NCERT Exemplar for Class 11 Maths Chapter 16 - Probability (Book Solutions)

Examples

Example 1: An ordinary deck of cards contains $52$ cards divided into four suits. The red suits are diamonds and hearts and black suits are clubs and spades. The cards $J$, $Q$ and $K$ are called face cards. Suppose we pick one card from the deck at random.

(a). What is the sample space of the experiment?

Ans: Given: A deck of $52$ cards divided into four suits.

The red suits are diamonds and hearts.

The black suits are clubs and spades.

$J$, $Q$ and $K$ are the face cards.

Here ,find the number of ways to select $1$ card out of a total of $52$ cards in the deck.

Since $1$ card is selected at random from the deck which has a total of $52$ cards, the total number of sample space is the number of ways in which $1$ card can be selected.

Therefore, the number of sample space $={}^{52}{C}_1=52$.

(b) What is the event that the chosen card is a black face card?

Ans: Given: A deck of $52$ cards divided into four suits.

The red suits are diamonds and hearts.

The black suits are clubs and spades.

$J$, $Q$ and $K$ are the face cards.

Here, check the number of face cards that are included in the black suit.

Since the face cards are Jacks $\left(J\right)$, Kings $\left(K\right)$ and Queens $\left(Q\right)$. In each suit there is one of each face card.

Two suits (spades and clubs) are black, so the Jacks, Kings and Queens of these two suits are the required event.

Example 2: Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children.

(a) List the eight elements in the sample space whose outcomes are all possible genders of the three children.

Ans: Given, A family having exactly three children.

Chance of having a girl is equal to the chance of a boy.

Represent a boy with $B$ and a girl with $G$ and form all the combinations that can be made regarding the genders of the three children.

Let us represent a boy with $B$ and a girl with $G$.

The possible combinations of the genders of the three children in the family is given as $\left\{BBB,BBG,BGB,GBB,GGG,GGB,GBG,BGG\right\}$. There can be a total of $8$ outcomes, so the number of sample spaces is $n\left(S\right)=8$.

(b) Write each of the following events as a set and find its probability:

(i) The event that exactly one child is a girl.

Ans: Let us find the favourable sample spaces in each subpart.

The event in which there is exactly one girl child is given by $X=\left\{BBG,BGB,GBB\right\}$. So the number of favourable elements is $n\left(X\right)=3$.

Therefore, probability $={\displaystyle \frac{n\left(X\right)}{n\left(S\right)}}={\displaystyle \frac{3}{8}}$.

(ii) The event the at least two children are girls.

Ans: Let us find the favourable sample spaces in each subpart.

The event in which there are at least two girl child is given by $Y=\left\{GGG,GGB,GBG,BGG\right\}$. So the number of favourable elements is $n\left(Y\right)=4$.

Therefore, probability $={\displaystyle \frac{n\left(Y\right)}{n\left(S\right)}}={\displaystyle \frac{4}{8}}={\displaystyle \frac{1}{2}}$.

(iii) The event that no child is a girl.

Ans: Let us find the favourable sample spaces in each subpart.

The event in which there is no girl child is given by $Z=\left\{BBB\right\}$. So the number of favourable elements is $n\left(Z\right)=1$.

Therefore, probability $={\displaystyle \frac{n\left(Z\right)}{n\left(S\right)}}={\displaystyle \frac{1}{8}}$.

Example 3: (a) How many two – digit positive integers are multiple of $3$?

Ans: Consider all the numbers from $10$ to $99$ as the two – digit positive integers. The first integer divisible by $3$ is $12$ and the last is $99$ (among the two digit – integers).

All the integers from $10$ to $99$ are the positive two - digit integers. So, there are $90$ two - digit positive integers. Therefore, the total number of sample spaces is $n\left(S\right)=90$.

Among all the two – digit positive integers, the multiples of $3$ are $12$, $15$, $18$, …, $99$. So there are a total of $30$ integers that are divisible by $3$. Therefore, the number of favourable sample spaces is $n\left(E\right)=30$.

(b) What is the probability that a randomly chosen two – digit positive integer is a multiple of $3$?

Ans: Consider all the numbers from $10$ to $99$ as the two – digit positive integers. The first integer divisible by $3$ is $12$ and the last is $99$ (among the two digit – integers).

All the integers from $10$ to $99$ are the positive two – digit integers. So, there are $90$ two – digit positive integers. Therefore, the total number of sample spaces is $n\left(S\right)=90$.

Hence, the probability of choosing an integer that is a multiple of $3$

$={\displaystyle \frac{n\left(E\right)}{n\left(S\right)}}$

$={\displaystyle \frac{30}{90}}$

$={\displaystyle \frac{1}{3}}$ 

Example 4: A typical PIN (personal identification number) is a sequence of any four symbols chosen from the $26$ letters in the alphabet and the ten digits. If all PINs are equally likely, what is the probability that a randomly chosen PIN contains a repeated symbol?

