Differentiation (Part-38) | S N De

Probability – Q29 to Q32 and Long Answer Type Questions 1 to 7. The numerical problems are solved step by step, and the 5-mark theorem questions are given with full statements and proofs.

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)  |  \(P(A\cap B)=P(A)\,P(B\mid A)\)  |  \(P(\bar A)=1-P(A)\)

Q29

Out of 20 games of chess played between two players A and B, A won 12, B won 4 and 4 ended in a tie. In a tournament of three games find the probability that (i) B wins all three (ii) B wins at least one (iii) two games end in a tie.

Solution

From the record: \(P(A\text{ wins})=\frac{12}{20}=\frac35\), \(P(B\text{ wins})=\frac{4}{20}=\frac15\), \(P(\text{tie})=\frac{4}{20}=\frac15\). The three games are independent.

(i) B wins all three:

\[\left(\frac15\right)^3=\frac{1}{125}\]

(ii) B wins at least one \(=1-P(\text{B wins none})\), and \(P(\text{B does not win a game})=\frac45\):

\[1-\left(\frac45\right)^3=1-\frac{64}{125}=\frac{61}{125}\]

(iii) Exactly two games tied (choose which 2 of 3 games are ties; the other is not a tie):

\[\binom32\left(\frac15\right)^2\left(\frac45\right)=3\times\frac{4}{125}=\frac{12}{125}\]

(i) \(\dfrac{1}{125}\)   (ii) \(\dfrac{61}{125}\)   (iii) \(\dfrac{12}{125}\)

Q30

The probability that a teacher will give a surprise test during any class meeting is \(\frac15\). If a student is absent on two days, what is the probability that he will miss at least one test?

Solution

\(P(\text{test on a day})=\frac15\), so \(P(\text{no test on a day})=\frac45\).

The student misses at least one test unless there is no test on both days:

\[P=1-\frac45\times\frac45=1-\frac{16}{25}\]

Required probability \(=\dfrac{9}{25}\)

Q31

A number is chosen at random from the first \(n\) natural numbers, and suppose A and B denote the events that the chosen number is divisible by 2 and 3 respectively. Prove that the events A and B are independent if \(n=96\).

Solution

For \(n=96\): A = number divisible by 2, B = number divisible by 3, so \(A\cap B\) = number divisible by 6.

\[n(A)=\frac{96}{2}=48,\quad n(B)=\frac{96}{3}=32,\quad n(A\cap B)=\frac{96}{6}=16\] \[P(A)=\frac{48}{96}=\frac12,\quad P(B)=\frac{32}{96}=\frac13,\quad P(A\cap B)=\frac{16}{96}=\frac16\]

Now \(P(A)\,P(B)=\frac12\times\frac13=\frac16=P(A\cap B)\).

Since \(P(A\cap B)=P(A)\,P(B)\), the events A and B are independent when \(n=96\).  Proved.

Q32

(i) A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls are made without replacement. Find the probability that the first drawing will give 3 white balls and the second 3 red balls. (ii) A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red. [CBSE '02C]

Solution

(i) Total balls \(=13\). Let \(E_1\) = first draw gives 3 white balls, \(E_2\) = second draw gives 3 red balls.

\[P(E_1)=\frac{\binom53}{\binom{13}{3}}=\frac{10}{286}=\frac{5}{143}\]

After the first draw, 10 balls remain: 8 red and 2 white.

\[P(E_2\mid E_1)=\frac{\binom83}{\binom{10}{3}}=\frac{56}{120}=\frac{7}{15}\] \[P(E_1\cap E_2)=\frac{5}{143}\times\frac{7}{15}=\frac{7}{429}\]

(ii) Total balls \(=15\); non-red balls \(=5+3=8\). For none to be red, all three drawn balls are non-red:

\[P=\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}=\frac{336}{2730}=\frac{8}{65}\]

Check: \(\dfrac{\binom83}{\binom{15}{3}}=\dfrac{56}{455}=\dfrac{8}{65}\).

(i) \(\dfrac{7}{429}\)   (ii) \(\dfrac{8}{65}\)

Long Answer Type Questions (Each question carries 5 marks)

Long Q1

State and prove the theorem of total probability.

