Differentiation (Part-38) | S N De

S N De Probability Solution – Very Short Answer Type Questions

Welcome! This post gives a complete S N De Probability Solution for the Very Short Answer Type Questions (Q1–Q9) of the chapter Probability. Each question carries 2 marks, and every answer below is explained step by step so that you can write it directly in your exam.

Topics covered: conditional probability, mutually exclusive and exhaustive events, independent events, set-theoretic expressions of events, and axioms of probability.

• VSA Q1
• VSA Q2
• VSA Q3
• VSA Q4
• VSA Q5
• VSA Q6
• VSA Q7
• VSA Q8
• VSA Q9

VSA Q1Conditional probability

Define conditional probability of event \(A\), given that event \(B\) has occurred.

Solution

The probability of occurrence of event \(A\) when it is known that event \(B\) has already occurred (with \(P(B)\neq 0\)) is called the conditional probability of \(A\) given \(B\). It is denoted by \(P(A/B)\) and defined as

\[ P(A/B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0 \]

VSA Q2Conclusions about events \(A\) and \(B\)

Two events \(A\) and \(B\) are given. What conclusion can be made in each of the following cases?

Solution

Part	Given	Conclusion
(i)	\(P(A\cup B)=P(A)+P(B)\)	Since \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we get \(P(A\cap B)=0\). So \(A\) and \(B\) are mutually exclusive.
(ii)	\(P(A\cap B)=0\)	\(A\) and \(B\) are mutually exclusive events.
(iii)	\(P(A)=P(B)\)	\(A\) and \(B\) are equally likely events.
(iv)	\(P(A\cup B)=1\)	\(A\cup B=S\), so \(A\) and \(B\) are exhaustive events.
(v)	\(P(A\cap B)\neq 0\)	\(A\) and \(B\) are not mutually exclusive; they can occur together.
(vi)	\(P(A/B)=P(A)\)	Occurrence of \(B\) does not affect \(A\), so \(A\) and \(B\) are independent events.
(vii)	\(P(A)\neq P(B)\)	\(A\) and \(B\) are not equally likely events.
(viii)	\(P(A\cap B)=P(A)P(B)\)	\(A\) and \(B\) are independent events.

VSA Q3Prove the result

If \(\bar A\) and \(\bar B\) are the events complementary to the events \(A\) and \(B\) respectively, prove that \(P(\bar A \text{ or } \bar B)=1-P(A)P(B/A)\).

Solution

By De Morgan's law, \(\bar A\cup\bar B=\overline{A\cap B}\). Therefore

\[ P(\bar A\text{ or }\bar B)=P(\bar A\cup\bar B)=P\!\left(\overline{A\cap B}\right)=1-P(A\cap B) \]

By the multiplication theorem, \(P(A\cap B)=P(A)\,P(B/A)\). Hence

\[ P(\bar A\text{ or }\bar B)=1-P(A)\,P(B/A) \]

Hence proved.

VSA Q4Set-theoretic expressions

Let \(A, B, C\) be three arbitrary events. Find expressions for the following events using the usual set-theoretic notations: (i) Only \(A\) occurs (ii) Both \(A\) and \(B\) occur but not \(C\) (iii) All the three events occur (iv) At least one event occurs (v) At least two events occur.

Solution

1. Only \(A\) occurs: \(\;A\cap\bar B\cap\bar C\)
2. Both \(A\) and \(B\) occur but not \(C\): \(\;A\cap B\cap\bar C\)
3. All the three events occur: \(\;A\cap B\cap C\)
4. At least one event occurs: \(\;A\cup B\cup C\)
5. At least two events occur: \(\;(A\cap B)\cup(B\cap C)\cup(A\cap C)\)

VSA Q5Exhaustive and mutually exclusive events

Let \(A_1, A_2, A_3\) be 3 events related to a random experiment. Under what conditions will the events be exhaustive and mutually exclusive?

Solution

• Exhaustive: \(A_1\cup A_2\cup A_3=S\) (the sample space).
• Mutually exclusive: \(A_i\cap A_j=\varnothing\) for \(i\neq j\), i.e. \(A_1\cap A_2=A_2\cap A_3=A_1\cap A_3=\varnothing\).

When both conditions hold, the events form a partition of \(S\), and

\[ P(A_1)+P(A_2)+P(A_3)=1 \]

VSA Q6Exactly one of two events

Show that the probability that exactly one of the events \(A\) and \(B\) occurs is \(P(A)+P(B)-2P(AB)\).

