Probability

NCERT Exemplar Solutions Class 11 Maths Probability — True or False, Fill in the Blanks and Matching Q30–43

NCERT Exemplar Problems, Class 11 Mathematics — Probability, True or False, Fill in the Blanks and Matching questions 30 to 43. All 14 questions solved.

Fourteen short questions that all test the rules a legitimate probability assignment must obey: each probability lies in $[0,1]$, the probabilities of all outcomes in the sample space add to $1$, $P(A\cap B)\le P(A)$, and $P(A\cup B)=P(A)+P(B)-P(A\cap B)\le1$. Several statements are false because an impossible number slips through one of these.

Key insight. Before accepting any set of probabilities, run three checks: every value is between $0$ and $1$; an intersection never exceeds either event; and a union never exceeds $1$. Almost every True or False item here fails one of them.

Question 30

State whether the statement is True or False. The probability that a person visiting a zoo will see the giraffe is $0.72$, the probability that he will see the bears is $0.84$ and the probability that he will see both is $0.52$.

Solution. Let $G$ and $B$ be seeing the giraffe and the bears. Then $P(G\cup B)=0.72+0.84-0.52=1.04$. A probability cannot exceed $1$, so these three numbers cannot belong to the same experiment. The statement is false.

False

Question 31

The probability that a student will pass his examination is $0.73$, the probability of the student getting a compartment is $0.13$, and the probability that the student will either pass or get compartment is $0.96$.

Solution. Passing and getting a compartment cannot both happen, so they are mutually exclusive, and the probability of “either” is the sum: $0.73+0.13=0.86$. The statement claims $0.96$, so it is false.

False

Question 32

The probabilities that a typist will make $0,1,2,3,4,5$ or more mistakes in typing a report are, respectively, $0.12,\ 0.25,\ 0.36,\ 0.14,\ 0.08,\ 0.11$.

Solution. The six cases cover every possibility with no overlap, so their probabilities must total $1$. Here $0.12+0.25+0.36+0.14+0.08+0.11=1.06\ne1$, so the statement is false.

False

Question 33

If $A$ and $B$ are two candidates seeking admission in an engineering College. The probability that $A$ is selected is $.5$ and the probability that both $A$ and $B$ are selected is at most $.3$. Is it possible that the probability of $B$ getting selected is $0.7$?

Solution. Put $P(B)=0.7$. From $P(A\cup B)\le1$,

$$P(A\cap B)=P(A)+P(B)-P(A\cup B)\ge0.5+0.7-1=0.2.$$

So $P(A\cap B)$ must lie between $0.2$ and $0.3$, and such values exist (for example $0.25$), consistent with $P(A\cap B)\le\min\{P(A),P(B)\}=0.5$. The two conditions are compatible, so it is possible. The statement is true.

True (yes, it is possible)

Question 34

The probability of intersection of two events $A$ and $B$ is always less than or equal to those favourable to the event $A$.

Solution. $A\cap B\subseteq A$, and a subset can never have a larger probability than the set that contains it. So $P(A\cap B)\le P(A)$ always.

True

Question 35

The probability of an occurrence of event $A$ is $.7$ and that of the occurrence of event $B$ is $.3$ and the probability of occurrence of both is $.4$.

Solution. $P(A\cap B)$ cannot exceed $P(B)$, which is $0.3$, yet $0.4>0.3$. (The union would also be $0.7+0.3-0.4=0.6$, which looks harmless, so this is easy to miss.) The statement is false.

False

Question 36

The sum of probabilities of two students getting distinction in their final examinations is $1.2$.

Solution. These are the probabilities of two different events (one student, then another), not of the outcomes of one sample space, so they need not add to $1$. Each value only has to lie between $0$ and $1$; for example $0.6+0.6=1.2$ is perfectly possible. The statement is true.

True

Question 37

Fill in the blanks. The probability that the home team will win an upcoming football game is $0.77$, the probability that it will tie the game is $0.08$, and the probability that it will lose the game is _____.

Solution. Win, tie and lose are the only three outcomes and cannot overlap, so their probabilities add to $1$: $1-0.77-0.08=0.15$.

$0.15$

Question 38

If $e_1,e_2,e_3,e_4$ are the four elementary outcomes in a sample space and $P(e_1)=.1$, $P(e_2)=.5$, $P(e_3)=.1$, then the probability of $e_4$ is ______.

Solution. The four elementary outcomes make up the whole sample space, so $P(e_4)=1-(0.1+0.5+0.1)=0.3$.

