Find the probability of having 53 fridays in a year
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A leap year has =366 days=52\xa0weeks + 2 extra\xa0daysSo total outcomes will be [(Mon, Tue),(Tue, Wed),(Wed, Thu),(Thu, Fri),(Fri, Sat),(Sat, Sun),(Sun, Mon)]=7So Friday can fall on any one of the 2 extra days i.e,(Thu, Fri),(Fri, Sat) hence total no of Fridays will become 53{tex}Proabibilty\\;of\\;the\\;event=\\frac{Number\\;of\\;favourble\\;outcomes}{Total\\;number\\;of\\;possible\\;outcomes}{/tex}The probability of getting 53 Fridays in a leap year ={tex}2 \\over 7{/tex}