinations possible 8^,
combinations with NO PRIME 4^4,
P(no prime)=4^4/8^4=1/2^4
P(p>=1)=1-1/2^4=15/16

Answer D.
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Bunuel
bhushan252
my approach...
after three zeros...we left with 4 places which can be filled with 2.3,4,5,6,7,8,9
this means 8 digits with 4 prime & 4 non prime nos to choose from.....

P = [(1P+3NP) OR (2P+2NP) OR (3P+1NP) OR (4P+0NP)]/[8C4]

[4C1*4C3+4C2*4C2+4C3*4C1+4C4*4C0] / [8C4]
[4*4+6*6+4*4+1]/70 = 69/70...BUT I AM WRONG SOMEWHERE :(

First step of getting rid of 3 digits of containing zeros is right: we are left with 4 digit number each of which can be one of 8 digit (2,3,4,5,6,7,8,9 as 0 is already used in first three and 1 is not used).

Next step:
total number of combinations possible 8^,
combinations with NO PRIME 4^4,
P(no prime)=4^4/8^4=1/2^4
P(p>=1)=1-1/2^4=15/16

Answer D.

Can you explain the last step
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Prime - 2,3,5,7

Non - Prime - 1,4,6,8

Format = 0 0 0 _ _ _ _

All Non-Primes = (4 * 4 * 4 * 4)/(8 * 8 * 8 * 8)

All Non-Primes = 1/16

At least 1 prime = 1 - 1/16 = 15/16

Answer - D
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badgerboy
A certain telephone number has seven digits. If the telephone number has the digit zero exactly 3 times, and the number 1 is not used at all, what is the probability that the phone number contains one or more prime digits.

A. 1/24
B. 1/16
C. 1/2
D. 15/16
E. 23/24

Yeah actually I failed to realize early in the problem that we didn't need to care about the 3 digits with the zero on it

But anyways, this is how I did it

What we are trying to do here is work in an inverse probabilistic approach hence, 1 - (Prob of having no primes)

Now, every other digit if the 4 that remain will have 8 possible options so total number of options is (8^4). This is total probabilities and denominator

Then for the numerator we can only take the numbers that aren't prime so 4,6,8,9. Everything else stays the same so we end up with (4^4)

Therefore, (4^4)/(8^4) = (1/2)^4 = 1/16

Now as we are looking for 1 - P(X) then answer will be 15/16

Hence D

Hope it helps
Cheers!
J :)
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Bunuel
bhushan252
my approach...
after three zeros...we left with 4 places which can be filled with 2.3,4,5,6,7,8,9
this means 8 digits with 4 prime & 4 non prime nos to choose from.....

P = [(1P+3NP) OR (2P+2NP) OR (3P+1NP) OR (4P+0NP)]/[8C4]

[4C1*4C3+4C2*4C2+4C3*4C1+4C4*4C0] / [8C4]
[4*4+6*6+4*4+1]/70 = 69/70...BUT I AM WRONG SOMEWHERE :(

First step of getting rid of 3 digits of containing zeros is right: we are left with 4 digit number each of which can be one of 8 digit (2,3,4,5,6,7,8,9 as 0 is already used in first three and 1 is not used).

Next step:
total number of combinations possible 8^,
combinations with NO PRIME 4^4,
P(no prime)=4^4/8^4=1/2^4
P(p>=1)=1-1/2^4=15/16

Answer D.

Hi Bunuel,

Just wanted to confirm something. Why is the order of the zeroes not important here? I mean why can we assume that zeroes will be the first 3 digits and not in any other order? I guess that the position of the zeroes does not change our probability scenario but just wanted to be 100% sure.

Thanks
Cheers
J
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jlgdr
Bunuel
bhushan252
my approach...
after three zeros...we left with 4 places which can be filled with 2.3,4,5,6,7,8,9
this means 8 digits with 4 prime & 4 non prime nos to choose from.....

