NCERT Exemplar Solutions Class 11 Maths Probability — Objective Type Q18–29

NCERT Exemplar Problems, Class 11 Mathematics — Probability, Objective Type questions 18 to 29. All 12 questions solved.

Twelve multiple-choice questions. Half are counting questions answered as a ratio (18, 19, 20, 21, 22, 25, 26), and the other half test the laws of probability directly: $P(\bar A)=1-P(A)$, the addition law $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, and the fact that every probability lies between $0$ and $1$ (23, 24, 27, 28, 29).

One answer in the printed key is wrong, Question 18. It is explained where it occurs.

Key insight. Probability is favourable over total, and the laws of probability are just that statement applied to sets. When a question gives you two of $P(A)$, $P(B)$, $P(A\cap B)$, $P(A\cup B)$, the addition law gives the third.

Question 18

In a non-leap year, the probability of having $53$ Tuesdays or $53$ Wednesdays is

Solution. A non-leap year has $365=52\times7+1$ days: $52$ full weeks, which give every weekday exactly $52$ times, plus one extra day. A weekday occurs $53$ times only if it is that extra day. The extra day is equally likely to be any of the $7$ weekdays, so having $53$ Tuesdays needs the extra day to be Tuesday ($\frac17$) and $53$ Wednesdays needs it to be Wednesday ($\frac17$). These cannot both happen, so they are mutually exclusive and the probabilities add:

$$P=\frac17+\frac17=\frac27.$$

The printed key gives (A) $\frac17$, which is wrong. $\frac17$ is the probability of one of the two weekdays only; the question says “or”, which allows either.

Question 19

Three numbers are chosen from $1$ to $20$. Find the probability that they are not consecutive.

Solution. Three numbers can be chosen in $\binom{20}{3}=1140$ ways. Consecutive triples are $(1,2,3),(2,3,4),\dots,(18,19,20)$, which is $18$ of them. Probability of being consecutive is $\frac{18}{1140}=\frac3{190}$, so

$$P(\text{not consecutive})=1-\frac{3}{190}=\frac{187}{190}.$$

Option (D), $\frac{18}{{}^{20}C_3}$, is the probability that they are consecutive.

Question 20

While shuffling a pack of $52$ playing cards, $2$ are accidentally dropped. Find the probability that the missing cards to be of different colours.

Solution. Choose $2$ of the $52$ cards: $\binom{52}2=1326$ ways. Different colours means one red (of $26$) and one black (of $26$): $26\times26=676$ ways.

$$P=\frac{676}{1326}=\frac{26}{51}.$$

(The same result comes from drawing in order: whatever the first card, $26$ of the remaining $51$ are of the other colour.)

Question 21

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is

Solution. Total seatings: $7!$. Glue the two persons into one block, leaving $6$ objects, which can be arranged in $6!$ ways, and the pair can swap inside the block in $2!$ ways:

$$P=\frac{2\cdot6!}{7!}=\frac27.$$

Question 22

Without repetition of the numbers, four digit numbers are formed with the numbers $0,2,3,5$. The probability of such a number divisible by $5$ is

Solution. The numbers must be genuine four-digit numbers, so $0$ cannot lead. Total: first digit from $\{2,3,5\}$ ($3$ ways), then the rest in $3!$ ways, giving $3\times6=18$.

Divisible by $5$ means the last digit is $0$ or $5$.

• Last digit $0$: the other three digits fill the first three places in $3!=6$ ways.
• Last digit $5$: the first digit is $2$ or $3$ ($2$ ways), then the remaining two digits in $2!$ ways: $2\times2=4$.

Favourable $=6+4=10$, so $P=\frac{10}{18}=\frac59$. (Counting all $24$ arrangements of the four digits, including those starting with $0$, would give $\frac12$, which is not an option; that is the trap.)

Question 23

If $A$ and $B$ are mutually exclusive events, then

Solution. Mutually exclusive means $A\cap B=\varnothing$, so everything in $A$ lies outside $B$: $A\subseteq\bar B$. A subset cannot be more probable than the set containing it, so $P(A)\le P(\bar B)$. Equality is possible (take $A=\bar B$), which is why the strict inequality (C) fails.

