NCERT Exemplar Solutions Class 11 Maths Probability — Short and Long Answer Type Q1–17

NCERT Exemplar Problems, Class 11 Mathematics — Probability, Short Answer and Long Answer Type questions 1 to 17. All 17 questions solved.

Seventeen questions that all run on two ideas. Where outcomes are equally likely, probability is a ratio of counts (Q1, 2, 3, 12, 13, 14, 15, 17); where they are not, you build the probabilities from a ratio such as “twice as likely” and use the fact that they add to $1$ (Q5, 9, 10, 16). The addition law $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and the complement rule $P(\bar A)=1-P(A)$ do the rest. Question 14(b) can be read two ways; both are given where it occurs.

Key insight. Count the favourable arrangements as a ratio to all arrangements, and when the condition is “together”, glue those items into one block first. When a condition is awkward (“non-adjacent”, “not all together”, “at least one”), count its opposite and subtract from $1$.

Question 1

If the letters of the word ALGORITHM are arranged at random in a row, what is the probability that the letters GOR must remain together as a unit?

Solution. ALGORITHM has $9$ different letters, so there are $9!$ equally likely arrangements. If G, O, R must stay together as the block GOR, the block behaves as a single letter, leaving $7$ objects ($A,L,I,T,H,M$ and the block), which can be arranged in $7!$ ways. The block itself is fixed in the order G-O-R.

$$P=\frac{7!}{9!}=\frac{1}{9\cdot 8}=\frac1{72}$$

Question 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 nonadjacent desks? [Hint: First find the probability that the couple has adjacent desks, and then subtract it from $1$.]

Solution. Following the hint, count the arrangements in which the couple sits together. Glue them into one block: $5$ objects can be arranged in $5!$ ways, and the two people can swap inside the block in $2!$ ways. So the adjacent arrangements number $2\cdot 5!=240$ out of $6!=720$.

$$P(\text{adjacent})=\frac{240}{720}=\frac13,\qquad P(\text{non-adjacent})=1-\frac13=\frac23$$

Question 3

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

Solution. “Or” means union, so use the addition law. Multiples of $2$: $\lfloor 1000/2\rfloor=500$. Multiples of $9$: $\lfloor 1000/9\rfloor=111$. Numbers that are both are multiples of $\operatorname{lcm}(2,9)=18$: $\lfloor 1000/18\rfloor=55$. These are counted twice in $500+111$, so subtract them once.

$$P=\frac{500+111-55}{1000}=\frac{556}{1000}=\frac{139}{250}=0.556$$

Question 4

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

Question 4 (i)

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

Solution. The experiment stops at the first $2$, so an outcome is a string of rolls ending in $2$ and containing no earlier $2$. For the first $2$ to come on roll $k$, each of the first $k-1$ rolls must be one of the $5$ other faces, and roll $k$ must be the $2$ (only $1$ choice). Hence $5^{k-1}\cdot 1$ outcomes.

Question 4 (ii)

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

Solution. “Not later than the $k^{\text{th}}$ roll” means the first $2$ comes on roll $1$, or $2$, or …, or $k$. These cases cannot overlap, so add the counts from part (i) for $j=1,\dots,k$:

$$1+5+5^2+\dots+5^{k-1}=\frac{5^k-1}{5-1}=\frac{5^k-1}{4}$$

Question 5

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

Solution. Let each even face have probability $p$; each odd face then has $2p$. The six probabilities must total $1$: $3(2p)+3p=9p=1$, so $p=\frac19$. The event $G=\{4,5,6\}$ contains two even faces and one odd face.

$$P(G)=\frac19+\frac29+\frac19=\frac49$$

Question 6

In a large metropolitan area, the probabilities are $.87$, $.36$, $.30$ 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?

Solution. Let $C$ be “owns colour”, $B$ “owns black and white”. Then $P(C)=0.87$, $P(B)=0.36$ and $P(C\cap B)=0.30$. Owning “either one or both” is the union, and the addition law subtracts the overlap that would otherwise be counted twice.

$$P(C\cup B)=0.87+0.36-0.30=0.93$$

Question 7

If $A$ and $B$ are mutually exclusive events, $P(A)=0.35$ and $P(B)=0.45$, find (a) $P(A’)$ (b) $P(B’)$ (c) $P(A\cup B)$ (d) $P(A\cap B)$ (e) $P(A\cap B’)$ (f) $P(A’\cap B’)$.

Solution. Mutually exclusive means $A\cap B=\varnothing$, so there is no overlap to subtract.

