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NCERT Solutions · Class 11 Mathematics Probability

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EXERCISE 14.1 1–7 (part 1 of 4)

  1. Exercise 1

    A die is rolled. Let E be the event "die shows $\displaystyle 4$" and F be the event "die shows even number". Are E and F mutually exclusive?

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    NCERT’s answer
    No.
    Mutually exclusive test. Two events are mutually exclusive when they cannot happen on the same trial, that is, when \(\displaystyle E \cap F = \phi \).The sample space is \(\displaystyle S = \{1,2,3,4,5,6\} \), and\[E = \{4\}, \qquad F = \{2,4,6\} \]Therefore\[E \cap F = \{4\} \neq \phi \]The outcome $\displaystyle 4$ lies in both events, so whenever the die shows $\displaystyle 4$ both E and F occur together.No — E and F are not mutually exclusive, since \(\displaystyle E \cap F = \{4\} \neq \phi \).

  2. Exercise 2

    A die is thrown. Describe the following events:
    (i)
    A: a number less than $\displaystyle 7$
    (ii)
    B: a number greater than $\displaystyle 7$
    (iii)
    C: a multiple of $\displaystyle 3$
    (iv)
    D: a number less than $\displaystyle 4$
    (v)
    E: an even number greater than $\displaystyle 4$
    (vi)
    F: a number not less than $\displaystyle 3$ Also find \(\displaystyle \mathrm{A} \cup \mathrm{B}, \mathrm{A} \cap \mathrm{B}, \mathrm{B} \cup \mathrm{C}, \mathrm{E} \cap \mathrm{F}, \mathrm{D} \cap \mathrm{E}, \mathrm{A}-\mathrm{C}, \mathrm{D}-\mathrm{E}, \mathrm{E} \cap \mathrm{F}^{\prime}, \mathrm{F}^{\prime}\)

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    NCERT’s answer
    (i)
    \{$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$, $\displaystyle 6$\} (ii) $\displaystyle \phi$ (iii) $\displaystyle \{3,6\}$ (iv) \{$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$\} (v) \{$\displaystyle 6$\} (vi) $\displaystyle \{3,4,5,6\}, \mathrm{A} \cup \mathrm{B}=\{1,2,3,4,5,6\}, \mathrm{A} \cap \mathrm{B}=\phi, \mathrm{B} \cup \mathrm{C}=\{3,6\}, \mathrm{E} \cap \mathrm{F}=\{6\}$, $\displaystyle \mathrm{D} \cap \mathrm{E}=\phi$, $\displaystyle \mathrm{A}-\mathrm{C}=\{1,2,4,5\}, \mathrm{D}-\mathrm{E}=\{1,2,3\}, \mathrm{E} \cap \mathrm{F}^{\prime}=\phi, \mathrm{F}^{\prime}=\{1,2\}$
    Listing the events as subsets of the sample space. For one throw of a die, \(\displaystyle S = \{1,2,3,4,5,6\} \). An event is described by listing the outcomes of \(\displaystyle S \) that make it happen.
    (i)
    A: a number less than $\displaystyle 7$ — every face qualifies, so \(\displaystyle A = \{1,2,3,4,5,6\} = S \) (a sure event).
    (ii)
    B: a number greater than $\displaystyle 7$ — no face qualifies, so \(\displaystyle B = \phi \) (an impossible event).
    (iii)
    C: a multiple of $\displaystyle 3$ — \(\displaystyle C = \{3,6\} \).
    (iv)
    D: a number less than $\displaystyle 4$ — \(\displaystyle D = \{1,2,3\} \).
    (v)
    E: an even number greater than $\displaystyle 4$ — the even faces are $\displaystyle 2$, $\displaystyle 4$, $\displaystyle 6$ and only $\displaystyle 6$ exceeds $\displaystyle 4$, so \(\displaystyle E = \{6\} \).
    (vi)
    F: a number not less than $\displaystyle 3$, i.e. \(\displaystyle \geq 3 \) — \(\displaystyle F = \{3,4,5,6\} \).
    Now the required combinations. Recall \(\displaystyle X' = S - X \).
    \[A \cup B = \{1,2,3,4,5,6\} = S \qquad (\text{since } B = \phi) \]
    \[A \cap B = \phi \]
    \[B \cup C = \phi \cup \{3,6\} = \{3,6\} \]
    \[E \cap F = \{6\} \cap \{3,4,5,6\} = \{6\} \]
    \[D \cap E = \{1,2,3\} \cap \{6\} = \phi \]
    \[A - C = \{1,2,3,4,5,6\} - \{3,6\} = \{1,2,4,5\} \]
    \[D - E = \{1,2,3\} - \{6\} = \{1,2,3\} \]
    Since \(\displaystyle F = \{3,4,5,6\} \), its complement is \(\displaystyle F' = \{1,2\} \), so
    \[E \cap F' = \{6\} \cap \{1,2\} = \phi \]
    \(\displaystyle A = S = \{1,2,3,4,5,6\},\; B = \phi,\; C = \{3,6\},\; D = \{1,2,3\},\; E = \{6\},\; F = \{3,4,5,6\} \); and \(\displaystyle A \cup B = \{1,2,3,4,5,6\},\; A \cap B = \phi,\; B \cup C = \{3,6\},\; E \cap F = \{6\},\; D \cap E = \phi,\; A - C = \{1,2,4,5\},\; D - E = \{1,2,3\},\; E \cap F' = \phi,\; F' = \{1,2\}. \)