Question 24

If $P(A\cup B)=P(A\cap B)$ for any two events $A$ and $B$, then

Solution. Always $A\cap B\subseteq A\subseteq A\cup B$ and likewise for $B$, so $P(A\cap B)\le P(A)\le P(A\cup B)$ and the same for $P(B)$. If the two extremes are equal, everything between them equals them too, so $P(A)=P(A\cap B)=P(B)$.

Question 25

$6$ boys and $6$ girls sit in a row at random. The probability that all the girls sit together is

Solution. Total arrangements of $12$ people: $12!$. Treat the six girls as one block: the block and $6$ boys make $7$ objects, arranged in $7!$ ways, and the girls arrange themselves inside the block in $6!$ ways.

$$P=\frac{7!\,6!}{12!}=\frac{5040\cdot720}{479\,001\,600}=\frac1{132}.$$

Question 26

A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is

Solution. PROBABILITY has $11$ letters: P, R, O, B, A, B, I, L, I, T, Y. The vowels are O, A, I, I, which is $4$ (the repeated I counts twice because each position is a separate possible pick). Hence $P=\frac4{11}$. Choosing option (D) by counting only the distinct vowels $O,A,I$ is the usual error.

Question 27

If the probabilities for $A$ to fail in an examination is $0.2$ and that for $B$ is $0.3$, then the probability that either $A$ or $B$ fails is

Solution. Let $F_A,F_B$ be “A fails”, “B fails”. Then $P(F_A\cup F_B)=0.2+0.3-P(F_A\cap F_B)=0.5-P(F_A\cap F_B)$. The question does not say the two events are independent or exclusive, so $P(F_A\cap F_B)$ is only known to be $\ge0$. Therefore $P(F_A\cup F_B)\le0.5$, with equality only when they cannot both fail. (Option (B), exactly $.5$, would need the extra assumption of mutual exclusivity.)

Question 28

The probability that at least one of the events $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.2$, then $P(\bar A)+P(\bar B)$ is

Solution. The addition law $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ gives $0.6=P(A)+P(B)-0.2$, so $P(A)+P(B)=0.8$. Then

$$P(\bar A)+P(\bar B)=(1-P(A))+(1-P(B))=2-0.8=1.2.$$

Question 29

If $M$ and $N$ are any two events, the probability that at least one of them occurs is

Solution. “At least one occurs” is $M\cup N$. Adding $P(M)+P(N)$ counts the outcomes in $M\cap N$ twice, so the overlap is subtracted once. Option (A) is the probability that exactly one occurs.

Common mistakes

• Question 18. Answering $\frac17$ by thinking of a single weekday. “Tuesdays or Wednesdays” has two favourable cases for the extra day, so the answer is $\frac27$, and that is also what the printed key misses.
• Question 19. Picking (D) because it looks familiar: $\frac{18}{{}^{20}C_3}$ is the probability that the three numbers are consecutive. The question asks for the complement.
• Question 22. Including numbers that start with $0$. A number like $0235$ is not a four-digit number, so the sample space has $18$ elements, not $24$.
• Question 23. Choosing the strict inequality (C). If $A=\bar B$ the two probabilities are equal, so only $\le$ is always safe.
• Question 26. Counting the vowels $O,A,I$ as $3$ because they are three different letters. Each position in the word is a separate equally likely pick, and PROBABILITY has two I’s.
• Question 27. Choosing $.5$ because $0.2+0.3=0.5$. That forgets the overlap term; with no information about whether the two events can occur together, only $\le .5$ is certain.

Practise next

• The rest of this unit: Short and Long Answer Type Q1–17 and True or False, Fill in the Blanks and Matching Q30–43.
• NCERT textbook: Exercise 14.1 for sample spaces and counting, Exercise 14.2 for the addition law, and the Miscellaneous Exercise for mixed problems.