Question Number 170584 by Shrinava last updated on 27/May/22

Commented by mr W last updated on 27/May/22
![i don′t think you mean [AMSQ] ∨ [NSPC].](https://www.tinkutara.com/question/Q170585.png)
$${i}\:{don}'{t}\:{think}\:{you}\:{mean} \\ $$$$\left[{AMSQ}\right]\:\vee\:\left[{NSPC}\right]. \\ $$
Commented by Shrinava last updated on 27/May/22

$$\mathrm{Sorry}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{yes} \\ $$
Commented by mr W last updated on 27/May/22
![S lies always in [AMSQ] ∨ [NSPC]! please check the question again!](https://www.tinkutara.com/question/Q170589.png)
$${S}\:{lies}\:{always}\:{in}\:\left[{AMSQ}\right]\:\vee\:\left[{NSPC}\right]! \\ $$$${please}\:{check}\:{the}\:{question}\:{again}! \\ $$
Commented by Shrinava last updated on 27/May/22

$$\mathrm{Dear}\:\mathrm{professor}, \\ $$$$\mathrm{can}\:\mathrm{by}\:\mathrm{in}\:\mathrm{DQSP}\:\mathrm{or}\:\mathrm{PSNC}\:\mathrm{not}\:\mathrm{only}\:\mathrm{in} \\ $$$$\mathrm{AMSQ}\:\mathrm{or}\:\mathrm{NSPC}, \\ $$$$\mathrm{or}\:\mathrm{can}\:\mathrm{be}\:\mathrm{in}\:\mathrm{MBNS}, \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{in}\:\mathrm{4}\:\mathrm{quadrilaterals} \\ $$
Commented by mr W last updated on 27/May/22

Commented by mr W last updated on 27/May/22
![you can ask for the probability that S lies in [AMTQ].](https://www.tinkutara.com/question/Q170604.png)
$${you}\:{can}\:{ask}\:{for}\:{the}\:{probability}\:{that}\:{S} \\ $$$${lies}\:{in}\:\left[{AMTQ}\right]. \\ $$
Commented by mr W last updated on 27/May/22

Commented by mr W last updated on 27/May/22
![you can not ask for the probability that S lies in [ABSQ]. because it′s always 100%. since S lies always in [AMSQ].](https://www.tinkutara.com/question/Q170606.png)
$${you}\:{can}\:{not}\:{ask}\:{for}\:{the}\:{probability} \\ $$$${that}\:{S}\:{lies}\:{in}\:\left[{ABSQ}\right].\:{because}\:{it}'{s} \\ $$$${always}\:\mathrm{100\%}.\:{since}\:{S}\:{lies}\:{always}\:{in} \\ $$$$\left[{AMSQ}\right]. \\ $$
Commented by Shrinava last updated on 27/May/22
![If S∈Int[ABCD]-rectangle , S-fixed M∈(AB) , N∈(BC) , P∈(CD) , Q∈(DA) , AM=PC , BN=DQ then find the probability that a random point from Int[ABCD] lies in: [AMSQ] ∪ [NSPC]](https://www.tinkutara.com/question/Q170615.png)
$$\mathrm{If}\:\:\mathrm{S}\in\mathrm{Int}\left[\mathrm{ABCD}\right]-\mathrm{rectangle}\:,\:\mathrm{S}-\mathrm{fixed} \\ $$$$\mathrm{M}\in\left(\mathrm{AB}\right)\:,\:\mathrm{N}\in\left(\mathrm{BC}\right)\:,\:\mathrm{P}\in\left(\mathrm{CD}\right)\:, \\ $$$$\mathrm{Q}\in\left(\mathrm{DA}\right)\:,\:\mathrm{AM}=\mathrm{PC}\:,\:\mathrm{BN}=\mathrm{DQ} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{a} \\ $$$$\mathrm{random}\:\mathrm{point}\:\mathrm{from}\:\:\mathrm{Int}\left[\mathrm{ABCD}\right] \\ $$$$\mathrm{lies}\:\mathrm{in}:\:\:\left[\mathrm{AMSQ}\right]\:\cup\:\left[\mathrm{NSPC}\right] \\ $$
Commented by Shrinava last updated on 27/May/22

$$\mathrm{Dear}\:\mathrm{professor}\:\mathrm{sorry}\:\mathrm{was}\:\mathrm{corrected} \\ $$
Commented by mr W last updated on 27/May/22

Commented by mr W last updated on 27/May/22

Commented by mr W last updated on 27/May/22

$${no}\:{certain}\:{solution}! \\ $$
Commented by Shrinava last updated on 27/May/22

$$\mathrm{Dear}\:\mathrm{professor}, \\ $$$$\mathrm{so}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{again}? \\ $$
Commented by mr W last updated on 27/May/22

$${yes}. \\ $$$${the}\:{hatched}\:{area}\:{is}\:{not}\:{defined}. \\ $$
Commented by Shrinava last updated on 27/May/22

$$\mathrm{Dera}\:\mathrm{professor}, \\ $$$$\mathrm{exactly}\:\mathrm{how}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{given},\:\mathrm{please} \\ $$
Commented by mr W last updated on 28/May/22

$${for}\:{example}: \\ $$$${AM}={PC}=\frac{{AB}}{\mathrm{3}},\:{BN}={DQ}=\frac{{BC}}{\mathrm{2}} \\ $$
Commented by Shrinava last updated on 28/May/22

$$\mathrm{Dear}\:\mathrm{professor}, \\ $$$$\mathrm{if}\:\mathrm{possible}\:\mathrm{you}\:\mathrm{would}\:\mathrm{share}\:\mathrm{your} \\ $$$$\mathrm{wonderful}\:\mathrm{solution} \\ $$
Commented by mr W last updated on 28/May/22
![p=(([AMSQ]+[NSPC])/([ABCD])) =(([ABQ]+[BPC])/([ABCD])) =(1/2)×(1/2)+(1/2)×(1/3) =(5/(12))](https://www.tinkutara.com/question/Q170642.png)
$${p}=\frac{\left[{AMSQ}\right]+\left[{NSPC}\right]}{\left[{ABCD}\right]} \\ $$$$=\frac{\left[{ABQ}\right]+\left[{BPC}\right]}{\left[{ABCD}\right]} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{12}} \\ $$
Commented by Shrinava last updated on 28/May/22

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor} \\ $$