  3. Exercise 3

    An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than $\displaystyle 8$, B: $\displaystyle 2$ occurs on either die C: the sum is at least $\displaystyle 7$ and a multiple of $\displaystyle 3$ . Which pairs of these events are mutually exclusive?

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    NCERT’s answer
    $\displaystyle \mathrm{A}=\{(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)\}$ $\displaystyle \mathrm{B}=\{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(2,1),(2,3),(2,4),(2,5),(2,6)\}$ $\displaystyle \mathrm{C}=\{(3,6),(6,3),(5,4),(4,5),(6,6)\}$ A and $\displaystyle \mathrm{B}, \mathrm{B}$ and C are mutually exclusive.
    Listing the events, then testing intersections. A pair of dice gives the $\displaystyle 36$ ordered pairs \[S = \{(x,y) : x,y = 1,2,3,4,5,6\} \]A: the sum is greater than $\displaystyle 8$, i.e. the sum is $\displaystyle 9$, $\displaystyle 10$, $\displaystyle 11$ or 12. \[A = \{(3,6),(4,5),(5,4),(6,3),\;(4,6),(5,5),(6,4),\;(5,6),(6,5),\;(6,6)\} \] ($\displaystyle 4$ + $\displaystyle 3$ + $\displaystyle 2$ + $\displaystyle 1$ = $\displaystyle 10$ outcomes.)B: $\displaystyle 2$ occurs on either die — every pair containing at least one 2. \[B = \{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2)\} \] ($\displaystyle 11$ outcomes; \(\displaystyle (2,2) \) is counted once.)C: the sum is at least $\displaystyle 7$ and a multiple of 3. The possible sums run from $\displaystyle 2$ to $\displaystyle 12$; those that are multiples of $\displaystyle 3$ are $\displaystyle 3$, $\displaystyle 6$, $\displaystyle 9$, $\displaystyle 12$, and of these only $\displaystyle 9$ and $\displaystyle 12$ are \(\displaystyle \geq 7 \). \[C = \{(3,6),(4,5),(5,4),(6,3),(6,6)\} \] ($\displaystyle 5$ outcomes.)Now test each pair for a common outcome.\(\displaystyle A \cap B \): every outcome of B contains a $\displaystyle 2$, so its largest possible sum is \(\displaystyle 2 + 6 = 8 \), while every outcome of A has sum \(\displaystyle \geq 9 \). Hence \(\displaystyle A \cap B = \phi \) — mutually exclusive.\(\displaystyle A \cap C \): every sum in C is $\displaystyle 9$ or $\displaystyle 12$, both greater than $\displaystyle 8$, so in fact \(\displaystyle C \subset A \) and \(\displaystyle A \cap C = C \neq \phi \) — not mutually exclusive.\(\displaystyle B \cap C \): none of the five pairs in C contains a $\displaystyle 2$, so \(\displaystyle B \cap C = \phi \) — mutually exclusive.A and B are mutually exclusive, and B and C are mutually exclusive; A and C are not (indeed \(\displaystyle C \subset A \)).

  4. Exercise 4

    Three coins are tossed once. Let A denote the event 'three heads show", B denote the event "two heads and one tail show", C denote the event" three tails show and D denote the event 'a head shows on the first coin". Which events are
    (i)
    mutually exclusive?
    (ii)
    simple?
    (iii)
    Compound?

