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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 61–70 (part 7 of 11)

  1. Choose the correct answer from the given four options in each of the exercises from $\displaystyle 56$ to 82.

    Exercise 61

    If P(A)=0.4,P(B)=0.8\displaystyle \mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.8 and P(B∣A)=0.6\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})=0.6, then P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B}) is equal to
    (A)
    0.24\displaystyle 24
    (B)
    0.3\displaystyle 3
    (C)
    0.48\displaystyle 48
    (D)
    0.96\displaystyle 96

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    NCERT’s answer
    D
    (D) \(\displaystyle 0.96\)\[P(A\cap B)=P(B\mid A)\,P(A)=0.6\times0.4=0.24 \] \[P(A\cup B)=P(A)+P(B)-P(A\cap B) \] \[=0.4+0.8-0.24=0.96 \]

  2. Exercise 62

    If A and B are two events and A≠ϕ,B≠ϕ\displaystyle \mathrm{A} \neq \phi, \mathrm{B} \neq \phi, then
    (A)
    P(A∣B)=P(A)⋅P(B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})
    (B)
    P(A∣B)=P(A∩B)P(B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}
    (C)
    P(A∣B)⋅P(B∣A)=1\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B}) \cdot \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1
    (D)
    P(A∣B)=P(A)∣P(B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A}) \mid \mathrm{P}(\mathrm{B})

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    NCERT’s answer
    B
    (B) This is the definition of conditional probability.\[P(A\mid B)=\frac{P(A\cap B)}{P(B)} \]For disjoint events of positive probability, \(\displaystyle P(A\mid B)=0\), so (A) and (C) fail; (D) is not a probability statement.

  3. Exercise 63

    A and B are events such that P(A)=0.4,P(B)=0.3\displaystyle \mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3 and P(A∪B)=0.5\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5. Then P(B′∩A)\displaystyle \mathrm{P}\left(\mathrm{B}^{\prime} \cap \mathrm{A}\right) equals
    (A)
    23\displaystyle \frac{2}{3}
    (B)
    12\displaystyle \frac{1}{2}
    (C)
    310\displaystyle \frac{3}{10}
    (D)
    15\displaystyle \frac{1}{5}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{1}{5}\)\[P(A\cap B)=P(A)+P(B)-P(A\cup B) \] \[=0.4+0.3-0.5=0.2 \] \[P(B'\cap A)=P(A)-P(A\cap B) \] \[=0.4-0.2=0.2=\frac{1}{5} \]

  4. Exercise 64

    You are given that A and B are two events such that P(B)=35,P(A∣B)=12\displaystyle \mathrm{P}(\mathrm{B})=\frac{3}{5}, \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{2} and P(A∪B)=45\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{4}{5}, then P(A)\displaystyle \mathrm{P}(\mathrm{A}) equals
    (A)
    310\displaystyle \frac{3}{10}
    (B)
    15\displaystyle \frac{1}{5}
    (C)
    12\displaystyle \frac{1}{2}
    (D)
    35\displaystyle \frac{3}{5}

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{1}{2}\)\[P(A\cap B)=P(A\mid B)\,P(B)=\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10} \] \[P(A\cup B)=P(A)+P(B)-P(A\cap B) \] \[\frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10} \] \[P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{2} \]

  5. Exercise 65

    In Exercise 64\displaystyle 64 above, P(B∣A′)\displaystyle \mathrm{P}\left(\mathrm{B} \mid \mathrm{A}^{\prime}\right) is equal to
    (A)
    15\displaystyle \frac{1}{5}
    (B)
    310\displaystyle \frac{3}{10}
    (C)
    12\displaystyle \frac{1}{2}
    (D)
    35\displaystyle \frac{3}{5}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{3}{5}\)\[P(A\cap B)=P(A\mid B)\,P(B)=\frac12\cdot\frac35=\frac{3}{10} \] \[P(A)=P(A\cup B)-P(B)+P(A\cap B)=\frac45-\frac35+\frac{3}{10}=\frac12 \] \[P(A'\cap B)=P(B)-P(A\cap B)=\frac35-\frac{3}{10}=\frac{3}{10} \] \[P(B\mid A')=\frac{P(A'\cap B)}{P(A')}=\frac{3/10}{1-\frac12}=\frac35 \]

