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

The probability that at least one of  the events A and B occurs is 0.7 and  they occur simultaneously with  probability 0.2. Then P(A^− )+P(B^� ) =

$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{occurs}\:\mathrm{is}\:\mathrm{0}.\mathrm{7}\:\mathrm{and} \\ $$$$\mathrm{they}\:\mathrm{occur}\:\mathrm{simultaneously}\:\mathrm{with} \\ $$$$\mathrm{probability}\:\mathrm{0}.\mathrm{2}.\:\mathrm{Then}\:\mathrm{P}\left(\overset{−} {\mathrm{A}}\right)+\mathrm{P}\left(\bar {\mathrm{B}}\right)\:= \\ $$

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

p(A∪B)=0.7  p(A∩B)=0.2  p(A∪B)=p(A)+p(B)−p(A∩B)⇒  0.7=p(A)+p(B)−0.2⇒  p(A)+p(B)=0.9  (1−p(A))+(1−p(B))=−0.9+2  ⇒p(A^− )+p(B^− )=1.1

$${p}\left({A}\cup{B}\right)=\mathrm{0}.\mathrm{7} \\ $$$${p}\left({A}\cap{B}\right)=\mathrm{0}.\mathrm{2} \\ $$$${p}\left({A}\cup{B}\right)={p}\left({A}\right)+{p}\left({B}\right)−{p}\left({A}\cap{B}\right)\Rightarrow \\ $$$$\mathrm{0}.\mathrm{7}={p}\left({A}\right)+{p}\left({B}\right)−\mathrm{0}.\mathrm{2}\Rightarrow \\ $$$${p}\left({A}\right)+{p}\left({B}\right)=\mathrm{0}.\mathrm{9} \\ $$$$\left(\mathrm{1}−{p}\left({A}\right)\right)+\left(\mathrm{1}−{p}\left({B}\right)\right)=−\mathrm{0}.\mathrm{9}+\mathrm{2} \\ $$$$\Rightarrow{p}\left(\overset{−} {{A}}\right)+{p}\left(\overset{−} {{B}}\right)=\mathrm{1}.\mathrm{1} \\ $$

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

Thanks but the question asks us to find  P(A^− )+P(B^− ).

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

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

Yes.

$$\mathrm{Yes}. \\ $$

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

It is a sum. So I think it is accepted.

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

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

0≤p(A^− )≤1  0≤p(B^− )≤1  0≤p(A^− )+p(B^− )≤2

$$\mathrm{0}\leqslant{p}\left(\overset{−} {{A}}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant{p}\left(\overset{−} {{B}}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant{p}\left(\overset{−} {{A}}\right)+{p}\left(\overset{−} {{B}}\right)\leqslant\mathrm{2} \\ $$

Commented by Her_Majesty last updated on 31/Aug/20

P(A^− )+P(B^− )≠P(A^− ∧B^− )  the sum of 2 independent probabilities is  a senseless value.  i.e. the probability of a thrown dice showing  not 6 is (5/6), the one of a thrown coin showing  head is (1/2) ⇒ the sum is (4/3)≈133%

$${P}\left(\overset{−} {{A}}\right)+{P}\left(\overset{−} {{B}}\right)\neq{P}\left(\overset{−} {{A}}\wedge\overset{−} {{B}}\right) \\ $$$${the}\:{sum}\:{of}\:\mathrm{2}\:{independent}\:{probabilities}\:{is} \\ $$$${a}\:{senseless}\:{value}. \\ $$$${i}.{e}.\:{the}\:{probability}\:{of}\:{a}\:{thrown}\:{dice}\:{showing} \\ $$$${not}\:\mathrm{6}\:{is}\:\frac{\mathrm{5}}{\mathrm{6}},\:{the}\:{one}\:{of}\:{a}\:{thrown}\:{coin}\:{showing} \\ $$$${head}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{the}\:{sum}\:{is}\:\frac{\mathrm{4}}{\mathrm{3}}\approx\mathrm{133\%} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com