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    NCERT’s answer
    (i)
    A and B; A and C; B and C; C and D (ii) A and C (iii) B and D
    Listing the events, then classifying them. Tossing three coins once gives \[S = \{HHH,\; HHT,\; HTH,\; THH,\; HTT,\; THT,\; TTH,\; TTT\} \] with $\displaystyle 8$ equally likely outcomes. The four events are \[A = \{HHH\}, \quad B = \{HHT, HTH, THH\}, \quad C = \{TTT\}, \quad D = \{HHH, HHT, HTH, HTT\} \] (D collects every outcome whose first coin is a head.)(i) Mutually exclusive pairs. Two events are mutually exclusive when their intersection is empty.\[A \cap B = \phi, \quad A \cap C = \phi, \quad B \cap C = \phi, \quad C \cap D = \phi \] \[A \cap D = \{HHH\} \neq \phi, \qquad B \cap D = \{HHT, HTH\} \neq \phi \]So the mutually exclusive pairs are (A, B), (A, C), (B, C) and (C, D).(ii) Simple events. A simple (elementary) event contains exactly one sample point. \(\displaystyle A = \{HHH\} \) and \(\displaystyle C = \{TTT\} \) each have one outcome.(iii) Compound events. A compound event contains more than one sample point: B has $\displaystyle 3$ outcomes and D has 4.(i) A and B, A and C, B and C, C and D are mutually exclusive; (ii) A and C are simple events; (iii) B and D are compound events.

  5. Exercise 5

    Three coins are tossed. Describe
    (i)
    Two events which are mutually exclusive.
    (ii)
    Three events which are mutually exclusive and exhaustive.
    (iii)
    Two events, which are not mutually exclusive.
    (iv)
    Two events which are mutually exclusive but not exhaustive.
    (v)
    Three events which are mutually exclusive but not exhaustive.

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    NCERT’s answer
    (i)
    "Getting at least two heads", and "getting at least two tails" (ii) "Getting no heads", "getting exactly one head" and "getting at least two heads" (iii) "Getting at most two tails", and "getting exactly two tails" (iv) "Getting exactly one head" and "getting exactly two heads" (v) "Getting exactly one tail", "getting exactly two tails", and getting exactly three tails" - Note There may be other events also as answer to the above question.
    Constructing events from the sample space. For three tosses, \[S = \{HHH,\; HHT,\; HTH,\; THH,\; HTT,\; THT,\; TTH,\; TTT\} \] Events are mutually exclusive if no two of them share an outcome, and exhaustive if together they cover the whole of \(\displaystyle S \).(i) Two mutually exclusive events. \[A: \text{at least two heads} = \{HHH, HHT, HTH, THH\}, \qquad B: \text{at least two tails} = \{TTT, TTH, THT, HTT\} \] \(\displaystyle A \cap B = \phi \), so they are mutually exclusive.(ii) Three mutually exclusive and exhaustive events. \[A: \text{no head} = \{TTT\}, \quad B: \text{exactly one head} = \{HTT, THT, TTH\}, \quad C: \text{two or more heads} = \{HHT, HTH, THH, HHH\} \] No two meet, and \(\displaystyle A \cup B \cup C = S \) ($\displaystyle 1$ + $\displaystyle 3$ + $\displaystyle 4$ = $\displaystyle 8$ outcomes), so they are exhaustive.(iii) Two events which are not mutually exclusive. \[A: \text{a head on the first coin} = \{HHH, HHT, HTH, HTT\}, \qquad B: \text{at least two heads} = \{HHH, HHT, HTH, THH\} \] Here \(\displaystyle A \cap B = \{HHH, HHT, HTH\} \neq \phi \).(iv) Two events, mutually exclusive but not exhaustive. \[A: \text{three heads} = \{HHH\}, \qquad B: \text{three tails} = \{TTT\} \] \(\displaystyle A \cap B = \phi \), but \(\displaystyle A \cup B = \{HHH, TTT\} \neq S \).(v) Three events, mutually exclusive but not exhaustive. \[A: \text{exactly one head} = \{HTT, THT, TTH\}, \quad B: \text{exactly two heads} = \{HHT, HTH, THH\}, \quad C: \text{exactly three heads} = \{HHH\} \] These are pairwise disjoint, but \(\displaystyle A \cup B \cup C \) omits \(\displaystyle TTT \), so they are not exhaustive.(Answers are not unique — any events with the stated intersection and union properties will do; the sets above are one valid choice for each part.)