  6. Exercise 66

    If P(B)=35,P(A∣B)=12\displaystyle \mathrm{P}(\mathrm{B})=\frac{3}{5}, \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1}{2} and P(A∪B)=45\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{4}{5}, then P(A∪B)′+P(A′∪B)=\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=
    (A)
    15\displaystyle \frac{1}{5}
    (B)
    45\displaystyle \frac{4}{5}
    (C)
    12\displaystyle \frac{1}{2}

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    NCERT’s answer
    D
    (D) \(\displaystyle 1\)\[P(A\cap B)=P(A\mid B)\,P(B)=\frac12\cdot\frac35=\frac{3}{10} \] \[P(A)=P(A\cup B)-P(B)+P(A\cap B)=\frac45-\frac35+\frac{3}{10}=\frac12 \] \[P\big((A\cup B)'\big)=1-\frac45=\frac15 \] \[P(A'\cup B)=P(A')+P(B)-P(A'\cap B)=\frac12+\frac35-\frac{3}{10}=\frac45 \] \[P\big((A\cup B)'\big)+P(A'\cup B)=\frac15+\frac45=1 \]

  7. Exercise 67

    Let P(A)=713,P(B)=913\displaystyle \mathrm{P}(\mathrm{A})=\frac{7}{13}, \mathrm{P}(\mathrm{B})=\frac{9}{13} and P(A∩B)=413\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{13}. Then P(A′∣B)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right) is equal to
    (A)
    613\displaystyle \frac{6}{13}
    (B)
    413\displaystyle \frac{4}{13}
    (C)
    49\displaystyle \frac{4}{9}
    (D)
    59\displaystyle \frac{5}{9}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{5}{9}\)\[P(A'\cap B)=P(B)-P(A\cap B)=\frac{9}{13}-\frac{4}{13}=\frac{5}{13} \] \[P(A'\mid B)=\frac{P(A'\cap B)}{P(B)}=\frac{5/13}{9/13}=\frac59 \]

  8. Exercise 68

    If A and B are such events that P(A)>0\displaystyle \mathrm{P}(\mathrm{A})>0 and P(B)≠1\displaystyle \mathrm{P}(\mathrm{B}) \neq 1, then P(A′∣B′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right) equals.
    (A)
    1−P(A∣B)\displaystyle 1-\mathrm{P}(\mathrm{A} \mid \mathrm{B})
    (B)
    1−P(A′∣B)\displaystyle 1-\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)
    (C)
    1−P(A∪B)P(B′)\displaystyle \frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}
    (D)
    P(A′)∣P(B′)\displaystyle P\left(A^{\prime}\right) \mid P\left(B^{\prime}\right)

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{1-P(A\cup B)}{P(B')}\)\[A'\cap B'=(A\cup B)' \quad \text{(De Morgan)} \] \[P(A'\mid B')=\frac{P(A'\cap B')}{P(B')}=\frac{P\big((A\cup B)'\big)}{P(B')}=\frac{1-P(A\cup B)}{P(B')} \]

  9. Exercise 69

    If A and B are two independent events with P(A)=35\displaystyle \mathrm{P}(\mathrm{A})=\frac{3}{5} and P(B)=49\displaystyle \mathrm{P}(\mathrm{B})=\frac{4}{9}, then P(A′∩B′)\displaystyle P\left(A^{\prime} \cap B^{\prime}\right) equals
    (A)
    415\displaystyle \frac{4}{15}
    (B)
    845\displaystyle \frac{8}{45}
    (C)
    13\displaystyle \frac{1}{3}
    (D)
    29\displaystyle \frac{2}{9}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{2}{9}\)A and B independent, so A' and B' are independent. \[P(A')=1-\frac35=\frac25,\qquad P(B')=1-\frac49=\frac59 \] \[P(A'\cap B')=P(A')\,P(B')=\frac25\cdot\frac59=\frac29 \]

  10. Exercise 70

    If two events are independent, then
    (A)
    they must be mutually exclusive
    (B)
    the sum of their probabilities must be equal to 1\displaystyle 1
    (C) (A) and (B) both are correct
    (D)
    None of the above is correct

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    NCERT’s answer
    D
    (D) None of the above is correct.Example: a coin shows head (A), a die shows six (B). \[P(A)=\frac12,\quad P(B)=\frac16,\quad P(A\cap B)=\frac1{12}=P(A)\,P(B) \] Independent, yet \[P(A\cap B)=\frac1{12}\neq 0 \quad\Rightarrow\quad \text{not mutually exclusive} \] \[P(A)+P(B)=\frac23\neq 1 \]