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NCERT Exemplar · Class 11 Mathematics Probability

43 questions · 43 still being checked

EXERCISE 16.3 21–30 (part 3 of 4)

  1. Choose the correct answer out of four given options in each of the Exercises $\displaystyle 18$ to $\displaystyle 29$ (M.C.Q.).

    Exercise 21

    Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is
    (A)
    13\displaystyle \frac{1}{3}
    (B)
    16\displaystyle \frac{1}{6}
    (C)
    27\displaystyle \frac{2}{7}
    (D)
    12\displaystyle \frac{1}{2}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{2}{7}\)Treat the two persons as one block, which can be ordered in \(\displaystyle 2!\) ways: \[n(S)=7! \] \[n(E)=6!\cdot 2! \] \[P=\frac{6!\cdot 2!}{7!}=\frac27 \]

  2. Exercise 22

    Without repetition of the numbers, four digit numbers are formed with the numbers 0\displaystyle 0, 2\displaystyle 2, 3\displaystyle 3, 5. The probability of such a number divisible by 5\displaystyle 5 is
    (A)
    15\displaystyle \frac{1}{5}
    (B)
    45\displaystyle \frac{4}{5}
    (C)
    130\displaystyle \frac{1}{30}
    (D)
    59\displaystyle \frac{5}{9}

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    NCERT’s answer
    D
    (D) \(\displaystyle \frac{5}{9}\)The first digit cannot be $\displaystyle 0$: \[n(S)=3\cdot 3!=18 \] Divisible by $\displaystyle 5$ means the last digit is $\displaystyle 0$ or 5. \[\text{last digit }0:\quad 3!=6 \] \[\text{last digit }5:\quad 2\cdot 2!=4 \quad(\text{first digit }2\text{ or }3) \] \[P=\frac{6+4}{18}=\frac59 \]

  3. Exercise 23

    If A and B are mutually exclusive events, then
    (A)
    P(A)≤P(B‾)\displaystyle \mathrm{P}(\mathrm{A}) \leq \mathrm{P}(\overline{\mathrm{B}})
    (B)
    P(A)≥P(B‾)\displaystyle \mathrm{P}(\mathrm{A}) \geq \mathrm{P}(\overline{\mathrm{B}})
    (C)
    P(A)<P(B‾)\displaystyle \mathrm{P}(\mathrm{A})<\mathrm{P}(\overline{\mathrm{B}})
    (D)
    none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle \mathrm{P}(\mathrm{A})\le\mathrm{P}(\overline{\mathrm{B}})\)Mutually exclusive events: \[\mathrm{A}\cap\mathrm{B}=\varnothing \implies \mathrm{A}\subseteq\overline{\mathrm{B}} \] \[\mathrm{P}(\mathrm{A})\le\mathrm{P}(\overline{\mathrm{B}}) \] Equality holds when \(\displaystyle \mathrm{A}=\overline{\mathrm{B}}\), so the inequality cannot be strict.

  4. Exercise 24

    If P(A∪B)=P(A∩B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) for any two events A and B, then
    (A)
    P(A)=P(B)\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})
    (B)
    P(A)>P(B)\displaystyle \mathrm{P}(\mathrm{A})>\mathrm{P}(\mathrm{B})
    (C)
    P(A)<P(B)\displaystyle \mathrm{P}(\mathrm{A})<\mathrm{P}(\mathrm{B})
    (D)
    none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})\)\[\mathrm{A}\cap\mathrm{B}\subseteq\mathrm{A}\subseteq\mathrm{A}\cup\mathrm{B} \] \[\mathrm{P}(\mathrm{A}\cap\mathrm{B})\le\mathrm{P}(\mathrm{A})\le\mathrm{P}(\mathrm{A}\cup\mathrm{B})=\mathrm{P}(\mathrm{A}\cap\mathrm{B}) \] \[\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A}\cap\mathrm{B}) \] The same chain for \(\displaystyle \mathrm{B}\) gives \(\displaystyle \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A}\cap\mathrm{B})\), so \(\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})\).

