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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 91–100 (part 10 of 11)

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

    Exercise 91

    In a college, 30\displaystyle 30% students fail in physics, 25\displaystyle 25% fail in mathematics and 10\displaystyle 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is
    (A)
    110\displaystyle \frac{1}{10}
    (B)
    25\displaystyle \frac{2}{5}
    (C)
    920\displaystyle \frac{9}{20}
    (D)
    13\displaystyle \frac{1}{3}

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    NCERT’s answer
    B
    (B) \(\displaystyle \tfrac25\)\(\displaystyle A\): fails physics, \(\displaystyle B\): fails mathematics. \[P(A)=0.30,\quad P(B)=0.25,\quad P(A\cap B)=0.10 \] \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.10}{0.25}=\frac{2}{5} \]

  2. Exercise 92

    A and B are two students. Their chances of solving a problem correctly are 13\displaystyle \frac{1}{3} and 14\displaystyle \frac{1}{4}, respectively. If the probability of their making a common error is, 120\displaystyle \frac{1}{20} and they obtain the same answer, then the probability of their answer to be correct is
    (A)
    112\displaystyle \frac{1}{12}
    (B)
    140\displaystyle \frac{1}{40}
    (C)
    13120\displaystyle \frac{13}{120}
    (D)
    1013\displaystyle \frac{10}{13}

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    NCERT’s answer
    D
    (D) \(\displaystyle \tfrac{10}{13}\)Common error is taken as \(\displaystyle \tfrac1{20}\) of the cases in which both are wrong. \[P(\text{both correct})=\tfrac13\cdot\tfrac14=\tfrac1{12} \] \[P(\text{both wrong})=\tfrac23\cdot\tfrac34=\tfrac12 \] \[P(\text{same wrong answer})=\tfrac12\cdot\tfrac1{20}=\tfrac1{40} \] \[P(\text{same answer})=\tfrac1{12}+\tfrac1{40}=\tfrac{13}{120} \] \[P(\text{correct}\mid\text{same answer})=\frac{1/12}{13/120}=\frac{10}{13} \]

  3. Exercise 93

    A box has 100\displaystyle 100 pens of which 10\displaystyle 10 are defective. What is the probability that out of a sample of 5\displaystyle 5 pens drawn one by one with replacement at most one is defective?
    (A)
    (910)5\displaystyle \left(\frac{9}{10}\right)^5
    (B)
    12(910)4\displaystyle \frac{1}{2}\left(\frac{9}{10}\right)^4
    (C)
    12(910)5\displaystyle \frac{1}{2}\left(\frac{9}{10}\right)^5
    (D)
    (910)5+12(910)4\displaystyle \left(\frac{9}{10}\right)^5+\frac{1}{2}\left(\frac{9}{10}\right)^4

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    NCERT’s answer
    D
    (D) \(\displaystyle \left(\tfrac{9}{10}\right)^5+\tfrac12\left(\tfrac{9}{10}\right)^4\)Drawing with replacement gives a binomial distribution: \[n=5,\quad p=\tfrac{10}{100}=\tfrac1{10},\quad q=\tfrac{9}{10} \] \[P(X\le1)=P(X=0)+P(X=1)=\binom50q^5+\binom51pq^4 \] \[=\left(\tfrac{9}{10}\right)^5+5\cdot\tfrac1{10}\left(\tfrac{9}{10}\right)^4=\left(\tfrac{9}{10}\right)^5+\tfrac12\left(\tfrac{9}{10}\right)^4 \]
  4. State True or False for the statements in each of the Exercises $\displaystyle 94$ to 103.

    Exercise 94

    Let P(A)>0\displaystyle \mathrm{P}(\mathrm{A})>0 and P(B)>0\displaystyle \mathrm{P}(\mathrm{B})>0. Then A and B can be both mutually exclusive and independent.

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    NCERT’s answer
    False
    False. The two conditions give contradictory values of \(\displaystyle P(A\cap B)\). \[A,B\ \text{mutually exclusive}:\quad P(A\cap B)=0 \] \[A,B\ \text{independent}:\quad P(A\cap B)=P(A)\,P(B)>0 \quad\text{since } P(A)>0,\ P(B)>0 \]

  5. Exercise 95

    If A and B are independent events, then A′\displaystyle \mathrm{A}^{\prime} and B′\displaystyle \mathrm{B}^{\prime} are also independent.

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    NCERT’s answer
    True
    True. \[P(A\cap B)=P(A)\,P(B) \quad\text{(given)} \] \[P(A'\cap B')=P\big((A\cup B)'\big)=1-P(A\cup B) \quad\text{(De Morgan)} \] \[=1-P(A)-P(B)+P(A\cap B) \] \[=1-P(A)-P(B)+P(A)\,P(B) \] \[=\big(1-P(A)\big)\big(1-P(B)\big)=P(A')\,P(B') \]

  6. Exercise 96

    If A and B are mutually exclusive events, then they will be independent also.

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    NCERT’s answer
    False
    False. Mutually exclusive events have \(\displaystyle P(A\cap B)=0\), which differs from \(\displaystyle P(A)P(B)\) unless one of them is impossible.Toss a coin: \(\displaystyle A\) = head, \(\displaystyle B\) = tail. \[P(A\cap B)=0\ne\tfrac14=P(A)\,P(B) \]

  7. Exercise 97

    Two independent events are always mutually exclusive.

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    NCERT’s answer
    False
    False. Independent events of positive probability cannot be disjoint: \[P(A\cap B)=P(A)\,P(B)>0 \] \[\text{mutually exclusive} \Rightarrow P(A\cap B)=0 \] \[P(A)=P(B)=\tfrac12 \Rightarrow P(A\cap B)=\tfrac14\neq 0 \]

  8. Exercise 98

    If A and B are two independent events then P(A and B)=P(A).P(B)\displaystyle \mathrm{P}(\mathrm{A} \text { and } \mathrm{B})=\mathrm{P}(\mathrm{A}) . \mathrm{P}(\mathrm{B}).

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    NCERT’s answer
    True
    True. This is the multiplication rule that defines independence: \[P(A\cap B)=P(A)\,P(B) \]

  9. Exercise 99

    Another name for the mean of a probability distribution is expected value.

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    NCERT’s answer
    True
    True. The mean of a probability distribution is its expected value: \[E(X)=\sum_{i} x_i\,p_i \]

  10. Exercise 100

    If A and B′\displaystyle \mathrm{B}^{\prime} are independent events, then P(A′∪B)=1−P(A)P(B′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=1-\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right)

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    NCERT’s answer
    True
    True. \(\displaystyle A'\cup B\) is the complement of \(\displaystyle A\cap B'\) (De Morgan), and \(\displaystyle A,\,B'\) are independent: \[P(A'\cup B)=1-P(A\cap B') \] \[P(A'\cup B)=1-P(A)\,P(B') \]