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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
Q. Define natural numbers Ans. The natural numbers (also called the counting numbers) are the numbers that you "naturally" use for counting: 1,2,3,4,... The set of n
Find the generating function for the number of r-combinations of {3.a, 5.b, 2.c} Ans: Terms sequence is given as r-combinations of {3.a, 5.b, 2.c}. This can be writte
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How many integers satisfy (sqrt n- sqrt 8836)^2 Solution) sqrt 8836 = 94 , let sqrt n=x the equation becomes... (x-94)^2 (x-94)^2 - 1 (x-95)(x-93) hence 93 8649 the number o
In the figure, ABCD is a square inside a circle with centre O. The Centre of the square coincides with O & the diagonal AC is horizontal of AP, DQ are vertical & AP = 45 cm, DQ = 2
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Solve for x: 4 log x = log (15 x 2 + 16) Solution: x 4 - 15 x 2 - 16 = 0 (x 2 + 1)(x 2 - 16) = 0 x = ± 4 But log x is
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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