Alright boys and girls, if you had to choose between Pepsi and Coke, explain why you choose Pepsi?

Alright boys and girls, if you had to choose between Pepsi and Coke, explain why you choose Pepsi?
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1 Answer

Answer :

They run out of Coke
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Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

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Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

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Last Answer : Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the ... {4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)

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Last Answer : (b) \(rac{1}{462}\)Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the ... )= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).

6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

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Last Answer : (d) \(rac{1}{132}\)Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the ... ) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).

A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

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Last Answer : (c) \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : ... )2∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

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There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

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Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2