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John wrote a phone number on a note that was later lost [#permalink]
24 May 2010, 14:15

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Question Stats:

43% (04:03) correct
57% (03:00) wrong based on 151 sessions

John wrote a phone number 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?

25. John wrote a phone number 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?

A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8

I don't understand the explanation to this one (so I obviously got it wrong...).

Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this?

Thanks, folks.

The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}.

{X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \frac{4}{8}=\frac{1}{2} and probability of {X} will not be a prime is again \frac{1}{2}.

We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes.

Let's count the opposite probability and subtract it from 1.

Opposite probability of at least 2 primes is 0 or 1 prime:

So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}.

Scenario 1 prime - {P}{NP}{NP}{NP}: \frac{4!}{3!}*\frac{1}{2}*(\frac{1}{2})^3=\frac{4}{16}. We are multiplying by \frac{4!}{3!} as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same).

Scenario 0 prime - {NP}{NP}{NP}{NP}: (\frac{1}{2})^4=\frac{1}{16}.

Hence opposite probability = \frac{4}{16}+\frac{1}{16}=\frac{5}{16}.

So probability of at least 2 primes is: 1-(Opposite probability) = 1-\frac{5}{16}=\frac{11}{16}

Just a small confusion 2,3,5,7 are primes , and 1,4,6,8,9 non primes ,just because there are 3 one's in the last three places it doesn't mean that 1 cannot be at any other place, question says boy remembers last three places having one's and not that there are only 3 ones in the 7 digit number the phone number could be {1 2 3 6 1 1 1 } this has two primes and 5 non primes and 4 one's or { 8 2 2 6 1 1 1 } this has 2 primes and all the primes are same

first going the long way p( 2 primes ) +p( 3 primes )+p(4 primes )

\frac{4}{9} * \frac{4}{9} * \frac{5}{9} *\frac{5}{9} *1*1*1* \frac{4!}{2!2!} ( exactly two primes ) Multiplying by \frac{4!}{2!2!} as we can permutate only the first 4 digits , the last are fixed (1,1,1)

\frac{4}{9} * \frac{4}{9}* \frac{4}{9}*\frac{5}{9}*1*1*1 *\frac{4!}{3!} ( exactly 3 primes )( 3 of a kind )

Now for the case 2 primes, both the primes could be same or different then how does the notation \frac{4!}{2!2!} change , or does it remain the same ?

Similarly for the case of 3 primes the primes could be 2,2,2 all same or 2,3,5 all different then how does the notation \frac{4!}{3! } change or does it remain the same?

In this question we are taking primes as one kind and non primes as other kind , so it doesn't matter if the primes are all same or all different ? Is this statement correct?

1)Please could you show how to do this sum individual probability way as I have tried above ?

2)Also please consider the fact that 1 may have to be included as a non prime as the question does not explicitly state that there are only 3 ones in the phone number , he only remembers that the last three are ones.

Yes, it does seem that the language of the question is not clear. When I read the question, I also assumed 4 prime digits and 5 non prime (including 1). After all, the question only says that 1 appears in the last 3 places (and hence, we can ignore the last 3 places). It doesn't say that 1 does not appear anywhere else. But it seems that it might also imply that 1 appears only in the last 3 places (looking at the options, that was their intention). Anyway, I am sure that if this question actually appears and they mean to say that 1 cannot be at any other place, they will definitely mention it. Let's calculate the probability of different number of prime numbers:

0 primes Probability = (5/9)*(5/9)*(5/9)*(5/9)

1 prime Probability = (4/9)*(5/9)*(5/9)*(5/9)*4 (there are 4 positions for the prime)

2 primes Probability = (4/9)*(4/9)*(5/9)*(5/9)*4C2 (select 2 of the 4 positions for the primes)

3 primes Probability = (4/9)*(4/9)*(4/9)*(5/9)*4 (4 positions for the non prime)

4 primes Probability = (4/9)*(4/9)*(4/9)*(4/9)

Probability of at least 2 primes = 1 - (Probability of 0 prime + Probability of 1 prime) Probability of at least 2 primes = Probability of 2 primes + Probability of 3 primes + Probability of 4 primes

The calculations are painful so let's leave it here. As I said, their intention was probability of prime = 1/2, probability of composite = 1/2 which makes the calculations simple.

