Miscellaneous Example (Revised) - Chapter 16 - Probability - Ncert Solutions class 11 - Maths

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Chapter 14: Probability - NCERT Solutions for Class 11 Maths

Example 1 

On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits
(i) $\mathrm{A}$ before $\mathrm{B}$ ?
(ii) $\mathrm{A}$ before $\mathrm{B}$ and $\mathrm{B}$ before $\mathrm{C}$ ?
(iii) A first and $\mathrm{B}$ last?
(iv) A either first or second?
(v) A just before B?

Solution

The number of arrangements (orders) in which Veena can visit four cities A, $\mathrm{B}$, $\mathrm{C}$, or $\mathrm{D}$ is 4 ! i.e., 24.Therefore, $n(\mathrm{~S})=24$.
Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is
$
\begin{aligned}
\mathrm{S}= & \{\mathrm{ABCD}, \mathrm{ABDC}, \mathrm{ACBD}, \mathrm{ACDB}, \mathrm{ADBC}, \mathrm{ADCB} \\
& \mathrm{BACD}, \mathrm{BADC}, \mathrm{BDAC}, \mathrm{BDCA}, \mathrm{BCAD}, \mathrm{BCDA} \\
& \mathrm{CABD}, \mathrm{CADB}, \mathrm{CBDA}, \mathrm{CBAD}, \mathrm{CDAB}, \mathrm{CDBA} \\
& \mathrm{DABC}, \mathrm{DACB}, \mathrm{DBCA}, \mathrm{DBAC}, \mathrm{DCAB}, \mathrm{DCBA}\}
\end{aligned}
$
(i) Let the event 'she visits $\mathrm{A}$ before $\mathrm{B}$ ' be denoted by $\mathrm{E}$

Therefore, $\mathrm{E}=\{\mathrm{ABCD}, \mathrm{CABD}, \mathrm{DABC}, \mathrm{ABDC}, \mathrm{CADB}, \mathrm{DACB}$
$
\mathrm{ACBD}, \mathrm{ACDB}, \mathrm{ADBC}, \mathrm{CDAB}, \mathrm{DCAB}, \mathrm{ADCB}\}
$

Thus
$
\mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{12}{24}=\frac{1}{2}
$
(ii) Let the event 'Veena visits $\mathrm{A}$ before $\mathrm{B}$ and $\mathrm{B}$ before $\mathrm{C}$ ' be denoted by $\mathrm{F}$.
$
\text { Here } \quad \mathrm{F}=\{\mathrm{ABCD}, \mathrm{DABC}, \mathrm{ABDC}, \mathrm{ADBC}\}
$

Therefore, $\mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{4}{24}=\frac{1}{6}$
Students are advised to find the probability in case of (iii), (iv) and (v).

Example 2

 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings.

Solution

Total number of possible hands $={ }^{52} \mathrm{C}_7$
(i) Number of hands with 4 Kings $={ }^4 \mathrm{C}_4 \times{ }^{48} \mathrm{C}_3$ (other 3 cards must be chosen from the rest 48 cards)

Hence
$
\mathrm{P} \text { (a hand will have } 4 \text { Kings) }=\frac{{ }^4 \mathrm{C}_4 \times{ }^{48} \mathrm{C}_3}{{ }^{52} \mathrm{C}_7}=\frac{1}{7735}
$
(ii) Number of hands with 3 Kings and 4 non-King cards $={ }^4 \mathrm{C}_3 \times{ }^{48} \mathrm{C}_4$

Therefore
$
\mathrm{P}(3 \text { Kings })=\frac{{ }^4 \mathrm{C}_3 \times{ }^{48} \mathrm{C}_4}{{ }^{52} \mathrm{C}_7}=\frac{9}{1547}
$
(iii) $\mathrm{P}$ (atleast 3 King)
$
\begin{aligned}
& =\mathrm{P}(3 \text { Kings or } 4 \text { Kings }) \\
& =\mathrm{P}(3 \text { Kings })+\mathrm{P}(4 \text { Kings }) \\
& =\frac{9}{1547}+\frac{1}{7735}=\frac{46}{7735}
\end{aligned}
$

Example 3

 If A, B, C are three events associated with a random experiment, prove that
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{C}) \\
& -\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})
\end{aligned}
$

Solution

Consider $\mathrm{E}=\mathrm{B} \cup \mathrm{C}$ so that
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) & =\mathrm{P}(\mathrm{A} \cup \mathrm{E}) \\
& =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{A} \cap \mathrm{E})
\end{aligned}
$

Now
$
\begin{aligned}
\mathrm{P}(\mathrm{E}) & =\mathrm{P}(\mathrm{B} \cup \mathrm{C}) \\
& =\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{B} \cap \mathrm{C})
\end{aligned}
$

Also $\quad \mathrm{A} \cap \mathrm{E}=\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A} \cap \mathrm{C})$ [using distribution property of intersection of sets over the union]. Thus
$
\mathrm{P}(\mathrm{A} \cap \mathrm{E})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{C})-\mathrm{P}[(\mathrm{A} \cap \mathrm{B}) \cap(\mathrm{A} \cap \mathrm{C})]
$

$
=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{C})-\mathrm{P}[\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}]
$

Using (2) and (3) in (1), we get
$
\begin{aligned}
\mathrm{P}[\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}]= & \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{B} \cap \mathrm{C}) \\
& -\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})
\end{aligned}
$

Example 4

In a relay race there are five teams A, B, C, D and E.
(a) What is the probability that A, B and C finish first, second and third, respectively.
(b) What is the probability that $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are first three to finish (in any order) (Assume that all finishing orders are equally likely)

Solution

If we consider the sample space consisting of all finishing orders in the first three places, we will have ${ }^5 \mathrm{P}_3$, i.e., $\frac{5!}{(5-3)!}=5 \times 4 \times 3=60$ sample points, each with a probability of $\frac{1}{60}$.
(a) A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., $\mathrm{ABC}$.

Thus $\mathrm{P}\left(\mathrm{A}, \mathrm{B}\right.$ and $\mathrm{C}$ finish first, second and third respectively) $=\frac{1}{60}$.
(b) $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are the first three finishers. There will be 3 ! arrangements for $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. Therefore, the sample points corresponding to this event will be 3 ! in number.

$\text { So } \quad \mathrm{P}(\mathrm{A}, \mathrm{B} \text { and } \mathrm{C} \text { are first three to finish })=\frac{3!}{60}=\frac{6}{60}=\frac{1}{10}$