Differentiation (Part-38) | S N De

S N De Probability Solution – Independent & Mutually Exclusive Events

This post gives a complete S N De Probability Solution for Very Short Answer Type Questions Q10–Q12 (2 marks each) and Short Answer Type Questions Q1–Q9 (4 marks each) of the chapter Probability. Every answer is worked out step by step, ready to write in your exam.

Topics covered: independent events, mutually exclusive events, addition theorem, multiplication theorem, conditional probability and pairwise independence.

• VSA Q10
• VSA Q11
• VSA Q12
• SA Q1
• SA Q2
• SA Q3
• SA Q4
• SA Q5
• SA Q6
• SA Q7
• SA Q8
• SA Q9

Very Short Answer Type Questions

Q10 – Q12 • 2 marks each

VSA Q10 Probability of union of independent events

\(A\) and \(B\) are two independent events with \(P(A)=\dfrac{2}{5}\) and \(P(B)=\dfrac{1}{3}\); evaluate \(P(A\cup B)\).

Solution

Since \(A\) and \(B\) are independent, \(P(A\cap B)=P(A)P(B)=\dfrac{2}{5}\cdot\dfrac{1}{3}=\dfrac{2}{15}\). By the addition theorem,

\[ \begin{aligned} P(A\cup B)&=P(A)+P(B)-P(A\cap B)\\ &=\frac{2}{5}+\frac{1}{3}-\frac{2}{15}=\frac{6+5-2}{15}=\frac{9}{15} \end{aligned} \]

\(P(A\cup B)=\dfrac{3}{5}\)

VSA Q11 Finding probabilities of related events

\(A\) and \(B\) are two events, not mutually exclusive, connected with a random experiment \(E\). If \(P(A)=\dfrac14\), \(P(B)=\dfrac25\) and \(P(A\cup B)=\dfrac12\), find (i) \(P(A\cap B)\) (ii) \(P(A\cap B^c)\) (iii) \(P(A^c\cup B^c)\), where \(c\) stands for the complement.

Solution

(i) From \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\):

\[ P(A\cap B)=\frac14+\frac25-\frac12=\frac{5+8-10}{20}=\frac{3}{20} \]

(ii) \(A\) is the union of the disjoint events \(A\cap B\) and \(A\cap B^c\), so

\[ P(A\cap B^c)=P(A)-P(A\cap B)=\frac14-\frac{3}{20}=\frac{5-3}{20}=\frac{1}{10} \]

(iii) By De Morgan's law, \(A^c\cup B^c=(A\cap B)^c\). Hence

\[ P(A^c\cup B^c)=1-P(A\cap B)=1-\frac{3}{20}=\frac{17}{20} \]

(i) \(\dfrac{3}{20}\)   (ii) \(\dfrac{1}{10}\)   (iii) \(\dfrac{17}{20}\)

VSA Q12 Are the events independent?

If \(P(\bar A\cup\bar B)=\dfrac56\), \(P(A)=\dfrac12\) and \(P(\bar B)=\dfrac23\), are the events \(A\) and \(B\) independent?

Solution

Since \(\bar A\cup\bar B=\overline{A\cap B}\), we have \(P(A\cap B)=1-P(\bar A\cup\bar B)=1-\dfrac56=\dfrac16\).

Also \(P(B)=1-P(\bar B)=1-\dfrac23=\dfrac13\). Now

\[ P(A)\,P(B)=\frac12\cdot\frac13=\frac16=P(A\cap B) \]

Since \(P(A\cap B)=P(A)P(B)\), the events \(A\) and \(B\) are independent.

Short Answer Type Questions

Q1 – Q9 • 4 marks each

SA Q1 Independent vs mutually exclusive

Define independent and mutually exclusive events. Can two events be simultaneously independent and mutually exclusive? Give reasons.

Solution

• Mutually exclusive events: \(A\) and \(B\) are mutually exclusive if they cannot occur together, i.e. \(A\cap B=\varnothing\), so \(P(A\cap B)=0\).
• Independent events: \(A\) and \(B\) are independent if the occurrence of one does not affect the probability of the other, i.e. \(P(A\cap B)=P(A)\,P(B)\).

Suppose \(A\) and \(B\) are both mutually exclusive and independent. Then

\[ P(A)P(B)=P(A\cap B)=0 \;\Rightarrow\; P(A)=0 \ \text{or}\ P(B)=0 \]

Two events can be simultaneously independent and mutually exclusive only if at least one of them has probability zero. If both have non-zero probability, they cannot be both.

