Math, asked by neel6673, 1 year ago

what is the probability that a non leap year selected at random will contain 53 sundays but not 53 monday​

Answers

Answered by Blaezii
4

Answer:

Step-by-step explanation:

A non-leap year has 365 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364 days .

365– 364 = 1day extra.

In a non-leap year there will be 52 Sundays and 1day will be left.

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.

Of these total 7 outcomes, the favourable outcomes are 1.

Hence the probability of getting 53 sundays = 1 / 7.

∴ probability of getting 52 sundays = 1 - 1/ 7 = 6 / 7.

Answered by Anonymous
0

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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