S N De Probability Solution | Short Answer Type Questions 10–17 | Step-by-Step Solutions
S N De Probability Solution – Short Answer Type Questions 10–17
Complete, exam-ready step-by-step solutions from the chapter Probability, covering independent events, complements, conditional probability and odds.
Short Answer Type Questions
Q10–Q17VSA Q10Independence in throwing two dice
Prove that in throwing two dice, the events ‘4 on the first die’ and ‘5 on the second die’ are statistically independent.
Solution
Let \(A\) be getting 4 on the first die and \(B\) be getting 5 on the second die. There are \(36\) equally likely ordered outcomes.
\[P(A)=\frac16,\quad P(B)=\frac16\]Only the outcome \((4,5)\) belongs to both events.
\[P(A\cap B)=\frac1{36}=\frac16\times\frac16=P(A)P(B)\]VSA Q11Testing dependence of three events
Eight equally likely outcomes \(e_1,e_2,\ldots,e_8\) are given. Let \(A=\{e_1,e_2,e_3,e_4\}\), \(B=\{e_3,e_4,e_5,e_6\}\), and \(C=\{e_3,e_4,e_7,e_8\}\). Examine the dependence or independence of \(A,B,C\).
Solution
Each event contains four outcomes, so
\[P(A)=P(B)=P(C)=\frac48=\frac12\]Pairwise intersections are:
\[A\cap B=A\cap C=B\cap C=\{e_3,e_4\}\]\[P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac28=\frac14\]For every pair, \(\frac14=\frac12\times\frac12\). Thus all three pairs are independent.
But \(A\cap B\cap C=\{e_3,e_4\}\), and
\[P(A\cap B\cap C)=\frac14\ne\frac18=P(A)P(B)P(C)\]VSA Q12Probabilities using complements
Given \(P(A)=a\), \(P(B)=b\), and \(P(A\cap B)=c\), find (i) \(P(A^c\cup B^c)\), (ii) \(P(A^c\cup B)\), (iii) \(P(A^c\cap B^c)\).
Solution
(i) By De Morgan’s law:
\[P(A^c\cup B^c)=1-P(A\cap B)=1-c\](ii) Since \((A^c\cup B)^c=A\cap B^c\):
\[\begin{aligned}P(A^c\cup B)&=1-P(A\cap B^c)\\&=1-[P(A)-P(A\cap B)]\\&=1-a+c\end{aligned}\](iii) By De Morgan’s law and the addition theorem:
\[\begin{aligned}P(A^c\cap B^c)&=1-P(A\cup B)\\&=1-[a+b-c]\\&=1-a-b+c\end{aligned}\]VSA Q13Finding related event probabilities
If \(P(A)=\frac13\), \(P(B)=\frac12\), and \(P(A\cap B)=\frac14\), find \(P(B^c)\), \(P(A\cap B^c)\), \(P(A\cup B)\), \(P(B\mid A)\), \(P(A^c\cap B^c)\), and \(P(A\cup B^c)\). State whether \(A,B\) are mutually exclusive, independent, equally likely, or exhaustive.
Solution
(i) Complement of B:
\[P(B^c)=1-\frac12=\frac12\](ii) A but not B:
\[P(A\cap B^c)=\frac13-\frac14=\frac1{12}\](iii) Union:
\[P(A\cup B)=\frac13+\frac12-\frac14=\frac7{12}\](iv) Conditional probability:
\[P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{1/4}{1/3}=\frac34\](v) Neither A nor B:
\[P(A^c\cap B^c)=1-\frac7{12}=\frac5{12}\](vi) A or not B:
\[P(A\cup B^c)=P(A)+P(B^c)-P(A\cap B^c)=\frac13+\frac12-\frac1{12}=\frac34\]Since \(P(A\cap B)=\frac14\ne0\), they are not mutually exclusive. Since \(\frac14\ne\frac13\cdot\frac12\), they are not independent. Since \(\frac13\ne\frac12\), they are not equally likely. Since \(P(A\cup B)=\frac7{12}\ne1\), they are not exhaustive.