Ans: Given: A PIN that contains a sequence of any four symbols taken from the $26$ letters of the alphabet and the ten digits.

Find the total number of sample space by considering that each place of the symbol can be filled with $36$ choices (if repetition is allowed).

Now, find the number of combinations that can be made if none of the symbols is repeating. Hence, subtract this from the total number of sample spaces to get the number of favourable sample spaces.

There are $26$ alphabets and $10$ digits, so there are a total of $36$ symbols. If the repetition is allowed then each of the four symbols can be chosen in $36$ ways.

Therefore, the total number of sample spaces is $n\left(S\right)=36\times 36\times 36\times 36=1679616$.

Now, if none of the symbols is repeated then the number of ways in which the four symbols can be filled is $36\times 35\times 34\times 33=1413720$.

So the number of combinations if at least one symbol is repeated $=\left(1679616-1413720\right)=265896$. These favourable outcomes are $n\left(E\right)=265896$.

Hence, the required probability $={\displaystyle \frac{265896}{1679616}}$.

Example 5: An experiment has four possible outcomes $A$, $B$, $C$ and $D$, that are mutually exclusive. Explain why the following assignments of probabilities are not permissible:

(a) $P\left(A\right)=0.12$, $P\left(B\right)=0.63$, $P\left(C\right)=0.45$, $P\left(D\right)=-0.20$

Ans: Given: Four mutually exclusive outcomes of an experiment $A$, $B$, $C$ and $D$ and their probabilities to occur.

use the fact that the probability of an event to occur lies in the interval $\left[0,1\right]$.

Here the probability of event $D$ to occur is $P\left(D\right)=-0.20$ which is negative. The probability of an event to occur is never negative and always lies in the interval $\left[0,1\right]$. Therefore, this assignment of probabilities is not permissible.

(b) $P\left(A\right)={\displaystyle \frac{9}{120}}$, $P\left(B\right)={\displaystyle \frac{45}{120}}$, $P\left(C\right)={\displaystyle \frac{27}{120}}$, $P\left(D\right)={\displaystyle \frac{46}{120}}$

Ans: Given: Four mutually exclusive outcomes of an experiment $A$, $B$, $C$ and $D$ and their probabilities to occur.

use the fact that the sum of all the probabilities of an event in an experiment is $1$.

Here the sum of all the probabilities

$=P\left(A\right)+P\left(B\right)+P\left(C\right)+P\left(D\right)$

$={\displaystyle \frac{9}{120}}+{\displaystyle \frac{45}{120}}+{\displaystyle \frac{27}{120}}+{\displaystyle \frac{46}{120}}$

$={\displaystyle \frac{127}{120}}$ 

The sum of all the probabilities of the events in an experiment is always $1$. Therefore, this assignment of probabilities is not permissible.

Example 6: Probability that a truck stopped at a roadblock will have faulty brakes or badly worn tires are $0.23$ and $0.24$, respectively. Also, the probability is $0.38$ that a truck stopped at the roadblock will have faulty brakes and/or badly working tires. What is the probability that a truck stopped at this roadblock will have faulty brakes as well as badly worn tires?

Ans: Given: Probability that the truck stopped due to faulty brakes $=0.23$

Probability that the truck stopped due badly worn tires $=0.24$

Probability that the truck stopped due to faulty brakes and/or badly working tires $=0.38$

Assume $P\left(A\right)$ and $P\left(B\right)$ as the probabilities that the truck stopped due to faulty brakes and badly worn tires, respectively.

Use the formula $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$ to find the value of $$P\left(A\cap B\right)$$.

Probability that the truck stopped due to faulty brakes $=P\left(A\right)=0.23$

Probability that the truck stopped due badly worn tires $=P\left(B\right)=0.24$

Probability that the truck stopped due to faulty brakes and/or badly working tires $P\left(A\cup B\right)=0.38$

Using the formula $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$, the probability that the truck stopped de to both faulty brakes and badly worn tires is

  $=P\left(A\cap B\right)$

   $=P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)$

   $=0.23+0.24-0.38$

   $=0.09$ 

Example 7: If a person visits his dentist, suppose the probability that he will have his teeth cleaned is $0.48$, the probability that he will have a cavity filled is $0.25$, the probability that he will have a tooth extracted is $0.20$, the probability that he will have a teeth cleaned and a cavity filled is $0.09$, the probability that he will have his teeth cleaned and tooth extracted is $0.12$, the probability that he will have a cavity filled and a tooth extracted is $0.07$ and the probability that he will have his teeth cleaned, a cavity filled, and a tooth extracted is $0.03$. What is the probability that a person visiting his dentist will have at least one of these things done to him?

