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Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 04:42
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Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is A. 341/12^6 B. 352/12^6 C. 352/12^5 D. 341/12^5 E. 371/12^6
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Re: Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 09:05
subhashghosh wrote: @fluke, please explain this :
000000,111111(All Birthdays in the same month) 6 people have two favorable choices each; Let's say the choices were 0(JAN),1(FEB) for simplicity just as it would be for head and tail. Thus, total number of combinations would be: 2^6. 000000 000001 000010 000011 . . 111111 Total=2^6 As per the condition that all 6 must have EXACTLY two calender months as their birthdays, at least two people should have different birthdays. They all can't have 000000 i.e. JAN,JAN,JAN,JAN,JAN,JAN or 1,1,1,1,1,1 i.e. FEB,FEB,FEB,FEB,FEB,FEB. So, Favorable possibilities = 2^62=62.
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Re: Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 06:32
jamifahad wrote: Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is
A. 341/12^6 B. 352/12^6 C. 352/12^5 D. 341/12^5 E. 371/12^6 Sol: Assuming that any person having a birthday on any month to be \(\frac{1}{12}\). In reality; probability of a person having a birthday on Feb(28 days) is less than having a birthday on January(31 days). \(C^{12}_2*\frac{2^62}{12^6}=\frac{341}{12^5}\) \(C^{12}_2\)  Number of ways to choose 2 months out of 12\(2^62\)  2 favorable choices(months) available for every person excluding 000000,111111(All Birthdays in the same month)\(12^6\)  Every person has 12 Total possible months Ans: "D"
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Re: Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 08:54
@fluke, please explain this : 000000,111111(All Birthdays in the same month)
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Re: Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 17:13
jamifahad wrote: Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is
A. 341/12^6 B. 352/12^6 C. 352/12^5 D. 341/12^5 E. 371/12^6 Another slightly different way of approaching this (and a little more complicated than fluke's  which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before). Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F. Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc J  1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J) F  5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F) The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6  2 = 62\) ways (If you recognize it, great, else just calculate) So probability \(= 12C2 * 62/ 12^6 = 341/12^5\)
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Re: Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 22:43
VeritasPrepKarishma wrote: jamifahad wrote: Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is
A. 341/12^6 B. 352/12^6 C. 352/12^5 D. 341/12^5 E. 371/12^6 Another slightly different way of approaching this (and a little more complicated than fluke's  which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before). Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F. Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc J  1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J) F  5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F) The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6  2 = 62\) ways (If you recognize it, great, else just calculate) So probability \(= 12C2 * 62/ 12^6 = 341/12^5\) My understanding of your solution goes this way.Let me know if I'm wrong. 1. Selecting any two months out of the 12 available can be done in 12C2 ways. 2. The 6 days can be distributed in the 2 available months in the following ways M1 M2 0 6 1 5 2 4 3 3 4 2 5 1 6 0 which essentially is 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6, which is of the form (x+a)^6 = x^6 + x^5 6C1 a^1 + x^4 6C2 a^2 + x^3 6C3 a^3 + . . + a^6. Substituting 1 for 'x' and 'a' we arrive at (1+1)^6 = 1 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 1 2^6 = 2 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 i.e, 2^62 = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 62 Here, we subtract 2 because we are not to take into account either cases where all 6 birthdays fall in M1 or M2 Thus the probability becomes 12C2 * 62/ 12^6 = 341/12^5
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Re: Probability that the birthdays of 6 different persons will fall in exa
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28 Apr 2011, 22:59
Exactly as the binomial dist you have written! That is basically weeding out all the cases which have "all" 6 birthdays in one month i.e. 6C6 and 6C0



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Re: Probability that the birthdays of 6 different persons will fall in exa
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29 Apr 2011, 03:21
deepakh88 wrote: My understanding of your solution goes this way.Let me know if I'm wrong. 1. Selecting any two months out of the 12 available can be done in 12C2 ways. 2. The 6 days can be distributed in the 2 available months in the following ways
M1 M2 0 6 1 5 2 4 3 3 4 2 5 1 6 0
which essentially is 6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6, which is of the form (x+a)^6 = x^6 + x^5 6C1 a^1 + x^4 6C2 a^2 + x^3 6C3 a^3 + . . + a^6. Substituting 1 for 'x' and 'a' we arrive at (1+1)^6 = 1 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 1 2^6 = 2 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5 i.e, 2^62 = 6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 62 Here, we subtract 2 because we are not to take into account either cases where all 6 birthdays fall in M1 or M2
Thus the probability becomes 12C2 * 62/ 12^6 = 341/12^5
Yes, that's correct. Many of us have used binomial theorem extensively and know that: \(nC0 + nC1 + nC2 + ..... nCn = 2^n\) (an application of binomial as described by you) Hence, it can be done this way. On the other hand, you could obviously say that each of the six people have 2 options  either his bday falls in J or in F. So those are a total of \(2*2*2*2*2*2 = 2^6\) ways. Out of those, we need to remove 2 ways: All bdays fall in J and all bdays fall in F. Hence it can be done in \((2^6  2)\) ways.
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Re: Probability that the birthdays of 6 different persons will fall in exa
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04 Sep 2017, 23:15
jamifahad wrote: Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is
A. 341/12^6 B. 352/12^6 C. 352/12^5 D. 341/12^5 E. 371/12^6 Really mind boggling question, You can't solve it in one go unless and until you have solved such question before.. fluke & VeritasPrepKarishma very nice solution... It seems easy once we read the solution but it's a real tough question..
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Re: Probability that the birthdays of 6 different persons will fall in exa
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10 Sep 2017, 00:49
VeritasPrepKarishma wrote: jamifahad wrote: Probability that the birthdays of 6 different persons will fall in exactly 2 calendar months is
A. 341/12^6 B. 352/12^6 C. 352/12^5 D. 341/12^5 E. 371/12^6 Another slightly different way of approaching this (and a little more complicated than fluke's  which I think is a very appropriate solution. Nevertheless, it might be useful for people who have done permutation combination before). Select 2 months and then no of people for each month. You select two months in 12C2 ways. Say you get J and F. Then you say that J could have 1 person's bday, 2 people's bday, 3 people's bday etc J  1/2/3/4/5 (You select 1 or 2 or 3 or 4 or 5 people for J) F  5/4/3/2/1 (Corresponding to no of people selected for J, remaining people go in F) The distribution can be done in \(6C1 + 6C2 + 6C3 + 6C4 + 6C5 = 2^6  2 = 62\) ways (If you recognize it, great, else just calculate) So probability \(= 12C2 * 62/ 12^6 = 341/12^5\) Responding to a pm: Quote: How the calculation came to 341 as (12C2 ) 66*31 is not 341. Please help.
12C2 = 12*11/2 \(\frac{12C2 * 62}{12^6} = \frac{12*11*62}{2*12^6} = \frac{11*31}{12^5} = \frac{341}{12^5}\)
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Re: Probability that the birthdays of 6 different persons will fall in exa
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02 Jan 2018, 17:21
How did n(s) become 12*6
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