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NCERT Solutions · Class 12 Mathematics Probability

62 questions · 62 still being checked

EXERCISE 13.2 11–18 (part 4 of 7)

  1. Exercise 11

    Given two independent events A and B such that P(A)=0.3,P(B)=0.6\displaystyle \mathrm{P}(\mathrm{A})=0.3, \mathrm{P}(\mathrm{B})=0.6. Find
    (i)
    P(A\displaystyle \mathrm{P}(\mathrm{A} and B)\displaystyle )
    (ii)
    P(A\displaystyle \mathrm{P}(\mathrm{A} and not B)\displaystyle )
    (iii)
    P(A\displaystyle \mathrm{P}(\mathrm{A} or B)\displaystyle )
    (iv)
    P (neither A nor B )

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    NCERT’s answer
    (i)
    0.$\displaystyle 18$ (ii) $\displaystyle 0.12$ (iii) $\displaystyle 0.72$ (iv) $\displaystyle 0.28$
    A and B are independent with \(\displaystyle \mathrm{P}(\mathrm{A})=0.3\), \(\displaystyle \mathrm{P}(\mathrm{B})=0.6\), so also \(\displaystyle \mathrm{P}(\mathrm{A}')=0.7\) and \(\displaystyle \mathrm{P}(\mathrm{B}')=0.4\). Recall that if A and B are independent, then so are the pairs \(\displaystyle (\mathrm{A},\mathrm{B}')\) and \(\displaystyle (\mathrm{A}',\mathrm{B}')\).
    (i) \(\displaystyle \mathrm{P}(\mathrm{A}\text{ and }\mathrm{B})=\mathrm{P}(\mathrm{A}\cap\mathrm{B})=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})=0.3\times0.6=0.18.\)
    (ii) \(\displaystyle \mathrm{P}(\mathrm{A}\text{ and not }\mathrm{B})=\mathrm{P}(\mathrm{A}\cap\mathrm{B}')=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}')=0.3\times0.4=0.12.\)
    (iii) By the addition theorem,
    \[\mathrm{P}(\mathrm{A}\text{ or }\mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap\mathrm{B})=0.3+0.6-0.18=0.72. \]
    (iv) 'Neither A nor B' is \(\displaystyle \mathrm{A}'\cap\mathrm{B}'=(\mathrm{A}\cup\mathrm{B})'\), so
    \[\mathrm{P}(\text{neither A nor B})=1-\mathrm{P}(\mathrm{A}\cup\mathrm{B})=1-0.72=0.28, \]
    which checks against \(\displaystyle \mathrm{P}(\mathrm{A}')\,\mathrm{P}(\mathrm{B}')=0.7\times0.4=0.28\).
    (i)
    \(\displaystyle 0.18\) (ii) \(\displaystyle 0.12\) (iii) \(\displaystyle 0.72\) (iv) \(\displaystyle 0.28\).

  2. Exercise 12

    A die is tossed thrice. Find the probability of getting an odd number at least once.

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    NCERT’s answer
    \(\displaystyle \frac{7}{8}\)
    The three tosses of a die are independent, and on one toss the odd numbers are \(\displaystyle \{1,3,5\}\), so \[\mathrm{P}(\text{odd on a toss})=\frac{3}{6}=\frac{1}{2},\qquad \mathrm{P}(\text{even on a toss})=\frac{1}{2}. \] 'At least once' is awkward to count directly (it covers exactly one, exactly two and all three), so use the complement: the opposite of 'odd at least once' is 'no odd number at all', i.e. an even number on every toss. By independence, \[\mathrm{P}(\text{no odd number})=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}. \] Hence \[\mathrm{P}(\text{odd at least once})=1-\frac{1}{8}=\frac{7}{8}. \] The required probability is \(\displaystyle \dfrac{7}{8}\).

  3. Exercise 13

    Two balls are drawn at random with replacement from a box containing 10\displaystyle 10 black and 8\displaystyle 8 red balls. Find the probability that
    (i)
    both balls are red.
    (ii)
    first ball is black and second is red.
    (iii)
    one of them is black and other is red.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{16}{81}\), (ii) \(\displaystyle \frac{20}{81}\), (iii) \(\displaystyle \frac{40}{81}\)
    The box holds \(\displaystyle 10+8=18\) balls, and the drawing is with replacement, so the composition of the box is the same for both draws and the two draws are independent:
    \[\mathrm{P}(\text{black on a draw})=\frac{10}{18}=\frac{5}{9},\qquad \mathrm{P}(\text{red on a draw})=\frac{8}{18}=\frac{4}{9}. \]
    (i) Both red.
    \[\mathrm{P}=\frac{4}{9}\times\frac{4}{9}=\frac{16}{81}. \]
    (ii) First black and second red. The order is fixed here, so there is exactly one arrangement:
    \[\mathrm{P}=\frac{5}{9}\times\frac{4}{9}=\frac{20}{81}. \]
    (iii) One black and the other red. Now the order is not fixed, so both arrangements count:
    \[\mathrm{P}=\mathrm{P}(\text{black, red})+\mathrm{P}(\text{red, black})=\frac{5}{9}\times\frac{4}{9}+\frac{4}{9}\times\frac{5}{9}=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}. \]
    (i)
    \(\displaystyle \dfrac{16}{81}\) (ii) \(\displaystyle \dfrac{20}{81}\) (iii) \(\displaystyle \dfrac{40}{81}\).