The 4 positions are different but you can have the same prime on one or more positions. When you say 4/9, you are including all cases (2, 3, 5, 7) so you don't need to account for them separately. When you say, (4/9)*(4/9)*(4/9)*(4/9), you are including all cases e.g. (2222, 2353, 3577, 2357 etc). All you need to do it separate out the primes and the non primes. That you do by arranging primes and non primes as NNPP or NPNP or PPNN etc (as we did above) _________________

Great so now I know probability of one prime 4/9* 5/9 * 5/9 *5/9 * 4= 2000/6561

PROBABILITY of 0 prime's = 5/9*5/9*5/9*5/9= 625/6561

probability of at least 2 primes = 1- ( 2000/6561 + 625/6561) 1-(2625/6561) = 3936/6561 = .59

now lets check by doing long way

2 primes

4/9 * 4/9 * 5/9*5/9* 4!/2!2! = 2400/6561 ( PPNN : two primes could be same ,2 , 2 or different 2, 3 doesn't matter we can think of this as, two of a kind and two of another kind ,hence 4!/2!2!)

3 primes

4/9 *4/9 *4/9*5/9*4 = 1280/6561 (4!/3! = 4) Three of one kind again PPPN.

4 primes

4/9*4/9*4/9*4/9 = 256/6561 all of the same kind , hence only one way to select them. PPPP

total 3936/6561 = .59 Bingo!

hence we can see both ways we are getting the same result.

Karishma if there is any error in my understanding, please do point out, I have considered primes as one kind and non primes as another kind instead of the position logic which you have mentioned.

Everything seems to be in order. It doesn't matter whether you choose to think in terms of position or arrangement of Ps and Ns. The result would be the same. When I need all arrangements of PNNN, I can write it as 4!/3! (= 4) or I can say that I will select one place for the P in 4 different ways. Either ways, we are arranging a P and 3 Ns. _________________

Re: John wrote a phone number on a note that was later lost [#permalink]
14 Aug 2014, 23:58

1

This post received KUDOS

gbascurs wrote:

Hi LucyDang,

Why I am trying to do now is:

Probability of 2 primes = 4*4*8*8*1*1*1 /8*8*8*8*1*1*1 + Probability of 3 primes = 4*4*4*8*1*1*1 /8*8*8*8*1*1*1 + Probability of 4 primes = 4*4*4*4*1*1*1 /8*8*8*8*1*1*1 Probability = 7/16

I cannot see where is the mistake. Thanks!!

Hey gbascurs,

There are overlap/incorrect calculation here: Probability of 2 primes = 4*4*8*8*1*1*1 /8*8*8*8*1*1*1 --> 4*4*8*8*1*1*1: there are 4 primes each placed in 3rd and 4th digit Probability of 3 primes = 4*4*4*8*1*1*1 /8*8*8*8*1*1*1 --> 4*4*4*8*1*1*1: there are 4 primes placed in 4th digit. You have to calculate the case in which same two primes appear, then you need to subtract the primes from the total numbers. It takes lots of time to do so, so please use the P(desire) = 1-P(opposite) to get the correct answer under time pressure. _________________

Re: John wrote a phone number on a note that was later lost [#permalink]
19 Aug 2014, 21:16

1

This post received KUDOS

Bunuel Can you explain me this part:

Scenario 1 prime - {P}{NP}{NP}{NP}: \frac{4!}{3!}*\frac{1}{2}*(\frac{1}{2})^3=\frac{4}{16}. We are multiplying by \frac{4!}{3!} as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same).

Bunuel wrote:

scaredshikless wrote:

25. John wrote a phone number 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?

A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8

I don't understand the explanation to this one (so I obviously got it wrong...).

Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this?

Thanks, folks.

The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}.

{X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \frac{4}{8}=\frac{1}{2} and probability of {X} will not be a prime is again \frac{1}{2}.

We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes.

Let's count the opposite probability and subtract it from 1.

Opposite probability of at least 2 primes is 0 or 1 prime:

So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}.