Ans: Given: Probability that the person will: -

get the teeth cleaned $=0.48$

get the cavity filled $=0.25$

get the tooth extracted $=0.20$

get the teeth cleaned and cavity filled $=0.09$

get the teeth cleaned and tooth extracted $=0.12$

get the cavity filled and tooth extracted $=0.07$

get the teeth cleaned, tooth extracted and cavity filled $=0.03$

Assume the probabilities of the person getting his teeth cleaned, cavity filled and tooth extracted as $P\left(A\right)$, $P\left(B\right)$ and $P\left(C\right)$ respectively.

Use the formula $P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(A\cap C\right)+P\left(A\cap B\cap C\right)$ to solve for the value of $L.H.S$.

Let us assume the probabilities of the person getting his teeth cleaned, cavity filled and tooth extracted as $P\left(A\right)$, $P\left(B\right)$ and $P\left(C\right)$ respectively. Therefore, according to the question,

$P\left(A\right)=0.48$, $P\left(B\right)=0.25$, $P\left(C\right)=0.20$

$P\left(A\cap B\right)=0.09$, $P\left(B\cap C\right)=0.12$, $P\left(A\cap C\right)=0.07$

$P\left(A\cap B\cap C\right)=0.03$

Therefore, the probability of the person getting at least one the three things done to his is $P\left(A\cup B\cup C\right)$ which is

$=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(A\cap C\right)+P\left(A\cap B\cap C\right)$

$=0.48+0.25+0.20-0.09-0.12-0.07+0.03$

$=0.68$ 

Example 8: An urn contains twenty white slips of paper numbered from 1 through $20$, ten red slips of paper numbered from $1$ through $10$, forty yellow slips of paper numbered from $1$ through $40$, and ten blue slips of paper numbered from $1$ through $10$. If these $80$ slips of paper are thoroughly shuffled so that each slip has the same probability of being drawn. Find the probabilities of drawing a slip of paper that is:

(a) Blue or white.

Ans: Given: Number of white slips $=20$

Number of red slips $=10$

Number of yellow slips $=40$

Number of blue slips $=10$

Find the total number of favourable slips that are satisfying the conditions in each case one by one.

The total number of slips is the total number of sample spaces $n\left(S\right)=80$.

There are $20$ white slips and $10$ blue slips and any one can be chosen. So the number of favourable sample spaces is $n\left(A\right)=20+10=30$.

Therefore, the required probability

   $={\displaystyle \frac{n\left(A\right)}{n\left(S\right)}}$

   $={\displaystyle \frac{30}{80}}$

   $={\displaystyle \frac{3}{8}}$ 

(b) Numbered $1$, $2$, $3$, $4$ or $5$.

Ans: Given: Number of white slips $=20$

Number of red slips $=10$

Number of yellow slips $=40$

Number of blue slips $=10$

Find the total number of favourable slips that are satisfying the conditions in each case one by one.

The total number of slips is the total number of sample spaces $n\left(S\right)=80$.

In each set of the coloured papers there are five papers numbered $1$, $2$, $3$, $4$ or $5$. So the total number of papers satisfying the condition is $n\left(B\right)=\left(5+5+5+5\right)=20$.

Therefore, the required probability

   $={\displaystyle \frac{n\left(B\right)}{n\left(S\right)}}$

   $={\displaystyle \frac{20}{80}}$

   $={\displaystyle \frac{1}{4}}$ 

(c) Red or yellow and numbered $1$, $2$, $3$ or $4$.

Ans: Given: Number of white slips $=20$

Number of red slips $=10$

Number of yellow slips $=40$

Number of blue slips $=10$

Find the total number of favourable slips that are satisfying the conditions in each case one by one.

The total number of slips is the total number of sample spaces $n\left(S\right)=80$.

There are four papers each in the sets of both red and yellow coloured papers numbered $1$, $2$, $3$ or $4$. So the total number of papers satisfying the condition is $n\left(C\right)=\left(4+4\right)=8$.

Therefore, the required probability

$={\displaystyle \frac{n\left(C\right)}{n\left(S\right)}}$

$={\displaystyle \frac{8}{80}}$

$={\displaystyle \frac{1}{10}}$ 

(d) Numbered $5$, $15$, $25$ or $35$.

Ans: Given: Number of white slips $=20$

Number of red slips $=10$

Number of yellow slips $=40$

Number of blue slips $=10$

Find the total number of favourable slips that are satisfying the conditions in each case one by one.

The total number of slips is the total number of sample spaces $n\left(S\right)=80$.

