NCERT Class 11 Mathematics — Probability, Miscellaneous Exercise on Chapter 14. All 10 questions solved.

Almost every question here combines Chapter 6’s counting with Chapter 14’s definition:

$$\mathrm P(\mathrm E) = \frac{n(\mathrm E)}{n(\mathrm S)}$$

The hard half is counting $n(\mathrm S)$ and $n(\mathrm E)$ correctly. Two habits carry most of the exercise:

• “At least one” is almost always easier as a complement. $\mathrm P(\text{at least one}) = 1 – \mathrm P(\text{none})$, and “none” is a single clean count.
• Selection uses $^n\mathrm C_r$; arrangement uses $^n\mathrm P_r$. Ask whether reordering the chosen objects gives a genuinely different outcome.

Key insight. A die need not have six distinct faces. In question 3 the faces read $1, 1, 2, 2, 2, 3$, so the outcomes are equally likely as faces but not as numbers — $\mathrm P(2) = \tfrac36$, not $\tfrac16$. Always count the physical outcomes, not the labels on them.

Question 1

A box contains $10$ red marbles, $20$ blue marbles and $30$ green marbles. $5$ marbles are drawn from the box. What is the probability that (i) all will be blue? (ii) at least one will be green?

Solution. There are $60$ marbles in all, and $5$ are drawn without regard to order, so

$$n(\mathrm S) = {}^{60}\mathrm C_5$$

(i) All blue. Choose $5$ from the $20$ blue:

$$\mathrm P = \frac{^{20}\mathrm C_5}{^{60}\mathrm C_5}$$

(ii) At least one green. Counting the cases with one, two, three, four or five greens would take five terms; the complement takes one. “No green” means all five come from the $30$ non-green marbles:

$$\mathrm P(\text{at least one green}) = 1 – \frac{^{30}\mathrm C_5}{^{60}\mathrm C_5}$$

Question 2

$4$ cards are drawn from a well-shuffled deck of $52$ cards. What is the probability of obtaining $3$ diamonds and one spade?

Solution. Four cards from $52$, order irrelevant:

$$n(\mathrm S) = {}^{52}\mathrm C_4$$

Choose $3$ of the $13$ diamonds and $1$ of the $13$ spades — two independent selections, so multiply:

$$\mathrm P = \frac{^{13}\mathrm C_3 \times {}^{13}\mathrm C_1}{^{52}\mathrm C_4}$$

Question 3

A die has two faces each with number ‘$1$’, three faces each with number ‘$2$’ and one face with number ‘$3$’. If the die is rolled once, determine (i) $\mathrm P(2)$ (ii) $\mathrm P(1$ or $3)$ (iii) $\mathrm P($not $3)$.

Solution. The six faces are equally likely, and they carry the numbers

$$1,\ 1,\ 2,\ 2,\ 2,\ 3$$

(i) Three faces show $2$:

$$\mathrm P(2) = \frac36 = \frac12$$

(ii) Two faces show $1$ and one shows $3$; these cannot happen together:

$$\mathrm P(1 \text{ or } 3) = \frac{2 + 1}{6} = \frac36 = \frac12$$

(iii) Only one face shows $3$:

$$\mathrm P(\text{not } 3) = 1 – \frac16 = \frac56$$

Question 4

In a certain lottery, $10{,}000$ tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket, (b) two tickets, (c) $10$ tickets?

Solution. Of the $10{,}000$ tickets, $10$ win and $9990$ do not.

(a) One ticket. Simply the fraction of losing tickets:

$$\mathrm P = \frac{9990}{10000} = \frac{999}{1000}$$

(b) Two tickets. Both must come from the $9990$ losers:

$$\mathrm P = \frac{^{9990}\mathrm C_2}{^{10000}\mathrm C_2}$$

(c) Ten tickets. All ten from the losers:

$$\mathrm P = \frac{^{9990}\mathrm C_{10}}{^{10000}\mathrm C_{10}}$$

Question 5

Out of $100$ students, two sections of $40$ and $60$ are formed. If you and your friend are among the $100$ students, what is the probability that (a) you both enter the same section? (b) you both enter different sections?