SA Q2 Positive-probability events

When are two events mutually exclusive? When are they mutually independent? Let \(A\) and \(B\) be any two events with \(P(A)\gt 0\) and \(P(B)\gt 0\). Show that \(A\) and \(B\) cannot be independent if they are mutually exclusive.

Solution

• \(A,B\) are mutually exclusive when \(A\cap B=\varnothing\), i.e. \(P(A\cap B)=0\).
• \(A,B\) are independent when \(P(A\cap B)=P(A)P(B)\).

Let \(A\) and \(B\) be mutually exclusive, so \(P(A\cap B)=0\). Since \(P(A)\gt 0\) and \(P(B)\gt 0\), we get \(P(A)P(B)\gt 0\). Therefore

\[ P(A\cap B)=0\neq P(A)P(B) \]

Hence mutually exclusive events with positive probabilities cannot be independent.

SA Q3 Multiplication rule for three events

\(A_1, A_2, A_3\) are three events. Show that \(P(A_1\cap A_2\cap A_3)=P(A_1)\,P(A_2/A_1)\,P[A_3/(A_1\cap A_2)]\). State under which condition \(P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3)\).

Solution

Let \(E=A_1\cap A_2\). By the multiplication theorem for two events,

\[ P(E\cap A_3)=P(E)\,P(A_3/E)=P(A_1\cap A_2)\,P[A_3/(A_1\cap A_2)] \]

and \(P(A_1\cap A_2)=P(A_1)\,P(A_2/A_1)\). Substituting,

\[ P(A_1\cap A_2\cap A_3)=P(A_1)\,P(A_2/A_1)\,P[A_3/(A_1\cap A_2)] \]

Condition: If \(A_1,A_2,A_3\) are mutually independent, then \(P(A_2/A_1)=P(A_2)\) and \(P[A_3/(A_1\cap A_2)]=P(A_3)\), and the result becomes

\[ P(A_1\cap A_2\cap A_3)=P(A_1)\,P(A_2)\,P(A_3) \]

SA Q4 Boole's inequality for two events

For any two events \(A\) and \(B\), prove that \(P(A\cup B)\le P(A)+P(B)\).

Solution

By the addition theorem,

\[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]

Since \(P(A\cap B)\ge 0\), subtracting it can only decrease the sum. Therefore

\[ P(A\cup B)\le P(A)+P(B) \]

Equality holds when \(P(A\cap B)=0\), i.e. when \(A\) and \(B\) are mutually exclusive.

SA Q5 Inequalities involving probabilities

For any two events \(A\) and \(B\) prove that:
(i) \(P(A)\ge P(A\cap B)\ge P(A)+P(B)-1\)
(ii) \(P(A\cap B)\le P(A)\le P(A\cup B)\le P(A)+P(B)\)
(iii) \(P(A/B)\lt P(B/A)\) if \(P(A)\lt P(B)\)

Solution

(i) Since \(A\cap B\subseteq A\), we get \(P(A)\ge P(A\cap B)\). Also \(P(A\cup B)\le 1\), so

\[ P(A\cap B)=P(A)+P(B)-P(A\cup B)\ \ge\ P(A)+P(B)-1 \]

Hence \(P(A)\ge P(A\cap B)\ge P(A)+P(B)-1\).

(ii) \(A\cap B\subseteq A\Rightarrow P(A\cap B)\le P(A)\), and \(A\subseteq A\cup B\Rightarrow P(A)\le P(A\cup B)\). By SA Q4, \(P(A\cup B)\le P(A)+P(B)\). Combining,

\[ P(A\cap B)\le P(A)\le P(A\cup B)\le P(A)+P(B) \]

(iii) We have

\[ \begin{aligned} P(A/B)&=\frac{P(A\cap B)}{P(B)}\\ P(B/A)&=\frac{P(A\cap B)}{P(A)} \end{aligned} \]

Given \(0\lt P(A)\lt P(B)\), we get \(\dfrac{1}{P(B)}\lt \dfrac{1}{P(A)}\). Multiplying by \(P(A\cap B)\gt 0\):

\[ P(A/B)\lt P(B/A) \]

Note: part (iii) needs \(P(A)\gt 0\) (so that \(P(B/A)\) exists) and \(P(A\cap B)\gt 0\) for the strict inequality. If \(P(A\cap B)=0\), both conditional probabilities are \(0\).

SA Q6 Complements of independent events

If \(A\) and \(B\) are two independent events, prove that the events (a) \(A^c\) and \(B^c\) (b) \(A^c\) and \(B\) (c) \(A\) and \(B^c\) are also independent.