The number $5$ paper is present in all the four sets of paper, the number $15$ paper is present in only white and yellow slips and the number $25$ and number $35$ paper is present only in the yellow set of paper. So the total number of papers satisfying the condition is $n\left(D\right)=\left(4+2+1+1\right)=8$.

Therefore, the required probability

$={\displaystyle \frac{n\left(D\right)}{n\left(S\right)}}$

$={\displaystyle \frac{8}{80}}$

$={\displaystyle \frac{1}{10}}$ 

(e) White and numbered higher than $12$ or yellow and numbered higher than $26$.

Ans: Given: Number of white slips $=20$

Number of red slips $=10$

Number of yellow slips $=40$

Number of blue slips $=10$

Find the total number of favourable slips that are satisfying the conditions in each case one by one.

The total number of slips is the total number of sample spaces $n\left(S\right)=80$.

The are $8$ white slips numbered greater than $12$ and $14$ yellow slips numbered greater than $26$. So the total number of papers satisfying the condition is $n\left(E\right)=\left(8+14\right)=22$.

Therefore, the required probability

$={\displaystyle \frac{n\left(E\right)}{n\left(S\right)}}$

$={\displaystyle \frac{22}{80}}$

$={\displaystyle \frac{11}{40}}$ 

Objective Type Questions:

Choose the correct answer from given four options in each of the Examples 9 to 15 (M.C.Q.):

Example 9: In a leap year the probability of having $53$ Sundays or $53$ Mondays is

(A) ${\displaystyle \frac{2}{7}}$

(B) ${\displaystyle \frac{3}{7}}$

(C) ${\displaystyle \frac{4}{7}}$

(D) ${\displaystyle \frac{5}{7}}$

Ans: Normal year has $365$ days and $52$ weeks and $1$ day. And a leap year has $366$ days and hence $52$

weeks and $2$ days. Then possibilities of other $2$ days can be 

1. Sunday, Monday

2. Monday, Tuesday

3. Tuesday, Wednesday

4. Wednesday, Thursday

5. Thursday, Friday

6.  Friday, Saturday

7.  Saturday, Sunday

Therefore, considering two events $P\left(A\right)$ and $P\left(\text{B}\right)$

Where, $P\left(A\right)$$=$ the leap year that contains $53$ Sundays

$P\left(B\right)$$=$ the leap year that contains $53$ Mondays

$P\left(A\cap \text{B}\right)$$=$ Probability that Sunday and Monday occur simultaneously 

So, there are $7$ possibilities of $53$ Sundays or $53$ Mondays

$\therefore n\left(\text{S}\right)=7$

$P\left(A\right)$$=$$2$

$P\left(B\right)$$=$$2$

Now,

$P\left(A\cup B\right)=\text{P}\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$={\displaystyle \frac{2}{7}}+{\displaystyle \frac{2}{7}}-{\displaystyle \frac{1}{7}}$

$={\displaystyle \frac{2+2-1}{7}}$

$={\displaystyle \frac{3}{7}}$ 

Hence, the probability of $53$ Sundays or $53$ Mondays is ${\displaystyle \frac{3}{7}}$

Correct Answer: B 

Example 10: Three digit numbers are formed using the digits 0, 2, 4, 6, 8. A number is

chosen at random out of these numbers. What is the probability that this number has

the same digits?

(A) ${\displaystyle \frac{1}{16}}$

 (B) ${\displaystyle \frac{16}{25}}$

(C) ${\displaystyle \frac{1}{645}}$

 (D) $${\displaystyle \frac{1}{25}}$$

Ans: $3$-digit number cannot start with digit $0$, but hundredth can have any of the $4$digits other than $0$. Now, all $5$ digits can be placed at unit and tens place then total possible $3$-digit numbers are $$4\times 5\times 5$$ i.e., $100$.

$\therefore n\left(S\right)=100$

The total possible $3$-digit number having all digits same $$=4$$ i.e., $n\left(\text{E}\right)$

Hence, Probability of $3$-digit number with same digits $={\displaystyle \frac{n\left(E\right)}{n\left(S\right)}}$

$={\displaystyle \frac{4}{100}}$

$={\displaystyle \frac{1}{25}}$ 

Correct Answer: D 

Example 11: Three squares of chess board are selected at random. The probability of getting $2$ squares of one colour and other of a different colour is:

(A) ${\displaystyle \frac{16}{21}}$

(B) ${\displaystyle \frac{8}{21}}$

(C) ${\displaystyle \frac{3}{32}}$

(D) ${\displaystyle \frac{3}{8}}$

Ans: A chess board has $64$ squares in which $32$ are Black and $32$ are White. Now, as per the question $3$ squares are to be selected in which $2$ can be of single colour and $1$can be of other colour so they can be selected in either of the two ways i.e., $\left(2\text{white or}1\text{black}\right)$ OR $\left(1\text{white or}2\text{black}\right)$ then these can be selected as:

${}^{32}{\text{C}}_2\times {}^{32}{\text{C}}_1\times 2$

Now the $3$ squares selected randomly is in ${}^{64}{\text{C}}_3$ way

$\therefore \text{n}\left(\text{E}\right)=$${}^{32}{\text{C}}_2\times {}^{32}{\text{C}}_1\times 2$

And $\text{n}\left(\text{S}\right)=$${}^{64}{\text{C}}_3$

So, $\text{P}\left(\text{E}\right)={\displaystyle \frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}}$

$={\displaystyle \frac{{}^{32}{\text{C}}_2\times {}^{32}{\text{C}}_1\times 2}{{}^{64}{\text{C}}_3}}$

$={\displaystyle \frac{16}{21}}$

Hence, probability is ${\displaystyle \frac{16}{21}}$

Correct Answer: A 

Example 12: If $\text{A}$ and $\text{B}$ are any two events having $P\left(A\cup B\right)={\displaystyle \frac{1}{2}}$ and $P\left(\overline{A}\right)={\displaystyle \frac{2}{3}}$ then the probability of $$\overline{A}\cap B$$ is:

(A) ${\displaystyle \frac{1}{2}}$

(B) $${\displaystyle \frac{2}{3}}$$

(C) ${\displaystyle \frac{1}{6}}$

(D) ${\displaystyle \frac{1}{3}}$

Ans: We know, $\text{P}\left(\text{A}\cup \text{B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)$

And $\text{P}\left(\text{A}\right)=1-\text{P}\left(\overline{A}\right)$

Given $\text{P}\left(\text{A}\cup \text{B}\right)={\displaystyle \frac{1}{2}}$ and $\text{P}\left(\overline{A}\right)={\displaystyle \frac{2}{3}}$

$\text{P}\left(\text{A}\right)=1-{\displaystyle \frac{2}{3}}={\displaystyle \frac{1}{3}}$

Therefore,

⇒$\text{P}\left(\text{A}\cup \text{B}\right)=\text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)$

${\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{3}}+\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)$

$\text{P}\left(\text{B}\right)-\text{P}\left(\text{A}\cap \text{B}\right)={\displaystyle \frac{1}{6}}$ 

And  

Then, $\text{P}\left(\overline{A}\cap \text{B}\right)={\displaystyle \frac{1}{6}}$

$$\left(\text{since A}\cap \text{ B}\equiv \text{ B}-\text{A}\right)$$

Correct Answer: C 

Example 13: Three of the six vertices of a regular hexagon are chosen at random. What is the probability that the triangle with these vertices is equilateral?

(A) ${\displaystyle \frac{3}{10}}$

(B) ${\displaystyle \frac{3}{20}}$

(C) ${\displaystyle \frac{1}{20}}$

(D) ${\displaystyle \frac{1}{10}}$

Ans: Regular hexagon $\text{ABCDEF}$

Then, total number of triangles are ${}^6{\text{C}}_3$

$={\displaystyle \frac{6!}{3!\left(6-3\right)!}}$

$={\displaystyle \frac{120}{6}}=20$ 

Because there are no three collinear points there are only two equilateral triangles i.e., $△\text{ACF}$ and $△\text{BDF}$

$\therefore \text{n}\left(\text{E}\right)=2$

$\text{n}\left(\text{S}\right)=20$

Therefore, required probability, $\text{P}\left(E\right)={\displaystyle \frac{n\left(\text{E}\right)}{n\left(S\right)}}$

$={\displaystyle \frac{2}{20}}={\displaystyle \frac{1}{10}}$

Correct Answer: D 

Example 14: If $A,B,C$ are three mutually exclusive and exhaustive events of an experiment such that $$3P\left(A\right)=2P\left(B\right)=P\left(C\right)$$ then $P\left(A\right)$ is equal to

(A) ${\displaystyle \frac{1}{11}}$

(B) ${\displaystyle \frac{2}{11}}$

(C) ${\displaystyle \frac{5}{11}}$

(D) ${\displaystyle \frac{6}{11}}$

Ans: Given, $$3\text{P}\left(\text{A}\right)=2\text{P}\left(\text{B}\right)=\text{P}\left(\text{C}\right)$$

Let $$3\text{P}\left(\text{A}\right)=2\text{P}\left(\text{B}\right)=\text{P}\left(\text{C}\right)=x$$

Then, 

$\text{P}\left(\text{A}\right)={\displaystyle \frac{x}{3}}$

$\text{P}\left(\text{B}\right)={\displaystyle \frac{x}{2}}$

And $\text{P}\left(\text{C}\right)=x$

$\therefore \text{P}\left(\text{A}\right)+\text{P}\left(\text{B}\right)+\text{P}\left(\text{C}\right)=1$