Solution. The cleanest route: fix your own place, then ask where your friend lands. Whatever section you are put in, your friend is one of the remaining $99$ students, all equally likely to occupy any remaining place.

(a) Same section.

• If you are in the section of $40$ (probability $\tfrac{40}{100}$), your friend must take one of the $39$ remaining places there, out of $99$.
• If you are in the section of $60$ (probability $\tfrac{60}{100}$), your friend must take one of the $59$ remaining places there.

$$\mathrm P = \frac{40}{100} \cdot \frac{39}{99} + \frac{60}{100} \cdot \frac{59}{99} = \frac{1560 + 3540}{9900} = \frac{5100}{9900} = \frac{17}{33}$$

(b) Different sections is the complement:

$$1 – \frac{17}{33} = \frac{16}{33}$$

Question 6

Three letters are dictated to three persons and an envelope is addressed to each of them. The letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Solution. The letters can be placed in $3! = 6$ ways, all equally likely.

Take the complement: “no letter in its proper envelope” is a derangement of three objects. Writing the correct assignment as $123$, the six permutations are

$$123,\ 132,\ 213,\ 231,\ 312,\ 321$$

Of these, $231$ and $312$ have no number in its own position — two derangements.

$$\mathrm P(\text{no letter correct}) = \frac26 = \frac13$$

$$\mathrm P(\text{at least one correct}) = 1 – \frac13 = \frac23$$

Question 7

$\mathrm A$ and $\mathrm B$ are two events such that $\mathrm P(\mathrm A) = 0.54$, $\mathrm P(\mathrm B) = 0.69$ and $\mathrm P(\mathrm A \cap \mathrm B) = 0.35$. Find

(i) $\mathrm P(\mathrm A \cup \mathrm B)$   (ii) $\mathrm P(\mathrm A’ \cap \mathrm B’)$   (iii) $\mathrm P(\mathrm A \cap \mathrm B’)$   (iv) $\mathrm P(\mathrm B \cap \mathrm A’)$

Solution.

(i)

$$\mathrm P(\mathrm A \cup \mathrm B) = 0.54 + 0.69 – 0.35 = 0.88$$

(ii) By De Morgan, $\mathrm A’ \cap \mathrm B’ = (\mathrm A \cup \mathrm B)’$:

$$1 – 0.88 = 0.12$$

(iii) $\mathrm A \cap \mathrm B’$ is “$\mathrm A$ but not $\mathrm B$”, which is $\mathrm A$ with the overlap removed:

$$0.54 – 0.35 = 0.19$$

(iv) Similarly, $\mathrm B$ with the overlap removed:

$$0.69 – 0.35 = 0.34$$

As a check, the four disjoint pieces must total $1$: $0.35 + 0.19 + 0.34 + 0.12 = 1$ ✓

Question 8

From the employees of a company, $5$ persons are selected to represent them in the managing committee:

S. No.	Name	Sex	Age (years)
$1$	Harish	M	$30$
$2$	Rohan	M	$33$
$3$	Sheetal	F	$46$
$4$	Alis	F	$28$
$5$	Salim	M	$41$

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over $35$ years?

Solution. Let $\mathrm M$ be “male” and $\mathrm E$ be “over $35$”.

$$\mathrm M = \{\text{Harish, Rohan, Salim}\} \quad\Longrightarrow\quad \mathrm P(\mathrm M) = \frac35$$

$$\mathrm E = \{\text{Sheetal }(46),\ \text{Salim }(41)\} \quad\Longrightarrow\quad \mathrm P(\mathrm E) = \frac25$$

The overlap is Salim alone — male and over $35$:

$$\mathrm P(\mathrm M \cap \mathrm E) = \frac15$$

$$\mathrm P(\mathrm M \cup \mathrm E) = \frac35 + \frac25 – \frac15 = \frac45$$

(Checking by listing: everyone except Alis qualifies, so $4$ of $5$ ✓)

Question 9

If $4$-digit numbers greater than $5{,}000$ are randomly formed from the digits $0, 1, 3, 5$ and $7$, what is the probability of forming a number divisible by $5$ when (i) the digits are repeated? (ii) the repetition of digits is not allowed?

