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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Feb 2008 15:31:36 IST
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Q) Find the P , that an year chosen at random has 53 sundays...
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1 Feb 2008 15:40:49 IST
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srry.........replied foolishly !! :(
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Who says nothing is impossible.
I've been doing nothing for years !!..............
I know KUNG FU KARATE
and 47 other dangerous words.............
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1 Feb 2008 15:47:57 IST
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edited
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1 Feb 2008 15:48:32 IST
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Let E= a year has 53 sundays.
L = the year is a leap year
NL= the year is not leap.
P(E)= P(L)P(E/L) + P(NL)P(E/NL)
= (1/4)(2/7) + ( 3/4) ( 1/7) = 5/28
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1 Feb 2008 15:50:08 IST
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hii can u tell the final answer?
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1 Feb 2008 15:52:29 IST
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P(E)= P(L)P(E/L) + P(NL)P(E/NL)
= (1/4)(2/7) + ( 3/4) ( 1/7) = 5/28
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1 Feb 2008 15:52:42 IST
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Well done Nadeemoidu.
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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1 Feb 2008 15:53:34 IST
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52 complete weeks means 364 days, probability that the 365th day is a sunday = 1/7. If it is a leap yr then the prob that the remaining 2 days have a sunday = 2/7 Probability that a year chosen at random is a leap yr = 1/4 .: Total prob = 3/4(1/7)+(1/4)(2/7)=5/28....(ans)
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1 Feb 2008 15:53:55 IST
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thats not correct
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1 Feb 2008 15:58:12 IST
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hey nadeem please explain how did you derive that formula
P(E)= P(L)P(E/L) + P(NL)P(E/NL)
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1 Feb 2008 15:58:57 IST
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joyfrancis--thanx
thats the correct ans
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1 Feb 2008 16:05:12 IST
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the way in which joy francis did is just the correct way.. these type of quest were there in class x. i think.....
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1 Feb 2008 16:25:08 IST
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if the year is non leap year then it has 365 days divide this by 7 we get quotient 52 and remainder 1 . thus the year has 52 sundays. the extra 1 day can be anyone of the 7 days.
thus
n(S)=7
LET A b the event of getting the 53rd day sunday
n(A)=1
P(A)=n(A)/n(S)
= 1/7.
if the year is leap year then there would be 366 days. if u calculate as above the extra two days can be,
S={ sun mon, mon tue, tue wed,......... sat sun}
n(S)=7
let b be the event of geeting sunday
B={sun mon, sat sun}
n(B)=2
P(B)= n(B)/n(S)
P(B)=2/7.
THE EVENTS A AND B ARE MUTUALLY XCLUSIVE
P( AUB) = P(A)+P(B)
=3/7
hope the ans is correct..................................
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2 Feb 2008 22:57:58 IST
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defination of leap year century year is not always leap year it must be divisble by 400
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2 Feb 2008 23:07:55 IST
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That's a good point.
So the answer will be
P(E)= P(L)P(E/L) + P(NL)P(E/NL)
= (99/400)(2/7) + ( 301/400) ( 1/7) = 499/2800
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