Solution

Given \(P(A\cap B)=P(A)P(B)\).

(a) Using De Morgan's law, \(A^c\cap B^c=(A\cup B)^c\):

\[ \begin{aligned} P(A^c\cap B^c)&=1-P(A\cup B)\\ &=1-P(A)-P(B)+P(A)P(B)\\ &=[1-P(A)][1-P(B)]\\ &=P(A^c)P(B^c) \end{aligned} \]

(b) \(B=(A\cap B)\cup(A^c\cap B)\), a union of disjoint events, so

\[ \begin{aligned} P(A^c\cap B)&=P(B)-P(A\cap B)\\ &=P(B)-P(A)P(B)\\ &=P(B)[1-P(A)]=P(A^c)P(B) \end{aligned} \]

(c) Similarly, \(A=(A\cap B)\cup(A\cap B^c)\), so

\[ \begin{aligned} P(A\cap B^c)&=P(A)-P(A\cap B)\\ &=P(A)-P(A)P(B)\\ &=P(A)[1-P(B)]=P(A)P(B^c) \end{aligned} \]

Hence all three pairs are independent.

SA Q7 Converse of independence of complements

Prove that, if \(A^c\) and \(B^c\) are independent events, then so are \(A\) and \(B\).

Solution

Since \(A^c\) and \(B^c\) are independent, \(P(A^c\cap B^c)=P(A^c)P(B^c)\). That is,

\[ \begin{aligned} 1-P(A\cup B)&=[1-P(A)][1-P(B)]\\ &=1-P(A)-P(B)+P(A)P(B) \end{aligned} \] \[ \Rightarrow\ P(A\cup B)=P(A)+P(B)-P(A)P(B) \]

But by the addition theorem, \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Comparing the two results,

\[ P(A\cap B)=P(A)P(B) \]

Hence \(A\) and \(B\) are independent.

SA Q8 Three tosses of a balanced coin

A balanced coin is tossed three times in succession. Let \(A\) denote the event of getting 'head' in the first toss and \(B\) the event of getting 'tail' in the second toss. Prove that the events \(A\) and \(B\) are independent.

Solution

The sample space has \(8\) equally likely outcomes:

\[ S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\} \]

\(A=\{HHH,HHT,HTH,HTT\}\Rightarrow P(A)=\dfrac48=\dfrac12\)

\(B=\{HTH,HTT,TTH,TTT\}\Rightarrow P(B)=\dfrac48=\dfrac12\)

\(A\cap B=\{HTH,HTT\}\Rightarrow P(A\cap B)=\dfrac28=\dfrac14\)

\[ P(A)P(B)=\frac12\cdot\frac12=\frac14=P(A\cap B) \]

Hence \(A\) and \(B\) are independent events.

SA Q9 Pairwise independence

One fair coin is tossed twice. Show that the events \(A\): head occurs on the first toss; \(B\): head occurs on the second toss; \(C\): head occurs once only; are pairwise independent.

Solution

The sample space is \(S=\{HH,HT,TH,TT\}\) with each outcome of probability \(\dfrac14\).

Event	Outcomes	Probability
\(A\)	\(\{HH,HT\}\)	\(\dfrac12\)
\(B\)	\(\{HH,TH\}\)	\(\dfrac12\)
\(C\)	\(\{HT,TH\}\)	\(\dfrac12\)
\(A\cap B\)	\(\{HH\}\)	\(\dfrac14\)
\(A\cap C\)	\(\{HT\}\)	\(\dfrac14\)
\(B\cap C\)	\(\{TH\}\)	\(\dfrac14\)

\[ \begin{aligned} P(A)P(B)&=\tfrac14=P(A\cap B)\\ P(A)P(C)&=\tfrac14=P(A\cap C)\\ P(B)P(C)&=\tfrac14=P(B\cap C) \end{aligned} \]

Hence \(A\), \(B\), \(C\) are pairwise independent.

Remark: \(A\cap B\cap C=\varnothing\), so \(P(A\cap B\cap C)=0\neq P(A)P(B)P(C)\). Thus the events are pairwise independent but not mutually independent.

Key formulas to remember:

\[ \begin{aligned} P(A\cap B)&=P(A)P(B)\ \text{(independent)}\\ P(A\cap B)&=0\ \text{(mutually exclusive)} \end{aligned} \] \[ \begin{aligned} P(A\cup B)&=P(A)+P(B)-P(A\cap B)\\ P(A/B)&=\frac{P(A\cap B)}{P(B)} \end{aligned} \]

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