VSA Q14Union and conditional probability
(i) Given \(P(E)=\frac13\), \(P(F)=\frac14\), \(P(E\cap F)=\frac16\), find \(P(E^c\cup F)\). (ii) If \(P(E)=0.6\), \(P(F)=0.3\), \(P(E\cap F)=0.2\), find \(P(E\mid F)\) and \(P(F\mid E)\).
Solution
(i) We use \(P(E^c\cup F)=1-P(E\cap F^c)\).
\[P(E\cap F^c)=\frac13-\frac16=\frac16\]\[P(E^c\cup F)=1-\frac16=\frac56\](ii) Using conditional probability:
\[P(E\mid F)=\frac{0.2}{0.3}=\frac23\]\[P(F\mid E)=\frac{0.2}{0.6}=\frac13\]VSA Q15Conditional probabilities and complements
Given \(P(A\mid B)=0.8\), \(P(B\mid A)=0.6\), and \(P(A^c\cup B^c)=0.7\), find \(P(A\mid B^c)\).
Solution
By De Morgan’s law,
\[P(A\cap B)=1-P(A^c\cup B^c)=0.3\]Using the two conditional probabilities:
\[P(B)=\frac{0.3}{0.8}=\frac38\]\[P(A)=\frac{0.3}{0.6}=\frac12\]Then
\[P(A\cap B^c)=P(A)-P(A\cap B)=\frac12-\frac3{10}=\frac15\]\[P(B^c)=1-\frac38=\frac58\]\[P(A\mid B^c)=\frac{1/5}{5/8}=\frac8{25}\]VSA Q16Testing independence and union probability
(i) Given \(P(A)=\frac37\), \(P(B)=\frac47\), and \(P(A\cup B)=\frac79\), find \(P(A\mid B)\), \(P(B\mid A)\). Are \(A,B\) independent? (ii) Evaluate \(P(A\cup B)\) if \(2P(A)=P(B)=\frac5{13}\) and \(P(A\mid B)=\frac25\).
Solution
(i) By the addition theorem,
\[\begin{aligned}P(A\cap B)&=\frac37+\frac47-\frac79\\&=1-\frac79=\frac29\end{aligned}\]\[P(A\mid B)=\frac{2/9}{4/7}=\frac7{18}\]\[P(B\mid A)=\frac{2/9}{3/7}=\frac{14}{27}\]Also \(P(A)P(B)=\frac{12}{49}\ne\frac29\), so the events are not independent.
(ii) We have
\[P(A)=\frac5{26},\qquad P(B)=\frac5{13}\]\[P(A\cap B)=P(A\mid B)P(B)=\frac25\cdot\frac5{13}=\frac2{13}\]\[\begin{aligned}P(A\cup B)&=\frac5{26}+\frac5{13}-\frac2{13}\\&=\frac5{26}+\frac3{13}=\frac{11}{26}\end{aligned}\]VSA Q17Odds and selection probabilities
(i) The odds against two events are \(2:7\) and \(7:5\), respectively. If the events are independent, find the probability that at least one occurs. (ii) Ramesh applies for posts A and B independently, with selection probabilities \(\frac16\) and \(\frac17\). Find the probability of selection for at least one post.
Solution
(i) Odds against \(A\) are \(2:7\), so \(P(A)=\frac79\) and \(P(A^c)=\frac29\). Odds against \(B\) are \(7:5\), so \(P(B)=\frac5{12}\) and \(P(B^c)=\frac7{12}\).
By independence,
\[\begin{aligned}P(A\cup B)&=1-P(A^c)P(B^c)\\&=1-\frac29\cdot\frac7{12}\\&=1-\frac7{54}=\frac{47}{54}\end{aligned}\](ii) Probability of not being selected for A is \(\frac56\), and for B it is \(\frac67\). Hence
\[\begin{aligned}P(\text{at least one selection})&=1-\frac56\cdot\frac67\\&=1-\frac57=\frac27\end{aligned}\]