 $\Rightarrow {\displaystyle \frac{x}{3}}+{\displaystyle \frac{x}{2}}+x=1$

 $\Rightarrow {\displaystyle \frac{2x+3x+6x}{6}}=1$

 ${\displaystyle \frac{11x}{6}}=1$

 $x={\displaystyle \frac{6}{11}}$ 

$\because \text{P}\left(\text{A}\right)={\displaystyle \frac{x}{3}}$

Then, $\text{P}\left(\text{A}\right)={\displaystyle \frac{x}{3}}={\displaystyle \frac{6}{11}}\times {\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{11}}$

Hence, $\text{P}\left(\text{A}\right)={\displaystyle \frac{2}{11}}$

Correct Answer: B 

Example 15: One mapping (function) is selected at random from all the mappings of the set $$A=\left\{1,2,3,...,n\right\}$$ into itself. The probability that the mapping selected is one to one is

(A) ${\displaystyle \frac{1}{n^n}}$

(B) ${\displaystyle \frac{1}{n!}}$

(C) ${\displaystyle \frac{\left(n-1\right)!}{n^{n-1}}}$

(D) none of these

Ans: $\left(\text{with }n\text{ elements}\right)$ number of mapping for set $\text{A}$$=n^n$

And, the number of $1\text{to}\text{1}$ mapping for the set $\text{A}$ itself is $n!$

So, the probability is $={\displaystyle \frac{n!}{n^n}}={\displaystyle \frac{\left(n-1\right)!}{n^{n-1}}}$

Correct Answer: C 

Long Type Question

Exercise:

1: If the letters of the word ALGORITHM are arranged in a row, what is the probability the letters GOR must remain together.

Ans: Given: The letters of the word ALGORITHM are arranged.

Find the total number of sample spaces by finding the number of words that can be formed with the given letters of the word.

Now, find the favourable number of arrangements by considering the letters GOR always together.

There are $9$ different letters in the word ALGORITHM, so the total number of words that can be formed is the number of sample spaces $n\left(S\right)=9!$.

Now, the letters GOR always remain together. These can be considered as a single unit, so the remaining $6$ letters and this single unit makes a total of $7$ things which can be arranged in $7!$ ways.

So the number of favourable arrangements is $n\left(E\right)=7!$.

Hence, the required probability 

$={\displaystyle \frac{n\left(S\right)}{n\left(E\right)}}$

$={\displaystyle \frac{7!}{9!}}$

$={\displaystyle \frac{1}{72}}$ 

2. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have non-adjacent desks?

Ans: Given: Total number of new employees $=6$

Total number of desks in the assigned row $=6$

A married couple is present among the $6$ employees.

Find the total number of sample space by arranging all the six employees without any condition.

Now, find the number of arrangements in which the couple always gets the adjacent seat. Subtract it from the total number of sample spaces to get the favourable sample spaces.

The $6$ new employees can be arranged at $6$ desks (without any condition) in $6!$ ways, so the number of sample space is $n\left(S\right)=6!$.

Let us consider that the couple gets the adjacent seat, so the two people in this couple can be arranged among themselves in $2!$ ways. This couple can be considered as a single unit, so the remaining four employees and this single unit can be considered as $5$ things which can be arranged in $5!$ ways.

The total number of arrangements where the couple is always together $=2!\times 5!$.

Therefore, the number of arrangements where the coupe in never together $n\left(E\right)=6!-\left(2!\times 5!\right)$.

Hence, the required probability

$={\displaystyle \frac{n\left(E\right)}{n\left(S\right)}}$

$={\displaystyle \frac{6!-\left(2!\times 5!\right)}{6!}}$

$={\displaystyle \frac{2}{3}}$ 

3. Suppose an integer from $1$ to $1000$ is chosen at random, find the probability that the integer is a multiple of $2$ or a multiple of $9$.

Ans: Given: An integer from $1$ to $1000$ is chosen at random.

Find the total number of sample space by selecting any one integer out of a total of $1000$ integers given.

Find the numbers that are divisible by $2$ or $9$ and count the number of integers which are divisible by both to get the favourable sample spaces.

The total number of integers is from $1$ to $1000$ is thousand, so the number of ways in which one can be selected is the total number of sample space $n\left(S\right)={}^{1000}{C}_1$.

Now, all the even integers are divisible by $2$, so there will be $500$ even numbers from $1$ to $1000$ and any one of them can be selected in ${}^{500}{C}_1$ ways. The integers that are divisible by $9$ are $9$, $18$, $27$, $36$, …, $999$. There are $111$ such integers divisible by $9$.

