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The probability that a leap year will have 53 sunday is ? and how please explain it ?(a)1/7 (b) 2/7 (c) 5/7 (d)6/7Sol)A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days. The remaining 2 days may be any of the following : (i) Sunday and Monday (ii) Monday and Tuesday (iii) Tuesday and Wednesday (iv) Wednesday and Thursday (v) Thursday and Friday (vi) Friday and Saturday (vii) Saturday and Sunday For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday. n(S) = 7 n(E) = 2 P(E) = n(E) / n(S) = 2 / 7
1) At a midway game at the state fair, the probability of winning an individual game is advertised to be 30% ( p = . 3). Suppose 50 people played the game (assume all 50 outcomes
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Solve the subsequent IVP. cos(x) y' + sin(x) y = 2 cos 3 (x) sin(x) - 1 y(p/4) = 3√2, 0 Solution : Rewrite the differential equation to determine the coefficient of t
Prove that A tree with n vertices has (n - 1) edges. Ans: From the definition of a tree a root comprise indegree zero and all other nodes comprise indegree one. There should
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Before going further, let us repeat an aspect of learning which is useful to keep in mind while formulating teaching strategies. A child who can add or subtract in the context of s
A non - leap year contains 365 days 52 weeks and 1 day more.i) We get 53 Sundays when the remaining day is Sunday.Number of days in week = 7∴ n(S) = 7Number of ways getting 53 Sundays.n(E) = 1n E 1n S 7=∴ Probability of getting 53 Sundays =1/7
A leap year consits of 366 days in those 364 days are completely 52 weeks so it contains 52 sundays remaining 2 days there are 2/7 probabilities are there.
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