Probability
If A and B are any two events in a sample space S then $$P\left ( A\cup B \right )$$ is
The probability that a leap year selected at random contains either $$53$$ Sundays or $$53$$ Mondays, is
Total number of days in a leap year is $$366$$.
It will contain $$52$$ weeks and $$2$$ days.
These two days can be
$$S=\{(Sun,Mon); (Mon,Tues); (Tues,Wed); (Wed, Thurs); (Thurs,Fri); (Fri,Sat); (Sat,Sun)\}$$
Therefore $$n(S)=7$$
Let $$A$$ be the event of getting $$53$$ sundays.
Therefore $$A=\{(Sun,Mon);(Sat,Sun)\}$$
For $$53$$ Sundays , probability is $$P(A)=\dfrac{2}{7}$$
Let $$B$$ be the event of getting $$53$$ mondays.
Therefore $$B=\{(Sun,Mon);(Mon,Tues)\}$$
For $$53$$ Mondays , probability is $$P(B)=\dfrac{2}{7}$$
=27
This includes one ways where sunday and monday simultaneously Occur
(i.e) $$A\cap B= \{Sun, Mon\}$$
Probability for this is $$P(A\cap B)=\dfrac{1}{7}$$.
=17
Hence required probability that a leap year selected at random contain $$53$$ sundays or $$53$$ mondays is
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$=\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{1}{7}=\dfrac{2+2-1}{7}$$
Therefore the required probability is $$\dfrac{3}{7}$$
If A and B are any two events in a sample space S then $$P\left ( A\cup B \right )$$ is
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