  5. Exercise 25

    6\displaystyle 6 boys and 6\displaystyle 6 girls sit in a row at random. The probability that all the girls sit together is
    (A)
    1432\displaystyle \frac{1}{432}
    (B)
    12431\displaystyle \frac{12}{431}
    (C)
    1132\displaystyle \frac{1}{132}
    (D)
    none of these

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{1}{132}\)Glue the $\displaystyle 6$ girls into one block: the $\displaystyle 7$ units arrange in \(\displaystyle 7!\) ways, the girls inside in \(\displaystyle 6!\) ways.\[n(S) = 12! \] \[n(E) = 7!\,6! \] \[P(E) = \frac{7!\,6!}{12!} = \frac{6!}{12\cdot 11\cdot 10\cdot 9\cdot 8} \] \[P(E) = \frac{720}{95040} = \frac{1}{132} \]

  6. Exercise 26

    A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel is
    (A)
    13\displaystyle \frac{1}{3}
    (B)
    411\displaystyle \frac{4}{11}
    (C)
    211\displaystyle \frac{2}{11}
    (D)
    311\displaystyle \frac{3}{11}

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{4}{11}\)PROBABILITY has $\displaystyle 11$ letters; the vowels are O, A, I, I.\[P(\text{vowel}) = \frac{4}{11} \]

  7. Exercise 27

    If the probabilities for A to fail in an examination is 0.2\displaystyle 0.2 and that for B is 0.3\displaystyle 0.3, then the probability that either A or B fails is
    (A)
    >.5\displaystyle >.5 (B) .5 (C) ≤.5\displaystyle \leq .5

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    NCERT’s answer
    C
    (C) \(\displaystyle \le .5\)\[P(A\cup B) = P(A)+P(B)-P(A\cap B) \] \[P(A\cup B) = 0.2+0.3-P(A\cap B) = 0.5-P(A\cap B) \] \[P(A\cap B)\ge 0 \;\Longrightarrow\; P(A\cup B)\le 0.5 \]

  8. Exercise 28

    The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2\displaystyle 0.2, then P(A‾)+P(B‾)\displaystyle \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}}) is
    (A)
    0.4\displaystyle 4
    (B)
    0.8\displaystyle 8
    (C)
    1.2\displaystyle 2
    (D)
    1.6\displaystyle 6

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    NCERT’s answer
    C
    (C) \(\displaystyle 1.2\)\[P(A\cup B) = P(A)+P(B)-P(A\cap B) \] \[0.6 = P(A)+P(B)-0.2 \;\Rightarrow\; P(A)+P(B) = 0.8 \] \[P(\bar A)+P(\bar B) = \bigl(1-P(A)\bigr)+\bigl(1-P(B)\bigr) = 2-0.8 = 1.2 \]

  9. Exercise 29

    If M and N are any two events, the probability that at least one of them occurs is
    (A)
    P(M)+P(N)−2P(M∩N)\displaystyle \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})-2 \mathrm{P}(\mathrm{M} \cap \mathrm{N})
    (B)
    P(M)+P(N)−P(M∩N)\displaystyle \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})-\mathrm{P}(\mathrm{M} \cap \mathrm{N})
    (C)
    P(M)+P(N)+P(M∩N)\displaystyle \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})+\mathrm{P}(\mathrm{M} \cap \mathrm{N})
    (D)
    P(M)+P(N)+2P(M∩N)\displaystyle \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})+2 \mathrm{P}(\mathrm{M} \cap \mathrm{N})

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    NCERT’s answer
    B
    (B) \(\displaystyle \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})-\mathrm{P}(\mathrm{M}\cap\mathrm{N})\)\(\displaystyle \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})\) counts \(\displaystyle \mathrm{M}\cap\mathrm{N}\) twice, so subtract it once.\[\mathrm{P}(\mathrm{M}\cup\mathrm{N}) = \mathrm{P}(\mathrm{M})+\mathrm{P}(\mathrm{N})-\mathrm{P}(\mathrm{M}\cap\mathrm{N}) \]
  10. State whether the statements are True or False in each of the Exercises $\displaystyle 30$ to 36.

    Exercise 30

    The probability that a person visiting a zoo will see the giraffee is 0.72\displaystyle 0.72, the probability that he will see the bears is 0.84\displaystyle 0.84 and the probability that he will see both is 0.52.

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    NCERT’s answer
    False
    FalseSeeing the giraffe (G) or the bears (B):\[P(G\cup B) = P(G)+P(B)-P(G\cap B) \] \[P(G\cup B) = 0.72+0.84-0.52 = 1.04 \] \[1.04 > 1 \]A probability cannot exceed 1.