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Question Number 147906 by Odhiambojr last updated on 24/Jul/21

4 bad apples are mixed accidentally with 20 good apples. Obtain the probability distribution of the numbers of bad apples in a draw of 2 apples at random

$$\mathrm{4}\:{bad}\:{apples}\:{are}\:{mixed}\:{accidentally} \\ $$$${with}\:\mathrm{20}\:{good}\:{apples}.\:{Obtain}\:{the} \\ $$$$\:{probability}\:{distribution}\:{of}\:{the}\: \\ $$$${numbers}\:{of}\:{bad}\:{apples}\:{in}\:{a}\:{draw}\:{of}\: \\ $$$$\mathrm{2}\:{apples}\:{at}\:{random} \\ $$

Answered by Olaf_Thorendsen last updated on 24/Jul/21

Let X be the probability of bad apples in a draw. Then X is a random variable with values 0, 1 and 2. Total apples are 4 Bad apples (B) + 20 Good apples (G) = 24. • P(X=0) is the probability of no bad apple in a draw of two apples. P(X=0) = (C_2 ^(20) /C_2 ^(24) ) = ((20×19)/(24×23)) = ((95)/(138)) • Similary P(X=1) = ((C_1 ^4 C_1 ^(20) )/C_2 ^(24) ) = ((40)/(138)) • P(X=2) = (C_2 ^4 /C_2 ^(24) ) = ((4×3)/(24×23)) = (3/(138)) Therefore, the probability distribution of X is : { ((P(X=0) = ((95)/(138)))),((P(X=1) = ((40)/(138)))),((P(X=2) = (3/(138)))) :}

$$\mathrm{Let}\:\mathrm{X}\:\mathrm{be}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{bad}\:\mathrm{apples} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{draw}.\:\mathrm{Then}\:\mathrm{X}\:\mathrm{is}\:\mathrm{a}\:\mathrm{random} \\ $$$$\mathrm{variable}\:\mathrm{with}\:\mathrm{values}\:\mathrm{0},\:\mathrm{1}\:\mathrm{and}\:\mathrm{2}. \\ $$$$ \\ $$$$\mathrm{Total}\:\mathrm{apples}\:\mathrm{are}\:\mathrm{4}\:\mathrm{Bad}\:\mathrm{apples}\:\left(\mathrm{B}\right)\:+ \\ $$$$\mathrm{20}\:\mathrm{Good}\:\mathrm{apples}\:\left(\mathrm{G}\right)\:=\:\mathrm{24}. \\ $$$$ \\ $$$$\bullet\:\mathrm{P}\left(\mathrm{X}=\mathrm{0}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{no}\:\mathrm{bad} \\ $$$$\mathrm{apple}\:\mathrm{in}\:\mathrm{a}\:\mathrm{draw}\:\mathrm{of}\:\mathrm{two}\:\mathrm{apples}. \\ $$$$\:\mathrm{P}\left(\mathrm{X}=\mathrm{0}\right)\:=\:\frac{\mathrm{C}_{\mathrm{2}} ^{\mathrm{20}} }{\mathrm{C}_{\mathrm{2}} ^{\mathrm{24}} }\:=\:\frac{\mathrm{20}×\mathrm{19}}{\mathrm{24}×\mathrm{23}}\:=\:\frac{\mathrm{95}}{\mathrm{138}} \\ $$$$ \\ $$$$\bullet\:\mathrm{Similary}\:\mathrm{P}\left(\mathrm{X}=\mathrm{1}\right)\:=\:\frac{\mathrm{C}_{\mathrm{1}} ^{\mathrm{4}} \mathrm{C}_{\mathrm{1}} ^{\mathrm{20}} }{\mathrm{C}_{\mathrm{2}} ^{\mathrm{24}} }\:=\:\frac{\mathrm{40}}{\mathrm{138}} \\ $$$$\bullet\:\mathrm{P}\left(\mathrm{X}=\mathrm{2}\right)\:=\:\frac{\mathrm{C}_{\mathrm{2}} ^{\mathrm{4}} }{\mathrm{C}_{\mathrm{2}} ^{\mathrm{24}} }\:=\:\frac{\mathrm{4}×\mathrm{3}}{\mathrm{24}×\mathrm{23}}\:=\:\frac{\mathrm{3}}{\mathrm{138}} \\ $$$$\mathrm{Therefore},\:\mathrm{the}\:\mathrm{probability}\:\mathrm{distribution} \\ $$$$\mathrm{of}\:\mathrm{X}\:\mathrm{is}\:: \\ $$$$\begin{cases}{\mathrm{P}\left(\mathrm{X}=\mathrm{0}\right)\:=\:\frac{\mathrm{95}}{\mathrm{138}}}\\{\mathrm{P}\left(\mathrm{X}=\mathrm{1}\right)\:=\:\frac{\mathrm{40}}{\mathrm{138}}}\\{\mathrm{P}\left(\mathrm{X}=\mathrm{2}\right)\:=\:\frac{\mathrm{3}}{\mathrm{138}}}\end{cases} \\ $$

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