mathmath333
  • mathmath333
Probabilities that Rajesh passes in Physics, Math and Chemistry are p, m and c respectively. Of these subjects, Rajesh has 75% chance of passing in at least one, 50% chance of passing in at least two and 40% chance of passing in exactly two. Find which of the following is true.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{Probabilities that Rajesh passes in Physics, Math and Chemistry }\hspace{.33em}\\~\\ & \normalsize \text{are p, m and c respectively. Of these subjects, Rajesh has }\hspace{.33em}\\~\\ & \normalsize \text{75% chance of passing in at least one, 50% chance of passing}\hspace{.33em}\\~\\ & \normalsize \text{in at least two and 40% chance of passing in exactly two. Find}\hspace{.33em}\\~\\ & \normalsize \text{which of the following is true.}\hspace{.33em}\\~\\ & a.)\ p+m+c=\dfrac{19}{20} \hspace{.33em}\\~\\ & b.)\ p+m+c=\dfrac{27}{20} \hspace{.33em}\\~\\ & c.)\ pmc=\dfrac{1}{20} \hspace{.33em}\\~\\ & d.)\ pmc=\dfrac18 \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
i found this but cant understand the answer https://math.stackexchange.com/questions/1422953/probability-problem-possibly-based-on-principle-of-inclusion-exclusion
mathmate
  • mathmate
Yes, principle of inclusion and exclusion is your answer. Let's start with a venn diagram so that we're on the same wavelength. |dw:1446339769586:dw|

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mathmate
  • mathmate
|dw:1446339800040:dw|
mathmate
  • mathmate
|dw:1446339939304:dw|
ybarrap
  • ybarrap
.
anonymous
  • anonymous
D
mathmate
  • mathmate
So from the diagram, we see that \(P(P\cup M\cup C)\)=1-0.25=0.75 \(P((P\cup M) \cup (M\cap C) \cup (C\cap P))\)=0.50 \(P(P\cap M \cap C)\)=0.5-0.4=0.10 Apply the principle of inclusion/exclusion, we have \(P(P\cup M\cup C)\)=P(P)+P(M)+P(C)-\(P((P\cup M)\)-\(P (M\cap C)\)-\(P (C\cap P))\)+\(P(P\cap M \cap C)\) Solve for P(P)+P(M)+P(C) P(P)+P(M)+P(C) =\(P(P\cup M\cup C)\)+\(P((P\cup M)\)+\(P (M\cap C)\)+\(P (C\cap P))\)-\(P(P\cap M \cap C)\) =0.75+\(P((P\cup M)\)+\(P (M\cap C)\)+\(P (C\cap P))\)-0.1 However, \(P((P\cup M)\)+\(P (M\cap C)\)+\(P (C\cap P))\) =\(P((P\cup M) \cup P (M\cap C) \cup P (C\cap P))\)+2(P(P∩M∩C) =0.5+2(0.1)=0.7 so P(P)+P(M)+P(C) =0.75+0.7-0.1 =1.35 =27/20 (note: I should have worked with fractions all along, but fortunately the decimal equivalents are exact!)
ybarrap
  • ybarrap
pmc 1-000 2-001 3-010 4-011 5-100 6-101 7-110 8-111 |dw:1446350159631:dw| $$ 4\in{p}\\ 1\in{c}\\ 2\in{m}\\ 6\in{p,m}\\ 5\in{p,c}\\ 3\in{c,m}\\ 7\in{p,m,c}\\ p=\cup_ {4,5,6,7}\\ m=\cup_{2,3,6,7}\\ c=\cup_{1,3,5,7} $$ When counting all members of p, m and c \(independently\), members that show up in only two of the sets were counted twice, members that show up in all three sets were counted three times. This justifies the fact that $$ p+m + c\\ =P(\text{member appears in only 1 set}) + \\ 2P(\text{member appears in only 2 sets}) + \\ 3P(\text{member appears in only 3 sets})\\ =0.25+2\times0.4+3\times0.1=\cfrac{27}{20} $$ The probability is greater than 1 because there is overcounting of members belonging to multiple sets. When we counted here (i.e. p+m+c), we did not care that a member could also be in another set; so when the member showed up in another set, we counted it again. If we wanted to exclude duplicates, we would have used the inclusion-exclusion principle instead.
ybarrap
  • ybarrap
Corrections- |dw:1446357581611:dw| $$ 5\in{p}\\ 2\in{c}\\ 3\in{m}\\ 7\in{p,m}\\ 6\in{p,c}\\ 4\in{c,m}\\ 8\in{p,m,c}\\ p=\cup_ {5,6,7,8}\\ m=\cup_{3,4,7,8}\\ c=\cup_{2,4,6,8} $$

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