Question and Answers Forum

All Questions      Topic List

Probability and Statistics Questions

Previous in All Question      Next in All Question      

Previous in Probability and Statistics      Next in Probability and Statistics      

Question Number 114875 by bobhans last updated on 21/Sep/20

$$Amansent7letterstohis7friend.$$$$thelettersarekeptinaddressedenvelopes$$$$atrandom.theprobabilitythat3friends$$$$receivecorrectlettersand4lettersgo$$$$towrongdestinationis\mathrm{\_}$$$$$$$$\left(oldquestionunanswered\right)$$

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

$$\mathrm{diarrangement}\mathrm{of}4\mathrm{letters}=!4$$$$3\mathrm{letters}\mathrm{are}\mathrm{in}\mathrm{correct}\mathrm{address}$$$$\mathrm{this}3\mathrm{letters}\mathrm{can}\mathrm{be}\mathrm{choose}\mathrm{in}{\mathrm{C}}_3^7\mathrm{ways}$$$$\therefore \mathrm{The}\mathrm{required}\mathrm{probability}$$$$=\frac{!4\times {\mathrm{C}}_3^7}{7!}=\frac{1}{16}$$$$\mathbf{please}\mathbf{check}$$

Commented by mr W last updated on 21/Sep/20

$$correct!$$

Commented by PRITHWISH SEN 2 last updated on 21/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

Commented by mr W last updated on 21/Sep/20

$$pleasetryasimilarquestion:$$$$Q111906$$

Commented by john santu last updated on 21/Sep/20

$$whatdefinitionof!4?$$$$andwhatformula!n$$

Commented by PRITHWISH SEN 2 last updated on 22/Sep/20

$$!\mathbf{n}=\mathbf{n}![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+.......+{(-1)}^{\mathrm{n}}\frac{1}{\mathrm{n}!}]$$

Commented by mr W last updated on 27/Sep/20

$$!nisthenumberofderangements$$$$fromnelements.$$

Commented by PRITHWISH SEN 2 last updated on 22/Sep/20

$$\mathrm{sir}\mathrm{please}\mathrm{check}\mathrm{Q}111906$$

Answered by soumyasaha last updated on 22/Sep/20

$$$$$${}^7{\mathrm{C}}_3\times 4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})\mathbf{÷}7!$$$$=35\times 9\mathbf{÷}5040=\frac{1}{16}$$$$=$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com