Solution. A number is divisible by $5$ when its last digit is $0$ or $5$.

(i) Repetition allowed.

Sample space. The number must exceed $5000$, so the leading digit is $5$ or $7$ — two choices — and the other three are free among the five digits:

$$2 \times 5 \times 5 \times 5 = 250$$

But this count includes $5000$ itself, which is not greater than $5000$, so

$$n(\mathrm S) = 250 – 1 = 249$$

Favourable. Leading digit $5$ or $7$ ($2$ choices), middle two digits free ($5 \times 5$), last digit $0$ or $5$ ($2$ choices):

$$2 \times 5 \times 5 \times 2 = 100$$

Again $5000$ must be removed, since it ends in $0$ and was counted:

$$n(\mathrm E) = 100 – 1 = 99$$

$$\mathrm P = \frac{99}{249} = \frac{33}{83}$$

(ii) Repetition not allowed.

Sample space. Leading digit $5$ or $7$ ($2$ choices), then the remaining three places filled from the four unused digits:

$$2 \times 4 \times 3 \times 2 = 48$$

(No exclusion is needed here — $5000$ repeats the digit $0$.)

Favourable. Split by the last digit.

• Last digit $0$: the leading digit is $5$ or $7$ ($2$ ways), and the middle two come from the three remaining digits: $3 \times 2 = 6$. Total $2 \times 6 = 12$.
• Last digit $5$: then the leading digit must be $7$ (since $5$ is used), and the middle two come from the three remaining: $3 \times 2 = 6$. Total $6$.

$$n(\mathrm E) = 12 + 6 = 18$$

$$\mathrm P = \frac{18}{48} = \frac38$$

Question 10

The number lock of a suitcase has $4$ wheels, each labelled with ten digits, i.e. from $0$ to $9$. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Solution. The four wheels are distinguishable and no digit repeats, so the number of possible sequences is a permutation:

$$^{10}\mathrm P_4 = 10 \times 9 \times 8 \times 7 = 5040$$

Exactly one of these opens the lock:

$$\mathrm P = \frac{1}{5040}$$

Common mistakes

• Question 1(ii), summing five cases. “At least one green” is far quicker as $1 – \mathrm P(\text{no green})$, and the complement is a single combination.
• Question 2, using $^{13}\mathrm C_3 + {}^{13}\mathrm C_1$. The two selections happen together, so the counts multiply.
• Question 3, taking $\mathrm P(2) = \tfrac16$. Three of the six faces show $2$. The faces are equally likely; the numbers on them are not.
• Question 4, computing the probability of winning. The question asks for not getting a prize, so the favourable tickets are the $9990$ losers.
• Question 5, treating the two sections as equally likely. They hold $40$ and $60$ students, so the split is $\tfrac{40}{100}$ and $\tfrac{60}{100}$, not $\tfrac12$ each.
• Question 6, counting the derangements as $0$. Two of the six arrangements leave every letter misplaced. Listing all six is quick and settles it.
• Question 8, forgetting to subtract Salim. He is both male and over $35$, so he is counted twice unless the overlap is removed.
• Question 9(i), forgetting to exclude $5000$. The numbers must be greater than $5000$, and $5000$ appears in both the total and the favourable count when repetition is allowed.
• Question 10, using $^{10}\mathrm C_4$. A lock combination is an ordered sequence, so it is $^{10}\mathrm P_4 = 5040$, not $^{10}\mathrm C_4 = 210$.

Practise next

• Exercise 14.2 — worth working again alongside question 7, since the addition rule and De Morgan’s laws carry both.
• Exercise 6.4 — the combination counting that questions 1, 2 and 4 depend on entirely.