• (a) $P(A’)=1-0.35=0.65$.
• (b) $P(B’)=1-0.45=0.55$.
• (c) $P(A\cup B)=0.35+0.45=0.80$.
• (d) $P(A\cap B)=0$, by definition of mutually exclusive.
• (e) Since $A$ lies wholly outside $B$, $A\cap B’=A$, so the probability is $0.35$.
• (f) $A’\cap B’=(A\cup B)’$ by De Morgan’s law, so $P=1-0.80=0.20$.

Question 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 or very simple are respectively $0.15,\ 0.20,\ 0.31,\ 0.26,\ .08$. Find the probabilities that a particular surgery will be rated

Solution. The five ratings are mutually exclusive and their probabilities add to $0.15+0.20+0.31+0.26+0.08=1$, so each part is a sum of the right ratings.

• (a) complex or very complex: $0.20+0.15=0.35$.
• (b) neither very complex nor very simple: remove those two from the total, $1-0.15-0.08=0.77$.
• (c) routine or complex: $0.31+0.20=0.51$.
• (d) routine or simple: $0.31+0.26=0.57$.

Question 9

Four candidates $A,B,C,D$ have applied for the assignment to coach a school cricket team. If $A$ is twice as likely to be selected as $B$, and $B$ and $C$ are given about the same chance of being selected, while $C$ is twice as likely to be selected as $D$, what are the probabilities that (a) $C$ will be selected? (b) $A$ will not be selected?

Solution. Express everything in terms of the weakest candidate. Let $P(D)=x$. Then $P(C)=2x$, $P(B)=P(C)=2x$ and $P(A)=2P(B)=4x$. Exactly one is selected, so $4x+2x+2x+x=9x=1$ and $x=\frac19$.

• (a) $P(C)=2x=\frac29$.
• (b) $P(A)=4x=\frac49$, so $P(\text{A not selected})=1-\frac49=\frac59$.

Question 10

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes $S=\{\text{John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}\}$. You are told that the chance of John’s promotion is the same as that of Gurpreet, Rita’s chances of promotion are twice as likely as John’s, Aslam’s chances are four times that of John.

(a) Determine $P(\text{John promoted})$, $P(\text{Rita promoted})$, $P(\text{Aslam promoted})$, $P(\text{Gurpreet promoted})$. (b) If $A=\{\text{John promoted or Gurpreet promoted}\}$, find $P(A)$.

Solution. Let $P(\text{John})=x$. Then Gurpreet is $x$, Rita is $2x$ and Aslam is $4x$. The four elementary outcomes exhaust the sample space, so $x+2x+4x+x=8x=1$, giving $x=\frac18$.

(a) John $\frac18$, Rita $\frac14$, Aslam $\frac12$, Gurpreet $\frac18$.

(b) $A$ is the union of two mutually exclusive outcomes: $P(A)=\frac18+\frac18=\frac14$.

Question 11

The accompanying Venn diagram (Fig. 16.2) shows three events $A$, $B$ and $C$, and also the probabilities of the various intersections (for instance, $P(A\cap B)=.07$). Determine (a) $P(A)$ (b) $P(B\cap\bar C)$ (c) $P(A\cup B)$ (d) $P(A\cap\bar B)$ (e) $P(B\cap C)$ (f) the probability of exactly one of the three occurs.

Reading the figure. The diagram has five regions: $A$ only $=0.13$; $A\cap B$ only $=0.07$; $B$ only $=0.10$; $B\cap C$ only $=0.15$; $C$ only $=0.28$. The circles for $A$ and $C$ do not meet, so $A\cap C$ and $A\cap B\cap C$ are empty.

Solution. Add the regions that make up each event.

• (a) $A$ is “$A$ only” plus “$A\cap B$”: $0.13+0.07=0.20$.
• (b) $B\cap\bar C$ is the part of $B$ outside $C$: $0.07+0.10=0.17$.
• (c) $A\cup B$ is every region touching $A$ or $B$: $0.13+0.07+0.10+0.15=0.45$.
• (d) $A\cap\bar B$ is “$A$ only”: $0.13$.
• (e) $P(B\cap C)=0.15$.
• (f) Exactly one event occurs in the “only” regions $A$, $B$, $C$: $0.13+0.10+0.28=0.51$.

Question 12

One urn contains two black balls (labelled $B_1$ and $B_2$) and one white ball. A second urn contains one black ball and two white balls (labelled $W_1$ and $W_2$). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.