  4. Exercise 14

    Probability of solving specific problem independently by A and B are 12\displaystyle \frac{1}{2} and 13\displaystyle \frac{1}{3} respectively. If both try to solve the problem independently, find the probability that
    (i)
    the problem is solved
    (ii)
    exactly one of them solves the problem.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{2}{3}\), (ii) \(\displaystyle \frac{1}{2}\)
    Let A be the event 'A solves the problem' and B the event 'B solves the problem', with
    \[\mathrm{P}(\mathrm{A})=\frac{1}{2},\quad \mathrm{P}(\mathrm{B})=\frac{1}{3},\qquad \mathrm{P}(\mathrm{A}')=\frac{1}{2},\quad \mathrm{P}(\mathrm{B}')=\frac{2}{3}, \]
    and A, B independent (given that they try independently).
    (i) The problem is solved means at least one of them solves it, i.e. \(\displaystyle \mathrm{A}\cup\mathrm{B}\). Using the complement 'neither solves it':
    \[\mathrm{P}(\mathrm{A}\cup\mathrm{B})=1-\mathrm{P}(\mathrm{A}')\,\mathrm{P}(\mathrm{B}')=1-\frac{1}{2}\times\frac{2}{3}=1-\frac{1}{3}=\frac{2}{3}. \]
    (The same value comes from \(\displaystyle \frac12+\frac13-\frac12\cdot\frac13=\frac{2}{3}\).)
    (ii) Exactly one of them solves it means A solves and B does not, or B solves and A does not — two disjoint cases:
    \[\mathrm{P}(\mathrm{A}\cap\mathrm{B}')+\mathrm{P}(\mathrm{A}'\cap\mathrm{B})=\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}. \]
    (i)
    \(\displaystyle \dfrac{2}{3}\) (ii) \(\displaystyle \dfrac{1}{2}\).

  5. Exercise 15

    One card is drawn at random from a well shuffled deck of 52\displaystyle 52 cards. In which of the following cases are the events E and F independent ?
    (i)
    E : 'the card drawn is a spade' F : 'the card drawn is an ace'
    (ii)
    E : 'the card drawn is black' F : 'the card drawn is a king'
    (iii)
    E : 'the card drawn is a king or queen' F : 'the card drawn is a queen or jack'.

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    NCERT’s answer
    (i)
    , (ii)
    One card is drawn from $\displaystyle 52$, so every probability below is (number of favourable cards)/52. In each part test whether \(\displaystyle \mathrm{P}(\mathrm{E}\cap\mathrm{F})=\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})\).(i) E: spade, F: ace. There are $\displaystyle 13$ spades and $\displaystyle 4$ aces, and \(\displaystyle \mathrm{E}\cap\mathrm{F}\) is the single card 'ace of spades': \[\mathrm{P}(\mathrm{E})=\frac{13}{52}=\frac{1}{4},\quad \mathrm{P}(\mathrm{F})=\frac{4}{52}=\frac{1}{13},\quad \mathrm{P}(\mathrm{E}\cap\mathrm{F})=\frac{1}{52}. \] \[\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})=\frac{1}{4}\times\frac{1}{13}=\frac{1}{52}=\mathrm{P}(\mathrm{E}\cap\mathrm{F})\ \Rightarrow\ \textbf{independent}. \](ii) E: black, F: king. There are $\displaystyle 26$ black cards and $\displaystyle 4$ kings; \(\displaystyle \mathrm{E}\cap\mathrm{F}\) is the two black kings: \[\mathrm{P}(\mathrm{E})=\frac{26}{52}=\frac{1}{2},\quad \mathrm{P}(\mathrm{F})=\frac{4}{52}=\frac{1}{13},\quad \mathrm{P}(\mathrm{E}\cap\mathrm{F})=\frac{2}{52}=\frac{1}{26}. \] \[\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})=\frac{1}{2}\times\frac{1}{13}=\frac{1}{26}=\mathrm{P}(\mathrm{E}\cap\mathrm{F})\ \Rightarrow\ \textbf{independent}. \](iii) E: king or queen, F: queen or jack. Each of E and F contains \(\displaystyle 4+4=8\) cards, and their overlap is exactly the $\displaystyle 4$ queens: \[\mathrm{P}(\mathrm{E})=\frac{8}{52}=\frac{2}{13},\quad \mathrm{P}(\mathrm{F})=\frac{8}{52}=\frac{2}{13},\quad \mathrm{P}(\mathrm{E}\cap\mathrm{F})=\frac{4}{52}=\frac{1}{13}. \] \[\mathrm{P}(\mathrm{E})\,\mathrm{P}(\mathrm{F})=\frac{2}{13}\times\frac{2}{13}=\frac{4}{169}\neq\frac{13}{169}=\frac{1}{13}=\mathrm{P}(\mathrm{E}\cap\mathrm{F})\ \Rightarrow\ \textbf{not independent}. \]So E and F are independent in cases (i) and (ii), and not independent in case (iii).

