What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? -Maths 9th

What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? -Maths 9th
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Answer :

A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week. Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}. ∴ Number of possible outcomes n(S) = 7 (As there are seven days in a week) Let A : Getting the extra day as Sunday or Tuesday or Thursday ⇒ A = {Sunday, Tuesday, Thursday} ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{7}.\)
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Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

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