what is the chance that a leap year selected at random contains 53 fridays?

• A leap year has 366 days, therefore 52 weeks(i.e. 52 fridays) and 2 days.
  
  The remaining 2 days may be any of the following :
  
  (i) Sunday and Monday
  
  (ii) Monday and Tuesday
  
  (iii) Tuesday and Wednesday
  
  (iv) Wednesday and Thursday
  
  (v) Thursday and Friday
  
  (vi) Friday and Saturday
  
  (vii) Saturday and Sunday
  
  For having 53 fridays in a year, one of the remaining 2 days must be a friday.
  
  n(S) = 7
  
  n(E) = 2
  
  P(E) = n(E) / n(S) = 2 / 7
• 10 years agoHelpfull: Yes(77) No(0)
• there are 366days in a leap year.so left 2 days may be friday out of 7 days hence 2/7 is the probablity
• 10 years agoHelpfull: Yes(4) No(2)
• no.of days in leap year=366=52weeks+2days.
  2days are consecutive and have 7 possibilities,2 combinations contain friday.
  so ans is 2/7.
• 10 years agoHelpfull: Yes(3) No(0)
• ans is 2/7
• 10 years agoHelpfull: Yes(2) No(2)
• Ans-n(S)=7,n(E)=2
  P(E)=n(E)/n(S)
  =2/7
• 10 years agoHelpfull: Yes(1) No(0)
• in one leap year ,there is 2 odd days
  and in one weeks 7 days
  chance of friday with two odd days(2)/no. of days in a weeks(7)
  ans 2/7
  
• 10 years agoHelpfull: Yes(1) No(0)