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Question Number 27235 by das47955@mail.com last updated on 03/Jan/18

(1) There are 60 tickets in a bag numbered 1 through 60.If a ticket is picked at random,find the probability that the number is divisible by 3 or 4 ?

$$\left(\mathrm{1}\right)\:\:\boldsymbol{\mathrm{There}}\:\boldsymbol{\mathrm{are}}\:\mathrm{60}\:\boldsymbol{\mathrm{tickets}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{bag}} \\ $$$$\boldsymbol{\mathrm{numbered}}\:\mathrm{1}\:\boldsymbol{\mathrm{through}}\:\mathrm{60}.\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{ticket}}\: \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{picked}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{random}},\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\: \\ $$$$\boldsymbol{\mathrm{probability}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{number}}\:\boldsymbol{\mathrm{is}} \\ $$$$\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}}\:\mathrm{3}\:\boldsymbol{\mathrm{or}}\:\mathrm{4}\:? \\ $$

Answered by Rasheed.Sindhi last updated on 03/Jan/18

Let A is the set of tickets numbered by multiple of 3 n(A)=60/3 =20 and B is the set of tickets numbered by multiple of 4 n(B)=60/4 =15 ∴ A∩B is the set of tickets numbered by multiple of 12 n(A∩B)=60/12 =5 ∴ A∪B is the set of the tickets numbered by multiples of 3 or 4 n(A∪B)=n(A)+n(B)−n(A∩B) =20+15−5=30 If U is theUniversal set(Set of all the tickets numbered by 1 to 60) n(U)=60 Probability=((n(A∪B))/(n(U))) Probability=((30)/(60))=(1/2)

$$\mathrm{Let}\:\mathrm{A}\:\mathrm{is}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{tickets}\:\mathrm{numbered} \\ $$$$\mathrm{by}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\:\:\:\:\:\mathrm{n}\left(\mathrm{A}\right)=\mathrm{60}/\mathrm{3}\:=\mathrm{20} \\ $$$$\mathrm{and}\:\mathrm{B}\:\mathrm{is}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{tickets}\:\mathrm{numbered} \\ $$$$\mathrm{by}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{n}\left(\mathrm{B}\right)=\mathrm{60}/\mathrm{4}\:=\mathrm{15} \\ $$$$\therefore\:\mathrm{A}\cap\mathrm{B}\:\mathrm{is}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{tickets}\:\mathrm{numbered} \\ $$$$\mathrm{by}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{12} \\ $$$$\:\:\:\:\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right)=\mathrm{60}/\mathrm{12}\:=\mathrm{5} \\ $$$$\therefore\:\mathrm{A}\cup\mathrm{B}\:\mathrm{is}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tickets} \\ $$$$\mathrm{numbered}\:\mathrm{by}\:\mathrm{multiples}\:\mathrm{of}\:\mathrm{3}\:\mathrm{or}\:\mathrm{4} \\ $$$$\:\:\:\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\right)=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)−\mathrm{n}\left(\mathrm{A}\cap\mathrm{B}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{20}+\mathrm{15}−\mathrm{5}=\mathrm{30} \\ $$$$\mathrm{If}\:\mathbb{U}\:\mathrm{is}\:\mathrm{theUniversal}\:\mathrm{set}\left(\mathrm{Set}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{tickets}\:\mathrm{numbered}\:\mathrm{by}\:\mathrm{1}\:\mathrm{to}\:\mathrm{60}\right) \\ $$$$\:\:\:\:\:\mathrm{n}\left(\mathbb{U}\right)=\mathrm{60} \\ $$$$\mathrm{Probability}=\frac{\mathrm{n}\left(\mathrm{A}\cup\mathrm{B}\right)}{\mathrm{n}\left(\mathbb{U}\right)} \\ $$$$\mathrm{Probability}=\frac{\mathrm{30}}{\mathrm{60}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by das47955@mail.com last updated on 03/Jan/18

Thank′s sir

$$\boldsymbol{\mathrm{T}}\mathrm{hank}'\mathrm{s}\:\mathrm{sir} \\ $$

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