Solution

Statement. Let \(E_1,E_2,\dots,E_n\) be mutually exclusive and exhaustive events of a sample space \(S\) with \(P(E_i)>0\) for all \(i\), and let \(A\) be any event associated with \(S\). Then \[P(A)=\sum_{i=1}^{n}P(E_i)\,P(A\mid E_i)\]

Proof. Since \(E_1,\dots,E_n\) are exhaustive, \(E_1\cup E_2\cup\dots\cup E_n=S\). Hence

\[A=A\cap S=(A\cap E_1)\cup(A\cap E_2)\cup\dots\cup(A\cap E_n)\]

Because the \(E_i\) are mutually exclusive, the events \(A\cap E_i\) are also mutually exclusive. By the addition theorem for mutually exclusive events,

\[P(A)=P(A\cap E_1)+P(A\cap E_2)+\dots+P(A\cap E_n)\]

By the multiplication theorem, \(P(A\cap E_i)=P(E_i)\,P(A\mid E_i)\). Therefore

\[P(A)=\sum_{i=1}^{n}P(E_i)\,P(A\mid E_i).\qquad\blacksquare\]

Long Q2

State and prove the "addition theorem" of probability for two mutually exclusive events.

Solution

Statement. If \(A\) and \(B\) are two mutually exclusive events, then \[P(A\cup B)=P(A)+P(B)\]

Proof. Let the sample space contain \(n\) equally likely outcomes. Let \(m_1\) outcomes be favourable to \(A\) and \(m_2\) outcomes be favourable to \(B\).

Since \(A\) and \(B\) are mutually exclusive, they have no outcome in common (\(A\cap B=\phi\)). So the number of outcomes favourable to \(A\cup B\) is \(m_1+m_2\). Hence

\[P(A\cup B)=\frac{m_1+m_2}{n}=\frac{m_1}{n}+\frac{m_2}{n}=P(A)+P(B).\qquad\blacksquare\]

Extension: for mutually exclusive events \(A_1,\dots,A_k\), \(\;P(A_1\cup\dots\cup A_k)=\sum P(A_i)\).

Long Q3

What is meant by compound event in probability? State and prove the theorem of compound probability.

Solution

Compound event. When two or more events occur together (simultaneously or in succession), their joint occurrence is called a compound event. For events \(A\) and \(B\), the compound event is \(A\cap B\) (also written \(AB\)).

Theorem of compound probability (multiplication theorem). For two events \(A\) and \(B\) of a sample space with \(P(A)>0\), \[P(A\cap B)=P(A)\,P(B\mid A)\]

Proof. Let the sample space have \(n\) equally likely outcomes. Let \(A\) contain \(m\) outcomes and \(A\cap B\) contain \(k\) outcomes. Then

\[P(A)=\frac{m}{n},\qquad P(A\cap B)=\frac{k}{n}\]

When \(A\) has already occurred, \(A\) itself becomes the reduced sample space with \(m\) outcomes, of which \(k\) are favourable to \(B\). So

\[P(B\mid A)=\frac{k}{m}\]

Multiplying,

\[P(A)\,P(B\mid A)=\frac{m}{n}\times\frac{k}{m}=\frac{k}{n}=P(A\cap B).\qquad\blacksquare\]

Long Q4(i) Show that for two non-mutually exclusive events A and B,

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

(ii) For any three events A, B and C, prove that

\[\begin{gathered}P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)\\-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)\end{gathered}\]

Solution

(i) Write \(A\cup B\) as a union of two mutually exclusive events:

\[A\cup B=A\cup(\bar A\cap B)\]

By the addition theorem for mutually exclusive events,

\[P(A\cup B)=P(A)+P(\bar A\cap B)\qquad\cdots(1)\]

Also \(B=(A\cap B)\cup(\bar A\cap B)\), and these two parts are mutually exclusive, so

\[P(B)=P(A\cap B)+P(\bar A\cap B)\;\Rightarrow\;P(\bar A\cap B)=P(B)-P(A\cap B)\qquad\cdots(2)\]

Substituting (2) in (1):

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

(ii) Let \(D=B\cup C\). Applying part (i) to \(A\) and \(D\):

\[P(A\cup B\cup C)=P(A)+P(D)-P(A\cap D)\qquad\cdots(3)\]

Applying (i) to \(B\) and \(C\):

\[P(D)=P(B)+P(C)-P(B\cap C)\qquad\cdots(4)\]

Now \(A\cap D=A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). Applying (i) again, and using \((A\cap B)\cap(A\cap C)=A\cap B\cap C\):

\[P(A\cap D)=P(A\cap B)+P(A\cap C)-P(A\cap B\cap C)\qquad\cdots(5)\]

Substituting (4) and (5) in (3):

\[P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)\\-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C).\qquad\blacksquare\]

Long Q5

State and prove the multiplicative theorem of probability. How is the result modified when the events are independent?

