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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

$$a)\frac{7}{12}\times \frac{6}{11}\times \frac{5}{10}=\frac{7}{44}$$$$b)\frac{1}{2}\times \frac{1}{2}\times \frac{1}{6}=\frac{1}{24}$$

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

$$a)totalnumberofarticles=12=n\left(s\right)$$$$numberofdefectivearticles=5$$$$\Rightarrow numberofnon-defectivearticles=7=n\left(a\right)$$$$becauseeverytimeweselectwithout$$$$insertingoneisreducedfromrthe$$$$previousnumber.$$$$firsttime:n\left(s\right)=12;n\left(a\right)=7\Rightarrow p=\frac{7}{12}$$$$secondtime:n\left(s\right)=11;n\left(a\right)=6\Rightarrow p=\frac{6}{11}$$$$thirdtime;n\left(s\right)=10;n\left(a\right)=5\Rightarrow p=\frac{6}{10}$$$$\Rightarrow answer=multiplication$$$$b)$$$$eachcoinhas2statesamdeach$$$$dicehas6states.$$$$firstcoin:p_1=\frac{1}{2}\mathrm{\&}secondcoin:p_2=\frac{1}{2}\mathrm{\&}coin:p_3=\frac{1}{6}$$$$\Rightarrow answer=multiplication$$

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