  6. Exercise 6

    Two dice are thrown. The events A, B and C are as follows: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice \(\displaystyle \leq 5\). Describe the events
    (i)
    A' (ii) not B
    (iii)
    A or B
    (iv)
    A and B
    (v)
    A but not C
    (vi)
    B or C
    (vii)
    B and C
    (viii)
    \(\displaystyle \mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\)

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    NCERT’s answer
    $\displaystyle \mathrm{A}=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$, $\displaystyle (6,1)$, $\displaystyle (6,2)$, $\displaystyle (6,3)$, $\displaystyle (6,4)$, $\displaystyle (6,5)$, $\displaystyle (6,6)$\} $\displaystyle \mathrm{B}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$, $\displaystyle (5,1)$, $\displaystyle (5,2)$, $\displaystyle (5,3)$, $\displaystyle (5,4)$, $\displaystyle (5,5)$, $\displaystyle (5,6)$\} $\displaystyle \mathrm{C}=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\}$ (i) $\displaystyle \mathrm{A}^{\prime}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$, $\displaystyle (5,1)$, $\displaystyle (5,2)$, $\displaystyle (5,3)$, $\displaystyle (5,4)$, $\displaystyle (5,5)$, $\displaystyle (5,6)$ \} = B (ii) $\displaystyle \mathrm{B}^{\prime}=\{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$, $\displaystyle (6,1)$, $\displaystyle (6,2)$, $\displaystyle (6,3)$, $\displaystyle (6,4)$, $\displaystyle (6,5)$, $\displaystyle (6,6)$\} =A (iii) $\displaystyle \mathrm{A} \cup \mathrm{B}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5)$, $\displaystyle (3,6)$, $\displaystyle (5,1)$, $\displaystyle (5,2)$, $\displaystyle (5,3)$, $\displaystyle (5,4)$, $\displaystyle (5,5)$, $\displaystyle (5,6)$, $\displaystyle (2,1)$, $\displaystyle (2,2)$, $\displaystyle (2,3)$, $\displaystyle (2,5)$, $\displaystyle (2,6)$, $\displaystyle (4,1)$, $\displaystyle (4,2)$, $\displaystyle (4,3)$, $\displaystyle (4,4)$, $\displaystyle (4,5)$, $\displaystyle (4,6)$, $\displaystyle (6,1)$, $\displaystyle (6,2)$, $\displaystyle (6,3)$, $\displaystyle (6,4)$, $\displaystyle (6,5),(6,6)\}=\mathrm{S}$ (iv) $\displaystyle \mathrm{A} \cap \mathrm{B}=\phi$ (v) $\displaystyle \mathrm{A}-\mathrm{C}=\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3)$, $\displaystyle (6,4)$, $\displaystyle (6,5)$, $\displaystyle (6,6)$\} (vi) $\displaystyle \mathrm{B} \cup \mathrm{C}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(3,1),(3,2)$, $\displaystyle (3,3)$, $\displaystyle (3,4)$, $\displaystyle (3,5)$, $\displaystyle (3,6)$, $\displaystyle (4,1)$, $\displaystyle (5,1)$, $\displaystyle (5,2)$, $\displaystyle (5,3)$, $\displaystyle (5,4)$, $\displaystyle (5,5)$, $\displaystyle (5,6)$\} (vii) $\displaystyle \mathrm{B} \cap \mathrm{C}=\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}$ (viii) $\displaystyle \mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}=\{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2)$, $\displaystyle (6,3)$, $\displaystyle (6,4)$, $\displaystyle (6,5)$, $\displaystyle (6,6)$\}
    Describing events as subsets of the $\displaystyle 36$-point sample space. Throwing two dice, \[S = \{(x,y): x,y = 1,\dots,6\}, \qquad n(S) = 36 \] where \(\displaystyle x \) is the first die. The three events are \[A = \{(x,y) : x \text{ even}\} \ (18 \text{ outcomes}), \quad B = \{(x,y): x \text{ odd}\} \ (18), \quad C = \{(x,y): x + y \leq 5\} \ (10) \] Explicitly, \(\displaystyle C = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)\} \).Note first that A and B split \(\displaystyle S \) by the parity of the first die, so \(\displaystyle B = A' \) and \(\displaystyle A = B' \).(i) \(\displaystyle A' \) = the first die is not even = an odd number on the first die = B: \[A' = \{(1,y),(3,y),(5,y) : y = 1,\dots,6\} \](ii) not B \(\displaystyle = B' = A \) = an even number on the first die: \[B' = \{(2,y),(4,y),(6,y): y = 1,\dots,6\} \](iii) A or B \(\displaystyle = A \cup B \): the first die is either even or odd, which is always true, so \(\displaystyle A \cup B = S \), all $\displaystyle 36$ outcomes — a sure event.(iv) A and B \(\displaystyle = A \cap B \): the first die cannot be both even and odd, so \(\displaystyle A \cap B = \phi \) — an impossible event.(v) A but not C \(\displaystyle = A - C = A \cap C' \). From A remove the outcomes with sum \(\displaystyle \leq 5 \), namely \(\displaystyle (2,1),(2,2),(2,3),(4,1) \): \[A - C = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \] which has \(\displaystyle 18 - 4 = 14 \) outcomes.(vi) B or C \(\displaystyle = B \cup C \): all of B together with the members of C not already in B, i.e. \(\displaystyle (2,1),(2,2),(2,3),(4,1) \): \[B \cup C = \{(1,y),(3,y),(5,y): y = 1,\dots,6\} \cup \{(2,1),(2,2),(2,3),(4,1)\} \] which has \(\displaystyle 18 + 4 = 22 \) outcomes.(vii) B and C \(\displaystyle = B \cap C \) = outcomes of C whose first die is odd: \[B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\} \qquad (6 \text{ outcomes}) \](viii) \(\displaystyle A \cap B' \cap C' \). Since \(\displaystyle B' = A \), this is \(\displaystyle A \cap A \cap C' = A \cap C' = A - C \), exactly the set found in (v): \[A \cap B' \cap C' = \{(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \]\(\displaystyle A' = B \) (odd on the first die); \(\displaystyle B' = A \) (even on the first die); \(\displaystyle A \cup B = S \); \(\displaystyle A \cap B = \phi \); \(\displaystyle A - C \) is the $\displaystyle 14$-element set listed in (v); \(\displaystyle B \cup C \) has $\displaystyle 22$ outcomes; \(\displaystyle B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\} \); and \(\displaystyle A \cap B' \cap C' = A - C \), the same $\displaystyle 14$ outcomes as (v).

