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1) if 10 men, among whom are A & B, stand in a row, what is the probability that there are exactly three men between A and B?
2) What is the probability that a leap year selected at random will contain 53 Tuesdays? 3) What is the probability that a leap year selected at random will contain either 53 Sundays or 53 Tuesdays? |
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For question 1 , ans is 2/15
Iam trying others
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Arushi :))
QUest 2....ans is 2/7
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In reply to this post by Arushi :))
quest 1 is 2/15
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In reply to this post by Arushi :))
For second , maybe 2/7
And for third , 2/7 as well. Iam not sure for the second one. Let me know if you have the answers.
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Arushi :))
For second it could be 12/49 as well. Just check .
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Arushi :))
for 2 it is 2/7
and 3 it is 3/7. And for 1st it is 2/15. |
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anjali , can u explain me the first one.
I got the rest. explanation: In a leap year there are 52 full weeks i.e 364 days and 2 extra days. Now these full weeks may start from any of the seven days. Monday to monday.......... to monday or tuesday to tuesday...........to tuesday So, if we want 53 tuesdays, then either the week should be from monday to monday and first extra day would be then tuesday , probability of this is 1/7 or the week should be from sunday to sunday.. then first extra day would be monday and second extra day would be tuesday... so againg probability would be 1/7 . then we get 2/7 as the answer. |
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For third,
what i did is p( either 53 sundays or 53 mondays)= p( 53 sundays)+ p(53 mondays)- p( 53 sundays & 53 mondays) Now again p of 53 sundays = 2/7 , because the full weeks should start from saturday or sunday. p of 53 mondays = 2/7 for p of 53 Sundays & 53 mondays we have just one way. i.e . the full week must start from saturday . and that has probability of 1/7 this gives, 2/7+2/7-1/7 = 3/7 I saw the answers and then made this explanation :P @ ron/ anjali, is it correct?? |
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@Arushi
quest 1.. 10 men stand in a row A can occupy any 1 of 10 seats n B any 1 of remaining 9 therfore exhaustive cases 10*9=90 case 1: A in 1st n B in 5th position case 2: A in second n B in 6th . . . . case 6:A in 6th and B in 10th therfore total =2*6=12 prob=12/90=2/15 |
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thanks Ron, i got it
But i have a small doubt here, why arent we considering the total possible arrangements here, i.e total no of ways why are we not taking as 10! And favorable cases why not, 6*2* 3! * 5!. Please explain this that why are we not taking other persons as well. why are we just taking A & B in the total number and favorable cases...??? |
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In reply to this post by Arushi :))
Hey !
For second it will be 2/7 ( pakka-pakka sure :-p ) However Iam not convinced with third - why have you taken intersection as 1/7 ? Sunday and Tuesday can never occur consecutively , then how can we take its intersection as 1/7 ? Also , shall I explain 1st ques , or you are through with it ?
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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hey m sorryyyyyyyyyy. i mistyped the question :/
its either 53 sundays or 53 mondays... then intersection would be 1/7 if week starts from saturday . sorry ! and i got ron's explaination for first but can u explain me the doubt i have asked??? |
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First thing - don't complicate the question - keep your focus on just A and B
Favourable options would be those which are concerned with the seating arrangements where 3 people are in between. Total would be the cases that involve all seating patterns of A and B , irrespective of any condition. Now go through ron's explanation again , and let me know in case of any doubt. ![]()
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Arushi :))
@Arushi....you can try with factorials also but it will complicate a quest which we can solve without much problem..
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In reply to this post by Arushi :))
Arushi for ecotrix which book are you referring??
MA Economics
DSE 2014-16 |
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Can anyone tell me that for which all papers ecotrix is required ?
"Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth."
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In reply to this post by Arushi :))
Only for dse
MA Economics
DSE 2014-16 |
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In reply to this post by The Villain
Even i tried doing it with factorials and all. I am not getting with that, its getting complicated.
It was like this 12*8C3*3!*5! / 10! I am getting the same answer as yours. |
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for iii.
shoudnt the answer be 4/7 ? @arushi:) it says sunday or tuesday (not monday) |
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