P = [(1P+3NP) OR (2P+2NP) OR (3P+1NP) OR (4P+0NP)]/[8C4]

[4C1*4C3+4C2*4C2+4C3*4C1+4C4*4C0] / [8C4]
[4*4+6*6+4*4+1]/70 = 69/70...BUT I AM WRONG SOMEWHERE :(

First step of getting rid of 3 digits of containing zeros is right: we are left with 4 digit number each of which can be one of 8 digit (2,3,4,5,6,7,8,9 as 0 is already used in first three and 1 is not used).

Next step:
total number of combinations possible 8^,
combinations with NO PRIME 4^4,
P(no prime)=4^4/8^4=1/2^4
P(p>=1)=1-1/2^4=15/16

Answer D.

Hi Bunuel,

Just wanted to confirm something. Why is the order of the zeroes not important here? I mean why can we assume that zeroes will be the first 3 digits and not in any other order? I guess that the position of the zeroes does not change our probability scenario but just wanted to be 100% sure.

Thanks
Cheers
J

The question involves probability. The number of arrangements that you make in the restricted case (using no primes in the numerator) will be the same as the number of arrangements you make when you have no restrictions (all 8 digits possible in the denominator). Hence, you don't need to worry about the arrangements even though a telephone number depends on the order of the digits.

With three 0s taking away 3 places, you are left with 8 digits - 4 prime and 4 non prime
2, 3, 5, 7 and 4, 6, 8, 9
Probability of not using any prime for all 4 digits = (4/8)*(4/8)*(4/8)*(4/8) = 1/16
Probability of using at least one prime = 1 - 1/16 = 15/16
Answer (D)
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badgerboy
A certain telephone number has seven digits. If the telephone number has the digit zero exactly 3 times, and the number 1 is not used at all, what is the probability that the phone number contains one or more prime digits.

A. 1/24
B. 1/16
C. 1/2
D. 15/16
E. 23/24

telephone: {000}{XXXX}… X≠{0,1}={2,3,4,5,6,7,8,9}=8… primes={2,3,5,7}=4… not_prime,0,1={4,6,8,9}=4

METHOD 1: find the probability of not getting any prime and subtract from 1.
unfavorable outcomes {XXXX} not_prime,0,1: {4*4*4*4}=\(4^4\)
total outcomes {XXXX}: {8*8*8*8}=\(8^4\)
1-unfavorable/total: \(1-\frac{4^4}{8^4}=\frac{15}{16}\)

Answer (D)

METHOD 2: find the probability of each case with primes and add them.
favorable outcomes {PNNN}: {4*4*4*4}=\(4^4\) • arrangements {PNNN}: \(4!/3!=4\)… \(=4^5\)
favorable outcomes {PPNN}: {4*4*4*4}=\(4^4\) • arrangements {PPNN}: \(4!/2!2!=6\)…\(=4^4•6\)
favorable outcomes {PPPN}: {4*4*4*4}=\(4^4\) • arrangements {PPPN}: \(4!/3!=4\)… \(=4^5\)
favorable outcomes {PPPP}: {4*4*4*4}=\(4^4\) • arrangements {PPPP}: \(4!/4!=1\)…\(=4^4\)
total outcomes {XXXX}: {8*8*8*8}=\(8^4\)
sum probability each case: \(\frac{4^5+4^4•6+4^5+4^4}{8^4}=15/16\)

Answer (D)
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There are 4 places for numbers selected from the set
{2,3,4,5,6,7,8,9}.
8^4 Ways we can select these four numbers.
4^4 ways in which the prime numbers are not selected.
So ,1- 4^4/8^4=15/16
Option D

Posted from my mobile device
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The question asks for a probability, so the answer will be a fraction. There is no simple way to calculate that “one or more” digits will be prime. It’s necessary to calculate that none of the digits will be prime and subtract that probability from 1. The telephone number has seven digits, and three of them are 0, so only four are potentially prime. Of the eight digits 2–9, four are prime (2, 3, 5, and 7) and four are not (4, 6, 8, 9). The probability that a randomly selected digit is not prime is therefore 4/8 = 1/2 . Since there are four possible digits, the probability that none of them is prime is (1/2)^4 = 16. The probability that at least one is prime is 1-(1/16) = 15/16.

Choice D
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