A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.

Let X be the random variable defined as the number of red balls. Then X = 0, 1 P(X = 0) = \(\frac{3}{4}\times\frac{2}{3}=\frac{6}{12}=\frac{1}{2}\) P(X = 1) = \(\frac{1}{4}\times\frac{3}{3}+\frac{3}{4}\times\frac{1}{3}=\frac{6}{12}=\frac{1}{2}\) Probability Distribution Table: X 0 1 P(X) \(\frac{1}{2}\) \(\frac{1}{2}\)

The required probability = P((The first is a red jack card and The second is a jack card) or (The first is a red non-jack card and The second is...

Correct option is (A) 9: 11 Mean of first 10 whole numbers \(=\frac{0+1+2+....+9}{10}\) \(=\frac{9\times10}{20}=\frac92\) Mean of first 10 natural numbers \(=\frac{1+2+3+....+10}{10}\) \(=\frac{10\times11}{20}=\frac{11}2\) \(\therefore\) Their ratio \(=\cfrac{\frac92}{\frac{11}2}\) \(=\frac9{11}\) = 9 : 11

I thought F or H = 80 is giving by n(F u H) = n(F) + n(H) - n(F n H)

Correct option is (A) 0 A month have at most 31 days in a calendar year. \(\therefore\) There are no month which have 32 days. \(\therefore\) Favourable outcomes = 0 Total month = 12 \(\therefore\) Probability that random...

Given that the first five natural numbers are: 1, 2, 3, 4, 5 Therefore, Mean \(=\frac{Sum\, of \, numbers}{Total\,numbers}\)  \(=\frac{1 + 2 + 3 + 4 + 5}{5}\) \(=\frac{15}{5}\) = 3

Given that the first 10 even natural numbers be 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 Mean \(=\frac{Sum\, of \,all \,numbers}{Total\,numbers}\)  \(=\frac{2 + 4 + 6 + 8 + 10...

For the given AP the first term a is 1, and common difference d is a difference of the second term and first term, which is 2 - 1 =...