Among these $111$ integers there will be integers like $18$, $36$, $54$, … etc. which are also divisible by $2$ and they are already considered earlier. There are $55$ such integers. So the number of integers that is to be considered is $\left(111-55\right)=56$ and any one of them can be chosen in ${}^{56}{C}_1$ ways.

Therefore, the number of favourable sample spaces $n\left(E\right)={}^{500}{C}_1+{}^{56}{C}_1$.

Hence, the required probability is

$={\displaystyle \frac{n\left(E\right)}{n\left(S\right)}}$

$={\displaystyle \frac{{}^{500}{C}_1+{}^{56}{C}_1}{{}^{1000}{C}_1}}$

$=0.556$ 

4: An experiment consists of rolling a die until a $2$ appears.

(i) How many elements of the sample space correspond to the event that $2$ appears on the $k^{th}$ roll of the die?

Ans: Given: A die is rolled until a $2$ appears.

consider that the first $\left(k-1\right)$ rolls of the die results in any of the five outcomes except $2$.

There are $6$ outcomes when a die is rolled.

In this case $2$ appears on the $k^{th}$ roll, so in the first $\left(k-1\right)$ roll there are $5$ possible outcomes other than $2$ on each roll.

Therefore, the number of elements $=5\times 5\times 5\times 5\times ...\left(k-1\right)\text{ times}=5^{k-1}$.

(ii) How many elements of the sample space correspond to the event that $2$ appears not later than the $k^{th}$ roll of the die?

Ans: Given: A die is rolled until a $2$ appears.

consider different situations where $2$ may appear in the first or second or third time etc. Add the number of elements of different situations.

In this case $2$ appears before the $k^{th}$ roll, so it may appear on the first roll or second roll or third role or … ${\left(k-1\right)}^{th}$ roll.

If it appears on the first roll then the number of elements $=1$.

If it appears on the second roll that means in the first roll there are $5$ outcomes and in the second role $1$, so the number of elements $=5\times 1$.

If it appears on the third roll that means in the first two rolls there are $5$ outcomes each and in the third role $1$, so the number of elements $=5\times 5\times 1$.

Observing the pattern, the number of elements in $\left(k-1\right)$ rolls $=5\times 5\times 5\times 5\times ...\left(k-1\right)\text{ times}\times 1$.

Hence, the total number of elements of all the different situations = $1+5+5^2+5^3+..{.5}^{\left(k-1\right)}$.

This is a geometric progression whose sum of the terms $=\left({\displaystyle \frac{5^k-1}{4}}\right)$.

5. A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find $P\left(G\right)$, where $G$ is the event that a number greater than $3$ occurs on a single roll of the die.

Ans: Given: Number of times the die is rolled $=1$

Probability of occurrence of an odd number is twice that of an even number.

$G$ is the event of occurrence of a number greater than $3$ on a single roll.

Find the total number of sample spaces when a die is rolled. Consider the odd numbers likely to occur twice.

Find the favourable sample spaces by considering the numbers $4$, $5$ or $6$ to occur.

When a die is rolled the number of outcomes is $6$ which are $\left\{1,2,3,4,5,6\right\}$. It is said that each odd number is likely to occur twice of the occurrence of each even number.

So the sample spaces can be $\left\{1,1,2,3,3,4,5,5,6\right\}$. Here $n\left(S\right)=9$.

The numbers greater than $3$ are $4$, $5$ and $6$, so the favourable sample spaces to occur are $\left\{4,5,5,6\right\}$. Here $n\left(E\right)=4$.

Hence, the required probability $P\left(G\right)={\displaystyle \frac{n\left(S\right)}{n\left(E\right)}}={\displaystyle \frac{4}{9}}$.

6. In a large metropolitan area, the probabilities are $0.87$, $0.36$, $0.30$ that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Ans: Given: Probability that the chosen family owns a colour television set $=0.87$

Probability that the chosen family owns a black and white television set $=0.36$

Probability that the chosen family owns both the sets $=0.30$

Assume $P\left(C\right)$ and $P\left(B\right)$ as the probabilities that the chosen family owns a colour television set and a black and white television set respectively.

Use the formula $P\left(B\cup C\right)=P\left(B\right)+P\left(C\right)-P\left(B\cap C\right)$.

Probability that the chosen family owns a colour television set $=P\left(C\right)=0.87$.

Probability that the chosen family owns a black and white television set $=P\left(B\right)=0.36$.

Probability that the chosen family owns both the sets $=P\left(B\cap C\right)=0.30$.

