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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 101–108 (part 11 of 11)

  1. State True or False for the statements in each of the Exercises $\displaystyle 94$ to 103.

    Exercise 101

    If A and B are independent, then P( exactly one of A, B occurs )=P( A)P( B′)+P( B)P( A′)\mathrm{P}(\text { exactly one of } \mathrm{A}, \mathrm{~B} \text { occurs })=\mathrm{P}(\mathrm{~A}) \mathrm{P}\left(\mathrm{~B}^{\prime}\right)+\mathrm{P}(\mathrm{~B}) \mathrm{P}\left(\mathrm{~A}^{\prime}\right)

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    NCERT’s answer
    True
    True. Exactly one occurs means \(\displaystyle A\cap B'\) or \(\displaystyle A'\cap B\), which are disjoint; independence of \(\displaystyle A,B\) gives independence of \(\displaystyle A,B'\) and \(\displaystyle A',B\). \[P=P(A\cap B')+P(A'\cap B) \] \[P=P(A)\,P(B')+P(A')\,P(B) \]

  2. Exercise 102

    If A and B are two events such that P(A)>0\displaystyle \mathrm{P}(\mathrm{A})>0 and P(A)+P(B)>1\displaystyle \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})>1, then P( B∣A)≥1−P( B′)P( A)\mathrm{P}(\mathrm{~B} \mid \mathrm{A}) \geq 1-\frac{\mathrm{P}\left(\mathrm{~B}^{\prime}\right)}{\mathrm{P}(\mathrm{~A})}

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    True. Since \(\displaystyle P(A\cup B)\le 1\): \[P(A\cap B)=P(A)+P(B)-P(A\cup B)\ge P(A)+P(B)-1 \] \[P(A\cap B)\ge P(A)-P(B') \] \[P(B\mid A)=\frac{P(A\cap B)}{P(A)}\ge 1-\frac{P(B')}{P(A)} \quad (P(A)>0) \]

  3. Exercise 103

    If A, B and C are three independent events such that P(A)=P(B)=P(C)=\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{C})= p\displaystyle p, then P( At least two of A, B,C occur )=3p2−2p3\mathrm{P}(\text { At least two of } \mathrm{A}, \mathrm{~B}, \mathrm{C} \text { occur })=3 p^2-2 p^3

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    NCERT’s answer
    True
    True. At least two means exactly two ($\displaystyle 3$ choices of the pair) or all three: \[P=3\,p^2(1-p)+p^3 \] \[P=3p^2-3p^3+p^3 \] \[P=3p^2-2p^3 \]
  4. Fill in the blanks in each of the following questions:

    Exercise 104

    If A and B are two events such that P( A∣B)=p,P( A)=p,P( B)=13\mathrm{P}(\mathrm{~A} \mid \mathrm{B})=p, \mathrm{P}(\mathrm{~A})=p, \mathrm{P}(\mathrm{~B})=\frac{1}{3} and P( A∪ B)=59, then p=\mathrm{P}(\mathrm{~A} \cup \mathrm{~B})=\frac{5}{9}, \text { then } p= ____

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}\)
    \(\displaystyle p=\tfrac13\) \[P(A\cap B)=P(A\mid B)\,P(B)=\frac{p}{3} \] \[P(A\cup B)=P(A)+P(B)-P(A\cap B) \] \[\frac59=p+\frac13-\frac p3 \] \[\frac{2p}{3}=\frac59-\frac13=\frac29 \] \[p=\frac13 \]

  5. Exercise 105

    If A and B are such that P( A′∪B′)=23 and P( A∪ B)=59, then P( A′)+P( B′)=…………….\begin{aligned} \mathrm{P}\left(\mathrm{~A}^{\prime} \cup \mathrm{B}^{\prime}\right) & =\frac{2}{3} \text { and } \mathrm{P}(\mathrm{~A} \cup \mathrm{~B})=\frac{5}{9}, \\ \text { then } \mathrm{P}\left(\mathrm{~A}^{\prime}\right)+\mathrm{P}\left(\mathrm{~B}^{\prime}\right) & =\ldots \ldots \ldots \ldots \ldots . \end{aligned}

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    NCERT’s answer
    \(\displaystyle \frac{10}{9}\)
    \(\displaystyle \dfrac{10}{9}\)\[\mathrm{P}(A' \cup B') = 1 - \mathrm{P}(A \cap B) \quad \text{(De Morgan)} \] \[\tfrac{2}{3} = 1 - \mathrm{P}(A \cap B) \;\Rightarrow\; \mathrm{P}(A \cap B) = \tfrac{1}{3} \] \[\mathrm{P}(A) + \mathrm{P}(B) = \mathrm{P}(A \cup B) + \mathrm{P}(A \cap B) = \tfrac{5}{9} + \tfrac{1}{3} = \tfrac{8}{9} \] \[\mathrm{P}(A') + \mathrm{P}(B') = 2 - \big[\mathrm{P}(A) + \mathrm{P}(B)\big] = 2 - \tfrac{8}{9} = \tfrac{10}{9} \]Answer: \(\displaystyle \dfrac{10}{9}\)

  6. Exercise 106

    If X follows binomial distribution with parameters n=5,p\displaystyle n=5, p and P(X=2)=9,P(X=3)\displaystyle \mathrm{P}(\mathrm{X}=2)=9, \mathrm{P}(\mathrm{X}=3), then p=\displaystyle p= ____

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    NCERT’s answer
    \(\displaystyle \frac{1}{10}\)
    \(\displaystyle p = \dfrac{1}{10}\)\[\mathrm{P}(X=r) = {}^5C_r\, p^r q^{5-r}, \quad q = 1-p \] \[{}^5C_2\, p^2 q^3 = 9\; {}^5C_3\, p^3 q^2 \] \[10\, p^2 q^3 = 90\, p^3 q^2 \;\Rightarrow\; q = 9p \] \[1 - p = 9p \;\Rightarrow\; p = \tfrac{1}{10} \]Answer: \(\displaystyle p = \dfrac{1}{10}\)

  7. Exercise 107

    Let X be a random variable taking values x1,x2,…,xn\displaystyle x_1, x_2, \ldots, x_{\mathrm{n}} with probabilities p1,p2,…,pn\displaystyle p_1, p_2, \ldots, p_n, respectively. Then var (X)=\displaystyle (\mathrm{X})= ____

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    NCERT’s answer
    \(\displaystyle \Sigma p_i x_i^2-\left(\Sigma p_i x_i\right)^2\)
    \(\displaystyle \displaystyle\sum_{i=1}^{n} p_i x_i^2 - \Big(\sum_{i=1}^{n} p_i x_i\Big)^2\)\[\mu = \mathrm{E}(X) = \sum_{i=1}^{n} p_i x_i \] \[\operatorname{var}(X) = \sum_{i=1}^{n} p_i (x_i - \mu)^2 \] \[= \sum p_i x_i^2 - 2\mu \sum p_i x_i + \mu^2 \sum p_i \qquad \Big(\textstyle\sum p_i = 1\Big) \] \[= \sum_{i=1}^{n} p_i x_i^2 - \mu^2 \]Answer: \(\displaystyle \operatorname{var}(X) = \displaystyle\sum p_i x_i^2 - \Big(\sum p_i x_i\Big)^2 = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\)

  8. Exercise 108

    Let A and B be two events. If P(A∣B)=P(A)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A}), then A is ____ of B.

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    NCERT’s answer
    independent
    independent\[\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} = \mathrm{P}(A) \] \[\Rightarrow\; \mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B) \]This is the condition for independent events.Answer: A is independent of B.