Solution

Exactly one of \(A\), \(B\) occurs means either \(A\) occurs and \(B\) does not, or \(B\) occurs and \(A\) does not. This is the event \((A\cap\bar B)\cup(\bar A\cap B)\). These two events are mutually exclusive, so

\[ P(\text{exactly one})=P(A\cap\bar B)+P(\bar A\cap B) \]

Now \(P(A\cap\bar B)=P(A)-P(A\cap B)\) and \(P(\bar A\cap B)=P(B)-P(A\cap B)\). Therefore

\[ P(\text{exactly one})=P(A)-P(AB)+P(B)-P(AB)=P(A)+P(B)-2P(AB) \]

Hence proved.

VSA Q7Valid measures of probability

Can the following represent measures of probability, where \((A\cup B\cup C)\) represents the sure event and \(A, B, C\) are mutually exclusive?

(i) \(P(A)=0.2,\,P(B)=0.7,\,P(C)=0.1\)   (ii) \(P(A)=0.4,\,P(B)=0.6,\,P(C)=0.2\)   (iii) \(P(A\cup B)=0.5,\,P(B)=0.6,\,P(C)=0.2\)   (iv) \(P(A)=0.4,\,P(B)=0.5,\,P(B\cap C)=0.1\)   (v) \(P(A)=0.32,\,P(B)=0.47,\,P(B\cup C)=0.68\)   (vi) \(P(A)=0.3,\,P(B')=0.5,\,P(C')=0.8\)

Solution

Since \(A,B,C\) are mutually exclusive and \(A\cup B\cup C=S\), a valid assignment needs every probability in \([0,1]\) and

\[ P(A)+P(B)+P(C)=1 \]

1. \(0.2+0.7+0.1=1\). Yes, it can represent probabilities.
2. \(0.4+0.6+0.2=1.2\neq 1\). No.
3. \(B\subseteq A\cup B\) requires \(P(B)\le P(A\cup B)\), but \(0.6>0.5\). (Also, \(P(A)=0.5-0.6=-0.1<0\).) No.
4. \(B\) and \(C\) are mutually exclusive, so \(P(B\cap C)\) must be \(0\), but it is given as \(0.1\). No.
5. \(P(B\cup C)=P(B)+P(C)\Rightarrow P(C)=0.68-0.47=0.21\). Then \(0.32+0.47+0.21=1\). Yes.
6. \(P(B)=1-0.5=0.5\) and \(P(C)=1-0.8=0.2\). Then \(0.3+0.5+0.2=1\). Yes.

Answer: (i), (v) and (vi) can represent probabilities; (ii), (iii) and (iv) cannot.

VSA Q8Value of \(P(X)\)

For three mutually exclusive events \(X, Y\) and \(Z\) it is given that \(P(X)=2P(Y)=3P(Z)\) and \(X\cup Y\cup Z=S\), where \(S\) denotes the sure event. Find the value of \(P(X)\).

Solution

Let \(P(X)=2P(Y)=3P(Z)=k\). Then

\[ P(X)=k,\qquad P(Y)=\frac{k}{2},\qquad P(Z)=\frac{k}{3} \]

Since \(X,Y,Z\) are mutually exclusive and exhaustive, \(P(X)+P(Y)+P(Z)=1\):

\[ k+\frac{k}{2}+\frac{k}{3}=1\;\Rightarrow\;\frac{6k+3k+2k}{6}=1\;\Rightarrow\;\frac{11k}{6}=1\;\Rightarrow\;k=\frac{6}{11} \]

\(P(X)=\dfrac{6}{11}\)

VSA Q9Find \(P(A)\) [CBSE '98C]

If \(A\) and \(B\) are two events associated with a random experiment such that \(P(B)=0.35\), \(P(A\text{ or }B)=0.85\) and \(P(A\text{ and }B)=0.15\), find \(P(A)\).

Solution

Using the addition theorem \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\):

\[ 0.85=P(A)+0.35-0.15 \] \[ P(A)=0.85-0.35+0.15=0.65 \]

\(P(A)=0.65\)

Key formulas to remember:

\[ P(A\cup B)=P(A)+P(B)-P(A\cap B),\qquad P(A/B)=\frac{P(A\cap B)}{P(B)},\qquad P(\bar A)=1-P(A) \]

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