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 110826 by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}$$$$\mathrm{the}\mathrm{events}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{occurs}\mathrm{is}0.7\mathrm{and}$$$$\mathrm{they}\mathrm{occur}\mathrm{simultaneously}\mathrm{with}$$$$\mathrm{probability}0.2.\mathrm{Then}\mathrm{P}\left(\stackrel{-}{\mathrm{A}}\right)+\mathrm{P}\left(\overline{\mathrm{B}}\right)=$$

Commented by kaivan.ahmadi last updated on 30/Aug/20

$$p(A\cup B)=0.7$$$$p(A\cap B)=0.2$$$$p(A\cup B)=p\left(A\right)+p\left(B\right)-p(A\cap B)\Rightarrow$$$$0.7=p\left(A\right)+p\left(B\right)-0.2\Rightarrow$$$$p\left(A\right)+p\left(B\right)=0.9$$$$(1-p\left(A\right))+(1-p\left(B\right))=-0.9+2$$$$\Rightarrow p\left(\stackrel{-}{A}\right)+p\left(\stackrel{-}{B}\right)=1.1$$

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Thanks}\mathrm{but}\mathrm{the}\mathrm{question}\mathrm{asks}\mathrm{us}\mathrm{to}\mathrm{find}$$$$\mathrm{P}\left(\stackrel{-}{\mathrm{A}}\right)+\mathrm{P}\left(\stackrel{-}{\mathrm{B}}\right).$$

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{Yes}.$$

Commented by Aina Samuel Temidayo last updated on 31/Aug/20

$$\mathrm{It}\mathrm{is}\mathrm{a}\mathrm{sum}.\mathrm{So}\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{is}\mathrm{accepted}.$$

Commented by kaivan.ahmadi last updated on 31/Aug/20

$$0⩽p\left(\stackrel{-}{A}\right)⩽1$$$$0⩽p\left(\stackrel{-}{B}\right)⩽1$$$$0⩽p\left(\stackrel{-}{A}\right)+p\left(\stackrel{-}{B}\right)⩽2$$

Commented by Her_Majesty last updated on 31/Aug/20

$$P\left(\stackrel{-}{A}\right)+P\left(\stackrel{-}{B}\right)\ne P(\stackrel{-}{A}\wedge \stackrel{-}{B})$$$$thesumof2independentprobabilitiesis$$$$asenselessvalue.$$$$i.e.theprobabilityofathrowndiceshowing$$$$not6is\frac{5}{6},theoneofathrowncoinshowing$$$$headis\frac{1}{2}\Rightarrow thesumis\frac{4}{3}\approx 133\mathrm{\%}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com