  6. Exercise 16

    In a hostel, 60\displaystyle 60% of the students read Hindi newspaper, 40\displaystyle 40% read English newspaper and 20%\displaystyle 20 \% read both Hindi and English newspapers. A student is selected at random.
    (a)
    Find the probability that she reads neither Hindi nor English newspapers.
    (b)
    If she reads Hindi newspaper, find the probability that she reads English newspaper.
    (c)
    If she reads English newspaper, find the probability that she reads Hindi newspaper.

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    Let H be the event 'the student reads the Hindi newspaper' and E the event 'the student reads the English newspaper'. Converting the percentages to probabilities,
    \[\mathrm{P}(\mathrm{H})=0.6=\frac{3}{5},\qquad \mathrm{P}(\mathrm{E})=0.4=\frac{2}{5},\qquad \mathrm{P}(\mathrm{H}\cap\mathrm{E})=0.2=\frac{1}{5}. \]
    (a) Neither newspaper. 'Neither H nor E' is \(\displaystyle \mathrm{H}'\cap\mathrm{E}'=(\mathrm{H}\cup\mathrm{E})'\) by De Morgan's law. First, by the addition theorem,
    \[\mathrm{P}(\mathrm{H}\cup\mathrm{E})=0.6+0.4-0.2=0.8, \]
    so \[\mathrm{P}(\mathrm{H}'\cap\mathrm{E}')=1-0.8=0.2=\frac{1}{5}. \]
    (b) Reads English given she reads Hindi. By the definition of conditional probability,
    \[\mathrm{P}(\mathrm{E}\mid\mathrm{H})=\frac{\mathrm{P}(\mathrm{E}\cap\mathrm{H})}{\mathrm{P}(\mathrm{H})}=\frac{0.2}{0.6}=\frac{1}{3}. \]
    (c) Reads Hindi given she reads English. Same definition, the other way round — the numerator is unchanged, only the conditioning denominator switches:
    \[\mathrm{P}(\mathrm{H}\mid\mathrm{E})=\frac{\mathrm{P}(\mathrm{H}\cap\mathrm{E})}{\mathrm{P}(\mathrm{E})}=\frac{0.2}{0.4}=\frac{1}{2}. \]
    (a)
    \(\displaystyle \dfrac{1}{5}\) (b) \(\displaystyle \dfrac{1}{3}\) (c) \(\displaystyle \dfrac{1}{2}\).
  7. Choose the correct answer in Exercises $\displaystyle 17$ and 18.

    Exercise 17

    The probability of obtaining an even prime number on each die, when a pair of dice is rolled is (A) 0\displaystyle 0 (B) 13\displaystyle \frac{1}{3} (C) 112\displaystyle \frac{1}{12} (D) 136\displaystyle \frac{1}{36}

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    NCERT’s answer
    D
    The key observation is that $\displaystyle 2$ is the only even prime number, so on each die the favourable outcome is the single face $\displaystyle 2$: \[\mathrm{P}(\text{even prime on one die})=\frac{1}{6}. \] The two dice are independent, so \[\mathrm{P}(\text{even prime on each die})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}. \] (Equivalently, of the \(\displaystyle 6\times6=36\) equally likely ordered pairs only \(\displaystyle (2,2)\) is favourable.)The correct answer is (D) \(\displaystyle \dfrac{1}{36}\).

  8. Exercise 18

    Two events A and B will be independent, if (A) A and B are mutually exclusive (B) P(A′B′)=[1−P(A)][1−P(B)]\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \mathrm{B}^{\prime}\right)=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] (C) P(A)=P(B)\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B}) (D) P(A)+P(B)=1\displaystyle \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1

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    NCERT’s answer
    B
    By definition, A and B are independent when \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})\). Test the options against this.
    (A)
    Mutually exclusive means \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=0\), which equals \(\displaystyle \mathrm{P}(\mathrm{A})\mathrm{P}(\mathrm{B})\) only in the trivial case that one of the events has probability 0. So this is not a condition for independence — in fact for events of nonzero probability mutually exclusive events are dependent.
    (C)
    and (D) are statements about the individual probabilities only; neither says anything about \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})\), so neither can force independence.
    (B)
    states \(\displaystyle \mathrm{P}(\mathrm{A}'\cap\mathrm{B}')=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]=\mathrm{P}(\mathrm{A}')\,\mathrm{P}(\mathrm{B}')\), i.e. \(\displaystyle \mathrm{A}'\) and \(\displaystyle \mathrm{B}'\) are independent. Check that this is equivalent to A and B being independent: by De Morgan's law and the addition theorem,
    \[\mathrm{P}(\mathrm{A}'\cap\mathrm{B}')=1-\mathrm{P}(\mathrm{A}\cup\mathrm{B})=1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}\cap\mathrm{B}), \]
    while
    \[[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]=1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B}). \]
    Equating the two sides cancels the common terms and leaves exactly \(\displaystyle \mathrm{P}(\mathrm{A}\cap\mathrm{B})=\mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})\).
    The correct answer is (B).