Scenario 1 prime - {P}{NP}{NP}{NP}: \frac{4!}{3!}*\frac{1}{2}*(\frac{1}{2})^3=\frac{4}{16}. We are multiplying by \frac{4!}{3!} as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same).

Scenario 0 prime - {NP}{NP}{NP}{NP}: (\frac{1}{2})^4=\frac{1}{16}.

Hence opposite probability = \frac{4}{16}+\frac{1}{16}=\frac{5}{16}.

So probability of at least 2 primes is: 1-(Opposite probability) = 1-\frac{5}{16}=\frac{11}{16}

25. John wrote a phone number 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?

A. 15/16 B. 11/16 C. 11/12 D. 1/2 E. 5/8

I don't understand the explanation to this one (so I obviously got it wrong...).

Also, at first, I read it as 1 only appears once in the last three places. If that were the case, how would you solve this?

Thanks, folks.

The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}.

{X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 non-primes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \frac{4}{8}=\frac{1}{2} and probability of {X} will not be a prime is again \frac{1}{2}.

We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes.

Let's count the opposite probability and subtract it from 1.

Opposite probability of at least 2 primes is 0 or 1 prime:

So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}.

Scenario 1 prime - {P}{NP}{NP}{NP}: \frac{4!}{3!}*\frac{1}{2}*(\frac{1}{2})^3=\frac{4}{16}. We are multiplying by \frac{4!}{3!} as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P} - 4 ways (basically the # of permutations of 4 objects out ow which 3 are the same).

Scenario 0 prime - {NP}{NP}{NP}{NP}: (\frac{1}{2})^4=\frac{1}{16}.

Hence opposite probability = \frac{4}{16}+\frac{1}{16}=\frac{5}{16}.

So probability of at least 2 primes is: 1-(Opposite probability) = 1-\frac{5}{16}=\frac{11}{16}

Answer: A.

Just a small confusion 2,3,5,7 are primes , and 1,4,6,8,9 non primes ,just because there are 3 one's in the last three places it doesn't mean that 1 cannot be at any other place, question says boy remembers last three places having one's and not that there are only 3 ones in the 7 digit number the phone number could be {1 2 3 6 1 1 1 } this has two primes and 5 non primes and 4 one's or { 8 2 2 6 1 1 1 } this has 2 primes and all the primes are same

first going the long way p( 2 primes ) +p( 3 primes )+p(4 primes )

\frac{4}{9} * \frac{4}{9} * \frac{5}{9} *\frac{5}{9} *1*1*1* \frac{4!}{2!2!} ( exactly two primes ) Multiplying by \frac{4!}{2!2!} as we can permutate only the first 4 digits , the last are fixed (1,1,1)

\frac{4}{9} * \frac{4}{9}* \frac{4}{9}*\frac{5}{9}*1*1*1 *\frac{4!}{3!} ( exactly 3 primes )( 3 of a kind )

Now for the case 2 primes, both the primes could be same or different then how does the notation \frac{4!}{2!2!} change , or does it remain the same ?

Similarly for the case of 3 primes the primes could be 2,2,2 all same or 2,3,5 all different then how does the notation \frac{4!}{3! } change or does it remain the same?

In this question we are taking primes as one kind and non primes as other kind , so it doesn't matter if the primes are all same or all different ? Is this statement correct?

1)Please could you show how to do this sum individual probability way as I have tried above ?

2)Also please consider the fact that 1 may have to be included as a non prime as the question does not explicitly state that there are only 3 ones in the phone number , he only remembers that the last three are ones.