Using the formula $P\left(B\cup C\right)=P\left(B\right)+P\left(C\right)-P\left(B\cap C\right)$, the probability that the chosen family owns either anyone or both kinds of sets

$=P\left(B\cup C\right)$

$=P\left(B\right)+P\left(C\right)-P\left(B\cap C\right)$

$=0.87+0.36-0.30$

$=0.93$ 

7. If $A$ and $B$ are mutually exclusive events, $P\left(A\right)=0.35$, $P\left(B\right)=0.45$, find

$\left(a\right)P\left(A^{\prime }\right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left(A\right)=0.35$

$P\left(B\right)=0.45$

Use the formulas: -

$P\left(A^{\prime }\right)=1-P\left(A\right)$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$P\left(A\cap B\right)=0$ (for mutually exclusive events)

$P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$

$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$

Here $P\left(A^{\prime }\right)$ means the probability that the event $A$ does not occur.

 $\Rightarrow P\left(A^{\prime }\right)=1-P\left(A\right)$

 $\Rightarrow P\left(A^{\prime }\right)=1-0.35$

 $\Rightarrow P\left(A^{\prime }\right)=0.65$ 

$\left(b\right)P\left(B^{\prime }\right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left(A\right)=0.35$

$P\left(B\right)=0.45$

Use the formulas: -

$P\left(A^{\prime }\right)=1-P\left(A\right)$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$P\left(A\cap B\right)=0$ (for mutually exclusive events)

$P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$

$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$

Here $P\left(B^{\prime }\right)$ means the probability that the event $B$ does not occur.

 $\Rightarrow P\left(B^{\prime }\right)=1-P\left(B\right)$

 $\Rightarrow P\left(B^{\prime }\right)=1-0.45$

 $\Rightarrow P\left(B^{\prime }\right)=0.55$ 

$\left(c\right)P\left(A\cup B\right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left(A\right)=0.35$

$P\left(B\right)=0.45$

Use the formulas: -

$P\left(A^{\prime }\right)=1-P\left(A\right)$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$P\left(A\cap B\right)=0$ (for mutually exclusive events)

$P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$

$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$

Here $P\left(A\cup B\right)$ is the probability of occurrence of either $A$ or $B$. Using the formula $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$, where $P\left(A\cap B\right)$ is the probability of occurrence of $A$ and $B$ simultaneously.

$\Rightarrow P\left(A\cup B\right)=0.45+0.35-P\left(A\cap B\right)$

$\Rightarrow P\left(A\cup B\right)=0.80-P\left(A\cap B\right)$ 

Since the events $A$ and $B$ are mutually exclusive that means they cannot occur simultaneously. So $P\left(A\cap B\right)=0$.

$\Rightarrow P\left(A\cup B\right)=0.80$

$\left(d\right)P\left(A\cap B\right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left(A\right)=0.35$

$P\left(B\right)=0.45$

Use the formulas: -

$P\left(A^{\prime }\right)=1-P\left(A\right)$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$P\left(A\cap B\right)=0$ (for mutually exclusive events)

$P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$

$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$

Since the events $A$ and $B$ are mutually exclusive that means they cannot occur simultaneously. So $P\left(A\cap B\right)=0$.

$\left(e\right)P\left(A\cap B^{\prime }\right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left(A\right)=0.35$

$P\left(B\right)=0.45$

Use the formulas: -

$P\left(A^{\prime }\right)=1-P\left(A\right)$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$P\left(A\cap B\right)=0$ (for mutually exclusive events)

$P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$

$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$

Using the formula $P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$,

$\Rightarrow P\left(A\cap B^{\prime }\right)=0.35-0$

$\Rightarrow P\left(A\cap B^{\prime }\right)=0.35$ 

$\left(f\right)P\left(A^{\prime }\cap B^{\prime }\right)$

Ans: Given: Two mutually exclusive events $A$ and $B$.

$P\left(A\right)=0.35$

$P\left(B\right)=0.45$

Use the formulas: -

$P\left(A^{\prime }\right)=1-P\left(A\right)$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

$P\left(A\cap B\right)=0$ (for mutually exclusive events)

$P\left(A\cap B^{\prime }\right)=P\left(A\right)-P\left(A\cap B\right)$

$P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$

Using the formula $P\left(A^{\prime }\cap B^{\prime }\right)=P{\left(A\cup B\right)}^{\prime }$,

 $\Rightarrow P\left(A^{\prime }\cap B^{\prime }\right)=1-P\left(A\cup B\right)$

 $\Rightarrow P\left(A^{\prime }\cap B^{\prime }\right)=1-0.80$

 $\Rightarrow P\left(A^{\prime }\cap B^{\prime }\right)=0.20$ 

8. A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple, very simple are respectively, $0.15$, $0.20$, $0.31$, $0.26$, $0.08$. Find the probabilities that a particular surgery will be rated

(a) Complex or very complex.

Ans: Given: Probabilities that the surgery is very complex $=0.15$

Probabilities that the surgery is complex $=0.20$

Probabilities that the surgery is routine $=0.31$

Probabilities that the surgery is simple $=0.26$