Question Number 193226 by byaw last updated on 07/Jun/23
$$ \\ $$(a) 5 out of 12 articles are known to be defective. If three articles are picked, one after the other, without replacement, find the probability that all the three articles are non-defective.
(b) Two coins are tossed and a dice is thrown. What is the probability of obtaining a head, a tail and a 4?
Answered by MM42 last updated on 10/Jun/23
$$\left.{a}\right)\:\frac{\mathrm{7}}{\mathrm{12}}×\frac{\mathrm{6}}{\mathrm{11}}×\frac{\mathrm{5}}{\mathrm{10}}=\frac{\mathrm{7}}{\mathrm{44}} \\ $$$$\left.{b}\right)\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$
Commented by byaw last updated on 09/Jun/23
$$\mathrm{my}\:\mathrm{dear}\:\mathrm{I}\:\mathrm{do}\:\mathrm{not}\:\mathrm{understand} \\ $$
Commented by MM42 last updated on 10/Jun/23
$$\left.{a}\right){total}\:{number}\:{of}\:{articles}\:=\:\mathrm{12}={n}\left({s}\right) \\ $$$${number}\:{of}\:{defective}\:{articles}=\mathrm{5} \\ $$$$\Rightarrow{number}\:{of}\:{non}−{defective}\:{articles}=\mathrm{7}={n}\left({a}\right) \\ $$$${because}\:{every}\:{time}\:{we}\:{select}\:{without}\: \\ $$$${inserting}\:{one}\:{is}\:{reduced}\:{fromr}\:{the}\: \\ $$$${previous}\:{number}\:. \\ $$$${first}\:{time}\::\:{n}\left({s}\right)=\mathrm{12}\:\:;\:\:{n}\left({a}\right)=\mathrm{7}\Rightarrow{p}=\frac{\mathrm{7}}{\mathrm{12}} \\ $$$${second}\:{time}\::\:{n}\left({s}\right)=\mathrm{11}\:;\:{n}\left({a}\right)=\mathrm{6}\Rightarrow{p}=\frac{\mathrm{6}}{\mathrm{11}} \\ $$$${third}\:{time}\:;\:{n}\left({s}\right)=\mathrm{10}\:\:;\:\:{n}\left({a}\right)=\mathrm{5}\Rightarrow{p}=\frac{\mathrm{6}}{\mathrm{10}} \\ $$$$\Rightarrow{answer}={multiplication}\: \\ $$$$\left.{b}\right) \\ $$$${each}\:{coin}\:{has}\:\mathrm{2}\:{states}\:{amd}\:\:{each} \\ $$$${dice}\:{has}\:\:\mathrm{6}\:{states}\:. \\ $$$${first}\:{coin}\::\:{p}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:\&\:{second}\:\:{coin}\::\:{p}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\&\:\:{coin}\::\:{p}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{answer}={multiplication}\: \\ $$