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Question Number 19393 by myintkhaing last updated on 10/Aug/17

$$\mathrm{Three}\mathrm{tennis}\mathrm{players}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{play}\mathrm{each}$$$$\mathrm{other}\mathrm{only}\mathrm{once}.\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{A}\mathrm{will}$$$$\mathrm{beat}\mathrm{B}\mathrm{is}\frac{3}{5},\mathrm{that}\mathrm{B}\mathrm{will}\mathrm{beat}\mathrm{C}\mathrm{is}\frac{2}{3},\mathrm{and}\mathrm{that}$$$$\mathrm{A}\mathrm{will}\mathrm{beat}\mathrm{C}\mathrm{is}\frac{5}{7}.\mathrm{Find}\left(1\right)\mathrm{the}\mathrm{probability}$$$$\mathrm{that}\mathrm{A}\mathrm{will}\mathrm{not}\mathrm{win}\mathrm{both}\mathrm{games}$$$$\left(2\right)\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{A}\mathrm{will}\mathrm{win}\mathrm{not}\mathrm{both}\mathrm{games}.$$

Answered by allizzwell23 last updated on 10/Aug/17

$$$$$$\mathrm{Pr}\left(\mathrm{A}\mathrm{win}\mathrm{B}\right)=\frac{3}{5}\mathrm{then}\mathrm{Pr}\left(\mathrm{A}\mathrm{lose}\mathrm{to}\mathrm{B}\right)=\frac{2}{5}$$$$\mathrm{Pr}\left(\mathrm{B}\mathrm{win}\mathrm{C}\right)=\frac{2}{3}\mathrm{then}\mathrm{Pr}\left(\mathrm{B}\mathrm{lose}\mathrm{to}\mathrm{C}\right)=\frac{1}{3}$$$$\mathrm{Pr}\left(\mathrm{A}\mathrm{win}\mathrm{C}\right)=\frac{5}{7}\mathrm{then}\mathrm{Pr}\left(\mathrm{A}\mathrm{lose}\mathrm{to}\mathrm{C}\right)=\frac{2}{7}$$$$\left(1\right)\mathrm{Pr}\left(\mathrm{A}\mathrm{lose}\mathrm{to}\mathrm{both}\mathrm{B}\mathrm{and}\mathrm{C}\right)=\frac{2}{5}\times \frac{2}{7}=\frac{4}{35}$$$$\left(1\right)\mathrm{Pr}\left(\mathrm{A}\mathrm{win}\mathrm{both}\mathrm{B}\mathrm{and}\mathrm{C}\right)=\frac{3}{5}\times \frac{5}{7}=\frac{3}{35}$$$$$$

Commented by myintkhaing last updated on 11/Aug/17

$$\mathrm{please}\mathrm{check}\mathrm{No}.\left(2\right)\mathrm{Tinkutara}$$

Commented by Tinkutara last updated on 11/Aug/17

$$\mathrm{Yes}\mathrm{your}\mathrm{solution}\mathrm{is}\mathrm{true}.\mathrm{Sorry}\mathrm{for}\mathrm{my}$$$$\mathrm{mistake}.$$

Commented by myintkhaing last updated on 11/Aug/17

$$\mathrm{Let}>\mathrm{means}\mathrm{beat}.(\mathrm{A}>\mathrm{B}\mathrm{means}\mathrm{A}\mathrm{will}\mathrm{beat}\mathrm{B})$$$$\mathrm{A}>\mathrm{B}\mid \mathrm{A}<\mathrm{B}$$$$\mathrm{B}>\mathrm{C}\mid \mathrm{B}<\mathrm{C}\mid \mathrm{B}>\mathrm{C}\mid \mathrm{B}<\mathrm{C}$$$$\mathrm{A}>\mathrm{C}\mid \mathrm{A}<\mathrm{C}\mid \mathrm{A}>\mathrm{C}\mid \mathrm{A}<\mathrm{C}\mid \mathrm{A}>\mathrm{C}\mid \mathrm{A}<\mathrm{C}\mid \mathrm{A}>\mathrm{C}\mid \mathrm{A}<\mathrm{C}$$$$\left(\mathrm{i}\right)\left(\mathrm{ii}\right)\left(\mathrm{iii}\right)\left(\mathrm{iv}\right)\left(\mathrm{v}\right)\left(\mathrm{vi}\right)\left(\mathrm{vii}\right)\left(\mathrm{viii}\right)$$$$\mathrm{number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}=8$$$$\mathrm{For}\mathrm{A}\mathrm{will}\mathrm{win}\mathrm{both}\mathrm{games},\mathrm{we}\mathrm{choose}\mathrm{event}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iii}\right)$$$$\mathrm{So},\mathrm{P}\left(\mathrm{A}\mathrm{will}\mathrm{win}\mathrm{both}\mathrm{games}\right)=\frac{3}{5}\times \frac{2}{3}\times \frac{5}{7}+\frac{3}{5}\times \frac{1}{3}\times \frac{5}{7}=\frac{3}{7}$$$$\therefore \mathrm{P}\left(\mathrm{A}\mathrm{will}\mathrm{not}\mathrm{win}\mathrm{both}\mathrm{games}\right)=1-\mathrm{P}\left(\mathrm{A}\mathrm{will}\mathrm{win}\mathrm{both}\mathrm{games}\right)$$$$=\frac{4}{7}$$$$\left(\mathrm{OR}\right)$$$$\mathrm{P}\left(\mathrm{A}\mathrm{will}\mathrm{win}\mathrm{not}\mathrm{both}\mathrm{games}\right)=1-\mathrm{P}\left(\mathrm{A}\mathrm{will}\mathrm{win}\mathrm{both}\mathrm{games}\right)$$$$=\frac{4}{7}$$$$\mathrm{Which}\mathrm{solution}\mathrm{is}\mathrm{true}?\mathrm{Please}.$$

Commented by myintkhaing last updated on 12/Aug/17

$$\mathrm{Thank}\mathrm{you}.\mathrm{Are}\mathrm{you}\mathrm{used}\mathrm{facebook},\mathrm{Tinkutara}?$$

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