(a) Write the sample space showing all possible outcomes. (b) What is the probability that two black balls are chosen? (c) What is the probability that two balls of opposite colour are chosen?

Solution. (a) Write each outcome as the ordered pair of balls drawn. From urn 1 (balls $B_1,B_2,W$) there are $3\cdot2=6$ ordered pairs, and likewise $6$ from urn 2 (balls $B,W_1,W_2$):

$$S=\{B_1B_2,\ B_1W,\ B_2B_1,\ B_2W,\ WB_1,\ WB_2,\ BW_1,\ BW_2,\ W_1B,\ W_1W_2,\ W_2B,\ W_2W_1\}$$

Each urn has probability $\frac12$ and each ordered pair within it $\frac16$, so all twelve outcomes are equally likely, with probability $\frac1{12}$ each.

(b) Two black balls: $B_1B_2$ and $B_2B_1$ only (urn 2 has just one black ball). So $P=\frac2{12}=\frac16$.

(c) Opposite colours: $B_1W,\ B_2W,\ WB_1,\ WB_2,\ BW_1,\ BW_2,\ W_1B,\ W_2B$, that is $8$ outcomes, so $P=\frac8{12}=\frac23$.

Question 13

A bag contains $8$ red and $5$ white balls. Three balls are drawn at random. Find the probability that (a) all the three balls are white (b) all the three balls are red (c) one ball is red and two balls are white.

Solution. Three balls from $13$ can be chosen in $\binom{13}{3}=286$ equally likely ways. Count favourable selections by choosing the right number from each colour.

• (a) All white: $\binom53=10$, so $P=\frac{10}{286}=\frac5{143}$.
• (b) All red: $\binom83=56$, so $P=\frac{56}{286}=\frac{28}{143}$.
• (c) One red and two white: $\binom81\binom52=8\cdot10=80$, so $P=\frac{80}{286}=\frac{40}{143}$.

Question 14

If the letters of the word ASSASSINATION are arranged at random, find the probability that (a) four S’s come consecutively in the word (b) two I’s and two N’s come together (c) all A’s are not coming together (d) no two A’s are coming together.

Solution. ASSASSINATION has $13$ letters: $A\times3,\ S\times4,\ I\times2,\ N\times2,\ T,\ O$. The total number of distinct arrangements is

$$\frac{13!}{3!\,4!\,2!\,2!}=10\,810\,800.$$

(a) Glue the four S’s into one block. The block and the other $9$ letters ($A\times3, I\times2, N\times2, T, O$) make $10$ objects, with the repeats $3!,2!,2!$ to divide out:

$$\frac{10!}{3!\,2!\,2!}=151\,200,\qquad P=\frac{151\,200}{10\,810\,800}=\frac{2}{143}.$$

(b) The book intends the four letters I, I, N, N to sit together as a single block of four (in any internal order). The block can be arranged internally in $\frac{4!}{2!2!}=6$ ways. The block plus $A\times3$, $S\times4$, $T$, $O$ gives $10$ objects: $\frac{10!}{3!\,4!}=25\,200$. Favourable $=6\times25\,200=151\,200$, so

$$P=\frac{151\,200}{10\,810\,800}=\frac{2}{143}.$$

A second reading. If instead the two I’s must be together and the two N’s must be together, but as two separate pairs, the blocks $II$ and $NN$ give $11$ objects, $\frac{11!}{3!\,4!}=277\,200$ arrangements and probability $\frac1{39}$. The printed answer, $\frac2{143}$, uses the single-block reading above, so that is the one we give.

(c) First find the probability that all three A’s are together. Block the A’s: the block plus $S\times4,\ I\times2,\ N\times2,\ T,\ O$ is $11$ objects, so $\frac{11!}{4!\,2!\,2!}=415\,800$ arrangements, probability $\frac{415\,800}{10\,810\,800}=\frac1{26}$. The complement is “not all together”:

$$P=1-\frac1{26}=\frac{25}{26}.$$

(d) No two A’s adjacent: arrange the $10$ non-A letters first ($S\times4,I\times2,N\times2,T,O$), in $\frac{10!}{4!\,2!\,2!}=37\,800$ ways. They leave $11$ gaps (including the two ends); choose $3$ different gaps for the A’s, $\binom{11}{3}=165$ ways. Favourable $=37\,800\times165=6\,237\,000$, so

$$P=\frac{6\,237\,000}{10\,810\,800}=\frac{15}{26}.$$

Question 15

A card is drawn from a deck of $52$ cards. Find the probability of getting a king or a heart or a red card.

