VeritasPrepKarishma wrote:
stne wrote:
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 )
\(\frac{4}{9} *\frac{4}{9} *\frac{4}{9} *\frac{4}{9} * 1,1,1\) ( exactly 4 primes )
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.
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