Question Number 7168 by Tawakalitu. last updated on 14/Aug/16
$${I}\:{have}\:{five}\:{envelopes}\:{numbered}\:\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7}\:{all}\:{hidden}\:{in}\:{a} \\ $$$${box},\:{i}\:{picked}\:{an}\:{envelope}\:.\:\:{If}\:{its}\:{prime}\:{then}\:{i}\:{get}\:{the}\: \\ $$$${square}\:{of}\:{that}\:{number}\:{in}\:{Naira}.\:{Otherwise}\:\left({without}\:\right. \\ $$$$\left.{replacement}\right)\:{i}\:{picked}\:{another}\:{envelope}\:{and}\:{then}\:{get}\:{the} \\ $$$${sum}\:{of}\:{the}\:{squares}\:{of}\:{the}\:{two}\:{numbers}\:{picked}\:\left({in}\:{Naira}\right) \\ $$$${what}\:{is}\:{the}\:{chance}\:{of}\:{me}\:{getting}\:\:{N}\mathrm{25}\:\:? \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 14/Aug/16
$${Yes}\:{correct}\:.....\:{Thank}\:{you}\:{so}\:{much}. \\ $$
Commented by Rasheed Soomro last updated on 14/Aug/16
![I can get N25 only in the following two cases: (i) If I picked 5 [∵ 5^2 =25] (ii) I f I picked 4 and 3 in succession[∵ 4^2 +3^2 =25] In case (i) the chance of getting 5 is (1/5) In case (ii) picking 4 has chance (1/5) and after picking 4, picking 3 has (1/4) probablity. Therefore picking 4 and 3 in succession has probablity (1/5)×(1/4)=(1/(20)) Probablity of (picking 5) or (4 and 3 in succession) will be (1/5)+(1/(20))=((4+1)/(20))=(5/(20))=(1/4)???](Q7179.png)
$${I}\:{can}\:{get}\:{N}\mathrm{25}\:\:{only}\:{in}\:{the}\:{following}\:{two}\:{cases}: \\ $$$$\left({i}\right)\:{If}\:{I}\:{picked}\:\mathrm{5}\:\:\:\left[\because\:\:\mathrm{5}^{\mathrm{2}} =\mathrm{25}\right] \\ $$$$\left({ii}\right)\:{I}\:{f}\:\:{I}\:{picked}\:\mathrm{4}\:{and}\:\mathrm{3}\:{in}\:{succession}\left[\because\:\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} =\mathrm{25}\right] \\ $$$${In}\:{case}\:\left({i}\right)\:{the}\:{chance}\:{of}\:{getting}\:\mathrm{5}\:{is}\:\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$ \\ $$$${In}\:{case}\:\left({ii}\right)\:{picking}\:\mathrm{4}\:{has}\:{chance}\:\frac{\mathrm{1}}{\mathrm{5}}\:{and}\:{after} \\ $$$${picking}\:\mathrm{4},\:{picking}\:\mathrm{3}\:{has}\:\frac{\mathrm{1}}{\mathrm{4}}\:{probablity}. \\ $$$${Therefore}\:{picking}\:\mathrm{4}\:{and}\:\mathrm{3}\:{in}\:{succession}\:{has} \\ $$$${probablity}\:\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{20}} \\ $$$$ \\ $$$${Probablity}\:{of}\:\left({picking}\:\mathrm{5}\right)\:{or}\:\:\left(\mathrm{4}\:{and}\:\mathrm{3}\:{in}\:{succession}\right) \\ $$$${will}\:{be}\:\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{20}}=\frac{\mathrm{4}+\mathrm{1}}{\mathrm{20}}=\frac{\mathrm{5}}{\mathrm{20}}=\frac{\mathrm{1}}{\mathrm{4}}??? \\ $$$$ \\ $$