Just a small confusion 2,3,5,7 are primes , and 1,4,6,8,9 non primes ,just because there are 3 one's in the last three places it doesn't mean that 1 cannot be at any other place, question says boy remembers last three places having one's and not that there are only 3 ones in the 7 digit number the phone number could be {1 2 3 6 1 1 1 } this has two primes and 5 non primes and 4 one's or { 8 2 2 6 1 1 1 } this has 2 primes and all the primes are same

first going the long way p( 2 primes ) +p( 3 primes )+p(4 primes )

\frac{4}{9} * \frac{4}{9} * \frac{5}{9} *\frac{5}{9} *1*1*1* \frac{4!}{2!2!} ( exactly two primes ) Multiplying by \frac{4!}{2!2!} as we can permutate only the first 4 digits , the last are fixed (1,1,1)

\frac{4}{9} * \frac{4}{9}* \frac{4}{9}*\frac{5}{9}*1*1*1 *\frac{4!}{3!} ( exactly 3 primes )( 3 of a kind )

Now for the case 2 primes, both the primes could be same or different then how does the notation \frac{4!}{2!2!} change , or does it remain the same ?

Similarly for the case of 3 primes the primes could be 2,2,2 all same or 2,3,5 all different then how does the notation \frac{4!}{3! } change or does it remain the same?

In this question we are taking primes as one kind and non primes as other kind , so it doesn't matter if the primes are all same or all different ? Is this statement correct?

1)Please could you show how to do this sum individual probability way as I have tried above ?

2)Also please consider the fact that 1 may have to be included as a non prime as the question does not explicitly state that there are only 3 ones in the phone number , he only remembers that the last three are ones.

Yes, it does seem that the language of the question is not clear. When I read the question, I also assumed 4 prime digits and 5 non prime (including 1). After all, the question only says that 1 appears in the last 3 places (and hence, we can ignore the last 3 places). It doesn't say that 1 does not appear anywhere else. But it seems that it might also imply that 1 appears only in the last 3 places (looking at the options, that was their intention). Anyway, I am sure that if this question actually appears and they mean to say that 1 cannot be at any other place, they will definitely mention it. Let's calculate the probability of different number of prime numbers:

0 primes Probability = (5/9)*(5/9)*(5/9)*(5/9)

1 prime Probability = (4/9)*(5/9)*(5/9)*(5/9)*4 (there are 4 positions for the prime)

2 primes Probability = (4/9)*(4/9)*(5/9)*(5/9)*4C2 (select 2 of the 4 positions for the primes)

3 primes Probability = (4/9)*(4/9)*(4/9)*(5/9)*4 (4 positions for the non prime)

4 primes Probability = (4/9)*(4/9)*(4/9)*(4/9)

Probability of at least 2 primes = 1 - (Probability of 0 prime + Probability of 1 prime) Probability of at least 2 primes = Probability of 2 primes + Probability of 3 primes + Probability of 4 primes

The calculations are painful so let's leave it here. As I said, their intention was probability of prime = 1/2, probability of composite = 1/2 which makes the calculations simple.

The 4 positions are different but you can have the same prime on one or more positions. When you say 4/9, you are including all cases (2, 3, 5, 7) so you don't need to account for them separately. When you say, (4/9)*(4/9)*(4/9)*(4/9), you are including all cases e.g. (2222, 2353, 3577, 2357 etc). All you need to do it separate out the primes and the non primes. That you do by arranging primes and non primes as NNPP or NPNP or PPNN etc (as we did above)

Thank you karishma, since there was no response for a while I thought my query was unreasonable, good to see at least some people appreciated what I wrote . Highly appreciate your response.

Great so now I know probability of one prime 4/9* 5/9 * 5/9 *5/9 * 4= 2000/6561

PROBABILITY of 0 prime's = 5/9*5/9*5/9*5/9= 625/6561

probability of at least 2 primes = 1- ( 2000/6561 + 625/6561) 1-(2625/6561) = 3936/6561 = .59

now lets check by doing long way

2 primes

4/9 * 4/9 * 5/9*5/9* 4!/2!2! = 2400/6561 ( PPNN : two primes could be same ,2 , 2 or different 2, 3 doesn't matter we can think of this as, two of a kind and two of another kind ,hence 4!/2!2!)

3 primes

4/9 *4/9 *4/9*5/9*4 = 1280/6561 (4!/3! = 4) Three of one kind again PPPN.

4 primes

4/9*4/9*4/9*4/9 = 256/6561 all of the same kind , hence only one way to select them. PPPP

total 3936/6561 = .59 Bingo!

hence we can see both ways we are getting the same result.

