Find the probability that a leap year selected at random will have 53 Sundays
sreedhar2:
i want the answer too..
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There are 366 days in a leap year that contain 52 weeks and 2 more days. So, 52 Sundays and 2 days.
These 2 days can be:
(Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat, Sun} and {Sun, Mon} (7 cases).
In order to have 53 Sundays we should have either {Sat, Sun} or {Sun, Mon} case.
No. of sample spaces = 7.
No. of event that gives 53rd Sunday in Leap Year = 2.
Required Probability = 2⁄7.
These 2 days can be:
(Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat, Sun} and {Sun, Mon} (7 cases).
In order to have 53 Sundays we should have either {Sat, Sun} or {Sun, Mon} case.
No. of sample spaces = 7.
No. of event that gives 53rd Sunday in Leap Year = 2.
Required Probability = 2⁄7.
In probability theory, the sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment.
mean all possible outcomes??
outcomes means results.
yep... Thanks. :))
for example when you toss a coin, the possible outcomes are HT, HH, TT, TH. These outcomes make a sample space of tossing a coin.
hope you understood.
thank u
Your welcome.
the above results come when we toss two coins
yes. two coins.
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Answer:
In a leap year there are 366 days. In 366 days, we have 522 weeks and 2 days, Thus we can say that leap year ah always 52 Sundays.
The remaining two days can be
(i) Sunday and Mondays
(ii) Mondays and Tuesdays
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(VI) Friday and Saturday
(vii) Saturady and Sunday.
From above it is clear that there are 7 elementary events associated with this random experiment.
Clearly the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.
∴ P (E) = n (E)/n (S) = 2/7
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