  7. Exercise 7

    Refer to question $\displaystyle 6$ above, state true or false: (give reason for your answer)
    (i)
    A and B are mutually exclusive
    (ii)
    A and B are mutually exclusive and exhaustive
    (iii)
    \(\displaystyle \mathrm{A}=\mathrm{B}^{\prime}\)
    (iv)
    A and C are mutually exclusive
    (v)
    A and B' are mutually exclusive.
    (vi)
    \(\displaystyle \mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}\) are mutually exclusive and exhaustive.

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    NCERT’s answer
    (i)
    True (ii) True (iii) True (iv) False (v) False (vi) False
    Deciding each statement from the sets of Question 6. There, \[A = \{(x,y): x \text{ even}\}, \quad B = \{(x,y): x \text{ odd}\}, \quad C = \{(x,y): x+y \leq 5\} \] with \(\displaystyle n(A) = n(B) = 18 \) and \(\displaystyle n(C) = 10 \).(i) A and B are mutually exclusive — TRUE. The first die cannot be even and odd at the same time, so \(\displaystyle A \cap B = \phi \).(ii) A and B are mutually exclusive and exhaustive — TRUE. Besides \(\displaystyle A \cap B = \phi \), every outcome has a first die that is either even or odd, so \(\displaystyle A \cup B = S \) (\(\displaystyle 18 + 18 = 36 \)).(iii) \(\displaystyle A = B' \) — TRUE. From (ii), A and B are disjoint and together fill \(\displaystyle S \); that is exactly the statement that each is the complement of the other.(iv) A and C are mutually exclusive — FALSE. For example \(\displaystyle (2,1) \) has an even first die and sum \(\displaystyle 3 \leq 5 \), so \(\displaystyle (2,1) \in A \cap C \). In fact \(\displaystyle A \cap C = \{(2,1),(2,2),(2,3),(4,1)\} \neq \phi \).(v) A and B' are mutually exclusive — FALSE. By (iii), \(\displaystyle B' = A \), so \(\displaystyle A \cap B' = A \cap A = A \neq \phi \). An event with outcomes can never be mutually exclusive with itself.(vi) \(\displaystyle A', B', C \) are mutually exclusive and exhaustive — FALSE. Here \(\displaystyle A' = B \) and \(\displaystyle B' = A \), so although \(\displaystyle A' \cap B' = B \cap A = \phi \), we have \[A' \cap C = B \cap C = \{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\} \neq \phi \] so the three events are not pairwise mutually exclusive. (They are exhaustive, since \(\displaystyle A' \cup B' = B \cup A = S \), but the mutual-exclusiveness fails, and the statement asks for both.)(i) True, (ii) True, (iii) True, (iv) False, (v) False, (vi) False.