A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th
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Answer :

(d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = n(red and diamond) = 13,n(B ∩ C) = n(diamond and jack) = 1,      (∵ There is only jack of diamonds) n(A ∩ C) = n(red and jack) = 2 (∵ There are two red jacks) n(A ∩ B ∩ C) = n(red, diamond, jack) = 1 ∴ P(A) = \(rac{26}{52}\),  P(B) = \(rac{13}{52}\),  P(C) = \(rac{4}{52}\), P(A\(\cap\)B) = \(rac{13}{52}\), P(B \(\cap\)C) = \(rac{1}{52}\),  P(A\(\cap\)C) = \(rac{2}{52}\), P(A\(\cap\)B\(\cap\)C) = \(rac{1}{52}\)∴ P(a red card or a diamond or a jack) = P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)= \(rac{26}{52}\) + \(rac{13}{52}\) + \(rac{4}{52}\) - \(\bigg(\)\(rac{13}{52}\) + \(rac{1}{52}\) + \(rac{2}{52}\)\(\bigg)\) + \(rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .
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What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Description : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).