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

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

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

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

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

$$\mathbf{T}{\mathrm{hank}}^{\prime }\mathrm{s}\mathrm{sir}$$

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