Karishma if there is any error in my understanding, please do point out, I have considered primes as one kind and non primes as another kind instead of the position logic which you have mentioned. _________________

Re: John wrote a phone number on a note that was later lost [#permalink]
31 Jul 2014, 06:14

jlgdr wrote:

Nice problem but I think we should change the language just to clarify that 1 can't be used again even if it appears on te last 4 digits.

Thanks mods

Cheers! J

I agree that we should change the language so that the question will be clearer than now. At first I also assume that 1 appears once within 3 last places. _________________

Re: John wrote a phone number on a note that was later lost [#permalink]
14 Aug 2014, 13:22

Hi! I have a question here. What does is wrong if we do this in this way? Outcome wanted/Total outcome. This means:

Outcome wanted: 4*4*8*8*1*1*1. I have in the first and in the second digit, 4 possibles numbers (just primes). For the third and fourth digit are 8 possible numbers (2,3,4,5,6,7,8 and 9). Total outcome: 8*8*8*8*1*1*1

Re: John wrote a phone number on a note that was later lost [#permalink]
14 Aug 2014, 18:38

gbascurs wrote:

Hi! I have a question here. What does is wrong if we do this in this way? Outcome wanted/Total outcome. This means:

Outcome wanted: 4*4*8*8*1*1*1. I have in the first and in the second digit, 4 possibles numbers (just primes). For the third and fourth digit are 8 possible numbers (2,3,4,5,6,7,8 and 9). Total outcome: 8*8*8*8*1*1*1

Please help!

Hey gbascurs, The question ask "What is the probability that the phone number contains at least two prime digits?". At least mean: 2, 3, or 4 primes digits appears in the phone number. You only include one case. Moreover, the two primes can be in the third and fourth digit, not just in the first and second digit.

Re: John wrote a phone number on a note that was later lost [#permalink]
14 Aug 2014, 19:35

LucyDang wrote:

gbascurs wrote:

Hi! I have a question here. What does is wrong if we do this in this way? Outcome wanted/Total outcome. This means:

Outcome wanted: 4*4*8*8*1*1*1. I have in the first and in the second digit, 4 possibles numbers (just primes). For the third and fourth digit are 8 possible numbers (2,3,4,5,6,7,8 and 9). Total outcome: 8*8*8*8*1*1*1

Please help!

Hey gbascurs, The question ask "What is the probability that the phone number contains at least two prime digits?". At least mean: 2, 3, or 4 primes digits appears in the phone number. You only include one case. Moreover, the two primes can be in the third and fourth digit, not just in the first and second digit.

Hope it clears.

Hi LucyDang,

Why I am trying to do now is:

Probability of 2 primes = 4*4*8*8*1*1*1 /8*8*8*8*1*1*1 + Probability of 3 primes = 4*4*4*8*1*1*1 /8*8*8*8*1*1*1 + Probability of 4 primes = 4*4*4*4*1*1*1 /8*8*8*8*1*1*1 Probability = 7/16

Re: John wrote a phone number on a note that was later lost [#permalink]
15 Aug 2014, 06:31

LucyDang wrote:

gbascurs wrote:

Hi LucyDang,

Why I am trying to do now is:

Probability of 2 primes = 4*4*8*8*1*1*1 /8*8*8*8*1*1*1 + Probability of 3 primes = 4*4*4*8*1*1*1 /8*8*8*8*1*1*1 + Probability of 4 primes = 4*4*4*4*1*1*1 /8*8*8*8*1*1*1 Probability = 7/16

I cannot see where is the mistake. Thanks!!

Hey gbascurs,

There are overlap/incorrect calculation here: Probability of 2 primes = 4*4*8*8*1*1*1 /8*8*8*8*1*1*1 --> 4*4*8*8*1*1*1: there are 4 primes each placed in 3rd and 4th digit Probability of 3 primes = 4*4*4*8*1*1*1 /8*8*8*8*1*1*1 --> 4*4*4*8*1*1*1: there are 4 primes placed in 4th digit. You have to calculate the case in which same two primes appear, then you need to subtract the primes from the total numbers. It takes lots of time to do so, so please use the P(desire) = 1-P(opposite) to get the correct answer under time pressure.

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