Solution

Statement. If \(A\) and \(B\) are two events with \(P(A)>0\) and \(P(B)>0\), then \[P(A\cap B)=P(A)\,P(B\mid A)=P(B)\,P(A\mid B)\]

Proof. Let the sample space have \(n\) equally likely outcomes; let \(A\) have \(m\) outcomes, \(B\) have \(r\) outcomes and \(A\cap B\) have \(k\) outcomes.

Given that \(A\) has occurred, the reduced sample space is \(A\) with \(m\) outcomes, of which \(k\) favour \(B\), so \(P(B\mid A)=\frac km\). Then

\[P(A)\,P(B\mid A)=\frac mn\cdot\frac km=\frac kn=P(A\cap B)\]

Similarly, given that \(B\) has occurred, \(P(A\mid B)=\frac kr\), so

\[P(B)\,P(A\mid B)=\frac rn\cdot\frac kr=\frac kn=P(A\cap B).\qquad\blacksquare\]

Independent events. Events \(A\) and \(B\) are independent if the occurrence of one does not affect the probability of the other, i.e. \(P(B\mid A)=P(B)\) and \(P(A\mid B)=P(A)\). The theorem then reduces to

\(P(A\cap B)=P(A)\,P(B)\)

More generally, for mutually independent events \(A_1,\dots,A_n\): \(\;P(A_1\cap A_2\cap\dots\cap A_n)=P(A_1)\,P(A_2)\cdots P(A_n)\).

Long Q6(i) If A, B and C are mutually independent events, prove that the events \((A\cup B)\) and C are also independent events.

(ii) Obtain \(P(\bar A\cup B)\) and \(P(A\cup\bar B)\) in terms of \(P(A)\), \(P(B)\) and \(P(AB)\), where \(\bar A\) is the event complementary to the event A.

Solution

(i) We must show \(P\big((A\cup B)\cap C\big)=P(A\cup B)\,P(C)\).

By the distributive law, \((A\cup B)\cap C=(A\cap C)\cup(B\cap C)\). Hence

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

Since \(A,B,C\) are mutually independent,

\[=P(A)P(C)+P(B)P(C)-P(A)P(B)P(C)\] \[=P(C)\big[P(A)+P(B)-P(A)P(B)\big]=P(C)\big[P(A)+P(B)-P(A\cap B)\big]\] \[=P(C)\,P(A\cup B).\]

Thus \((A\cup B)\) and \(C\) are independent.  Proved.

(ii) Using \(P(\bar A)=1-P(A)\) and \(P(\bar A\cap B)=P(B)-P(A\cap B)\):

\[P(\bar A\cup B)=P(\bar A)+P(B)-P(\bar A\cap B)=\big[1-P(A)\big]+P(B)-\big[P(B)-P(AB)\big]\] \[P(\bar A\cup B)=1-P(A)+P(AB)\]

Similarly, \(P(A\cap\bar B)=P(A)-P(AB)\), so

\[P(A\cup\bar B)=P(A)+\big[1-P(B)\big]-\big[P(A)-P(AB)\big]\] \[P(A\cup\bar B)=1-P(B)+P(AB)\]

\(P(\bar A\cup B)=1-P(A)+P(AB)\)  and  \(P(A\cup\bar B)=1-P(B)+P(AB)\)

Long Q7

If \(P(A)=a\), \(P(B)=b\), then show that \(P(A/B)\ge\dfrac{a+b-1}{b}\).

Solution

By the definition of conditional probability (with \(b>0\)),

\[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{b}\]

From the addition theorem, \(P(A\cap B)=P(A)+P(B)-P(A\cup B)=a+b-P(A\cup B)\).

Since any probability is at most 1, \(P(A\cup B)\le1\). Therefore

\[P(A\cap B)\ge a+b-1\]

Dividing by \(b>0\):

\[P(A\mid B)\ge\frac{a+b-1}{b}.\qquad\blacksquare\]

Note: the printed question uses ">". Strict inequality holds whenever \(P(A\cup B)<1\); equality occurs only in the special case \(P(A\cup B)=1\).

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