document.write( "Question 1128275: A large box contains 365 tickets, one for each day of an ordinary year. Suppose one ticket is selected at random.\r
\n" ); document.write( "\n" ); document.write( "a)Find the probability that the selected day is in December.
\n" ); document.write( "b)Find the probability that the selected day is your birthday. (If you were born on February 29 on a leap year, the probability is 0 since there is no ticket for your birthday.)\r
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Algebra.Com's Answer #744744 by Quadratic1600(28)\"\" \"About 
You can put this solution on YOUR website!
Hi\r
\n" ); document.write( "\n" ); document.write( "Ok so for\r
\n" ); document.write( "\n" ); document.write( "a) We have to know how many days are in December - There are 31 days in December so the probability of selecting a day in December is 31/365 \r
\n" ); document.write( "\n" ); document.write( "b) Ok on to the birthday question \r
\n" ); document.write( "\n" ); document.write( "We only have one birthday a year so surely the probability is as simple as 1/365 meaning one day in 365 days. \r
\n" ); document.write( "\n" ); document.write( "Hope this gives some help
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