Solution. Every heart is a red card, so “heart or red” is just “red”: $26$ cards. A king adds only those kings that are not already red, namely the $2$ black kings (king of spades and of clubs). Total favourable $=26+2=28$.

$$P=\frac{28}{52}=\frac{7}{13}$$

Question 16

A sample space consists of $9$ elementary outcomes $e_1,e_2,\dots,e_9$ whose probabilities are $P(e_1)=P(e_2)=.08$, $P(e_3)=P(e_4)=P(e_5)=.1$, $P(e_6)=P(e_7)=.2$, $P(e_8)=P(e_9)=.07$. Suppose $A=\{e_1,e_5,e_8\}$, $B=\{e_2,e_5,e_8,e_9\}$.

(a) Calculate $P(A)$, $P(B)$ and $P(A\cap B)$. (b) Using the addition law of probability, calculate $P(A\cup B)$. (c) List the composition of the event $A\cup B$, and calculate $P(A\cup B)$ by adding the probabilities of the elementary outcomes. (d) Calculate $P(\bar B)$ from $P(B)$, also calculate $P(\bar B)$ directly from the elementary outcomes of $\bar B$.

Solution. The probabilities do add to $1$: $0.16+0.30+0.40+0.14=1$.

(a) $P(A)=0.08+0.1+0.07=0.25$. $P(B)=0.08+0.1+0.07+0.07=0.32$. The outcomes common to $A$ and $B$ are $e_5,e_8$, so $P(A\cap B)=0.1+0.07=0.17$.

(b) $P(A\cup B)=0.25+0.32-0.17=0.40$.

(c) $A\cup B=\{e_1,e_2,e_5,e_8,e_9\}$, and $0.08+0.08+0.1+0.07+0.07=0.40$, agreeing with (b).

(d) $P(\bar B)=1-0.32=0.68$. Directly, $\bar B=\{e_1,e_3,e_4,e_6,e_7\}$ and $0.08+0.1+0.1+0.2+0.2=0.68$.

Question 17

Determine the probability $p$, for each of the following events. (a) An odd number appears in a single toss of a fair die. (b) At least one head appears in two tosses of a fair coin. (c) A king, $9$ of hearts, or $3$ of spades appears in drawing a single card from a well shuffled ordinary deck of $52$ cards. (d) The sum of $6$ appears in a single toss of a pair of fair dice.

Solution.

• (a) Odd faces $1,3,5$ out of $6$: $p=\frac36=\frac12$.
• (b) Easier by the complement. “No head” means TT, one of the $4$ outcomes, so $p=1-\frac14=\frac34$.
• (c) The $9$ of hearts and $3$ of spades are not kings, so the three kinds are disjoint: $4+1+1=6$ cards, $p=\frac6{52}=\frac3{26}$.
• (d) Of the $36$ ordered pairs, those with sum $6$ are $(1,5),(2,4),(3,3),(4,2),(5,1)$: $5$ of them, so $p=\frac5{36}$.

Common mistakes

• Question 1. Students count GOR as a block but forget it must stay in the order G-O-R; allowing all $3!$ internal orders would give $\frac{6\cdot7!}{9!}=\frac1{12}$ instead. Read what “as a unit” demands.
• Question 2. Forgetting the factor $2!$ for the couple swapping places inside their block halves the “adjacent” count and gives a wrong complement.
• Question 3. Adding $500+111$ without subtracting the $55$ multiples of $18$ counts those numbers twice, because “or” in probability is a union, not a sum.
• Question 4. Writing $5^k$ for part (i). The $k^{\text{th}}$ roll is forced to be the $2$, so only $k-1$ rolls carry the factor $5$.
• Question 9. “$A$ is twice as likely as $B$” gives $A=2B$, not $B=2A$. Setting the smallest quantity to $x$ (here $D$) keeps every other probability a plain multiple.
• Question 14(d). Counting arrangements of all $13$ letters first and then “removing” adjacent A’s double-counts. Place the non-A letters first and drop the A’s into separate gaps; there are $11$ gaps, not $10$, since the two ends count.
• Question 15. Adding $4+13+26$ treats king, heart and red as disjoint. They overlap heavily, and since every heart is red the answer is $26+2$, not $43$.

Practise next

• The rest of this unit: Objective Type Q18–29 and True or False, Fill in the Blanks and Matching Q30–43.
• NCERT textbook: Exercise 14.1 for sample spaces and events, Exercise 14.2 for the addition law and complements, and the Miscellaneous Exercise for mixed counting problems.