$0.3$

Question 39

Let $S=\{1,2,3,4,5,6\}$ and $E=\{1,3,5\}$, then $\bar E$ is _________.

Solution. $\bar E$ is everything in $S$ that is not in $E$: $\{2,4,6\}$.

$\bar E=\{2,4,6\}$

Question 40

If $A$ and $B$ are two events associated with a random experiment such that $P(A)=0.3$, $P(B)=0.2$ and $P(A\cap B)=0.1$, then the value of $P(A\cap\bar B)$ is _______.

Solution. The event $A$ splits into the part inside $B$ and the part outside $B$: $P(A)=P(A\cap B)+P(A\cap\bar B)$. So $P(A\cap\bar B)=0.3-0.1=0.2$.

$0.2$

Question 41

The probability of happening of an event $A$ is $0.5$ and that of $B$ is $0.3$. If $A$ and $B$ are mutually exclusive events, then the probability of neither $A$ nor $B$ is ________.

Solution. Mutually exclusive, so $P(A\cup B)=0.5+0.3=0.8$. “Neither $A$ nor $B$” is the complement of the union: $P(\bar A\cap\bar B)=1-0.8=0.2$.

$0.2$

Question 42

Match the proposed probability under Column $C_1$ with the appropriate written description under Column $C_2$.

$C_1$ Probability $C_2$ Written description
(a) $0.95$ (i) An incorrect assignment
(b) $0.02$ (ii) No chance of happening
(c) $-0.3$ (iii) As much chance of happening as not
(d) $0.5$ (iv) Very likely to happen
(e) $0$ (v) Very little chance of happening

Solution. A probability near $1$ means very likely, near $0$ means very little chance, exactly $0$ means impossible, exactly $0.5$ is an even chance, and a negative number can never be a probability.

(a) $\leftrightarrow$ (iv), (b) $\leftrightarrow$ (v), (c) $\leftrightarrow$ (i), (d) $\leftrightarrow$ (iii), (e) $\leftrightarrow$ (ii)

Question 43

Match the following.

Column 1 Column 2
(a) If $E_1$ and $E_2$ are the two mutually exclusive events (i) $E_1\cap E_2=E_1$
(b) If $E_1$ and $E_2$ are the mutually exclusive and exhaustive events (ii) $(E_1-E_2)\cup(E_1\cap E_2)=E_1$
(c) If $E_1$ and $E_2$ have common outcomes, then (iii) $E_1\cap E_2=\phi,\ E_1\cup E_2=S$
(d) If $E_1$ and $E_2$ are two events such that $E_1\subset E_2$ (iv) $E_1\cap E_2=\phi$

Solution.

  • (a) Mutually exclusive means no outcome in common: $E_1\cap E_2=\phi$, so (iv).
  • (b) Exhaustive adds that together they cover the sample space, so both $E_1\cap E_2=\phi$ and $E_1\cup E_2=S$: (iii).
  • (c) If they share outcomes, split $E_1$ into the part outside $E_2$ and the part inside it: $E_1=(E_1-E_2)\cup(E_1\cap E_2)$, which is (ii). (This identity is true for any pair of events, but of the four options it is the only one that does not forbid common outcomes.)
  • (d) If $E_1\subset E_2$ then every outcome of $E_1$ is also in $E_2$, so $E_1\cap E_2=E_1$: (i).

(a) $\leftrightarrow$ (iv), (b) $\leftrightarrow$ (iii), (c) $\leftrightarrow$ (ii), (d) $\leftrightarrow$ (i)

Common mistakes

  • Question 30. Checking only that each number is below $1$. The test is on the union: $0.72+0.84-0.52=1.04$ exceeds $1$.
  • Question 31. Treating the sum $0.73+0.13$ as the answer to something else. Pass and compartment are different outcomes of the same result, so the union is $0.86$, not $0.96$.
  • Question 33. Rejecting $0.7$ because it exceeds $P(A)=0.5$. The probabilities of two different events need not be ordered; only the intersection is limited by both.
  • Question 35. Checking the union ($0.6$, which is fine) and missing that $P(A\cap B)=0.4$ is larger than $P(B)=0.3$.
  • Question 36. Insisting that probabilities must sum to $1$. That is only true for the elementary outcomes of one sample space; for two separate events the sum can be anything from $0$ to $2$.
  • Question 40. Computing $0.3-0.2$ (subtracting $P(B)$) instead of $P(A)-P(A\cap B)$. The part of $A$ outside $B$ is $A$ minus the overlap.
  • Question 43(c). Matching “common outcomes” with (iv). That is the opposite situation: (iv) says there are none.

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