SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Probability

62 questions · 62 still being checked

EXERCISE 13.1 11–17 (part 2 of 7)

  1. Exercise 11

    A fair die is rolled. Consider events E={1,3,5},F={2,3}\displaystyle \mathrm{E}=\{1,3,5\}, \mathrm{F}=\{2,3\} and G={2,3,4,5}\displaystyle \mathrm{G}=\{2,3,4,5\} Find
    (i)
    P(E∣F)\displaystyle \mathrm{P}(\mathrm{E} \mid \mathrm{F}) and P(F∣E)\displaystyle \mathrm{P}(\mathrm{F} \mid \mathrm{E})
    (ii)
    P (E|G) and P(G|E)
    (iii)
    P((E∪F)∣G)\displaystyle \mathrm{P}((\mathrm{E} \cup \mathrm{F}) \mid \mathrm{G}) and P((E∩F)∣G)\displaystyle \mathrm{P}((\mathrm{E} \cap \mathrm{F}) \mid \mathrm{G})

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{2}, \frac{1}{3}\) (ii) \(\displaystyle \frac{1}{2}, \frac{2}{3}\) (iii) \(\displaystyle \frac{3}{4}, \frac{1}{4}\)
    The die is fair, so each of the six outcomes has probability \(\displaystyle \frac16\). With \(\displaystyle E=\{1,3,5\}\), \(\displaystyle F=\{2,3\}\), \(\displaystyle G=\{2,3,4,5\}\):
    \[P(E)=\frac{3}{6}=\frac12,\qquad P(F)=\frac{2}{6}=\frac13,\qquad P(G)=\frac{4}{6}=\frac23\]
    (i)
    \(\displaystyle E\cap F=\{3\}\), so \(\displaystyle P(E\cap F)=\frac16\).
    \[P(E\mid F)=\frac{1/6}{1/3}=\frac12,\qquad P(F\mid E)=\frac{1/6}{1/2}=\frac13\]
    (ii)
    \(\displaystyle E\cap G=\{3,5\}\), so \(\displaystyle P(E\cap G)=\frac{2}{6}=\frac13\).
    \[P(E\mid G)=\frac{1/3}{2/3}=\frac12,\qquad P(G\mid E)=\frac{1/3}{1/2}=\frac23\]
    (iii)
    \(\displaystyle E\cup F=\{1,2,3,5\}\), so \(\displaystyle (E\cup F)\cap G=\{2,3,5\}\) and \(\displaystyle P\big((E\cup F)\cap G\big)=\frac{3}{6}=\frac12\).
    \[P\big((E\cup F)\mid G\big)=\frac{1/2}{2/3}=\frac{3}{4}\]
    Also \(\displaystyle E\cap F=\{3\}\), so \(\displaystyle (E\cap F)\cap G=\{3\}\) and \(\displaystyle P\big((E\cap F)\cap G\big)=\frac16\).
    \[P\big((E\cap F)\mid G\big)=\frac{1/6}{2/3}=\frac{1}{4}\]
    The key step in (iii) is to intersect \(\displaystyle E\cup F\) (and \(\displaystyle E\cap F\)) with \(\displaystyle G\) first, and only then divide by \(\displaystyle P(G)\).

  2. Exercise 12

    Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
    (i)
    the youngest is a girl,
    (ii)
    at least one is a girl?

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{2}\) (ii) \(\displaystyle \frac{1}{3}\)
    Record the two children in order of birth (elder, younger); since each child is equally likely to be a boy or a girl, the four outcomes are equally likely:
    \[S=\{GG,\ GB,\ BG,\ BB\}\]
    Let \(\displaystyle E\) = 'both are girls' \(\displaystyle =\{GG\}\), so \(\displaystyle P(E)=\frac14\).
    (i)
    \(\displaystyle F\) = 'the youngest is a girl' = the second letter is \(\displaystyle G\) \(\displaystyle =\{GG,\ BG\}\), so \(\displaystyle P(F)=\frac24=\frac12\). Also \(\displaystyle E\cap F=\{GG\}\), so \(\displaystyle P(E\cap F)=\frac14\).
    \[P(E\mid F)=\frac{1/4}{1/2}=\frac{1}{2}\]
    (ii)
    \(\displaystyle H\) = 'at least one is a girl' \(\displaystyle =\{GG,\ GB,\ BG\}\), so \(\displaystyle P(H)=\frac34\); and \(\displaystyle E\cap H=\{GG\}\), so \(\displaystyle P(E\cap H)=\frac14\).
    \[P(E\mid H)=\frac{1/4}{3/4}=\frac{1}{3}\]
    The two answers differ because naming which child is a girl rules out one more outcome (\(\displaystyle GB\)) than merely knowing that some child is a girl.
    Hence the probabilities are \(\displaystyle \frac12\) and \(\displaystyle \frac13\).

  3. Exercise 13

    An instructor has a question bank consisting of 300\displaystyle 300 easy True / False questions, 200\displaystyle 200 difficult True / False questions, 500\displaystyle 500 easy multiple choice questions and 400\displaystyle 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

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    NCERT’s answer
    \(\displaystyle \frac{5}{9}\)
    Count the question bank: \[300+200+500+400=1400\ \text{questions in all}\] Let \(\displaystyle E\) = 'the question is easy' and \(\displaystyle M\) = 'the question is multiple choice'. The multiple choice questions number \(\displaystyle 500+400=900\), and those that are also easy number \(\displaystyle 500\). Since the question is drawn at random, every question is equally likely, so \[P(M)=\frac{900}{1400},\qquad P(E\cap M)=\frac{500}{1400}\] \[P(E\mid M)=\frac{P(E\cap M)}{P(M)}=\frac{500/1400}{900/1400}=\frac{500}{900}=\frac{5}{9}\] Hence the required probability is \(\displaystyle \dfrac{5}{9}\).

  4. Exercise 14

    Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4\displaystyle 4'.

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    NCERT’s answer
    \(\displaystyle \frac{1}{15}\)
    Throwing two dice gives \(\displaystyle 36\) equally likely ordered pairs. Let \[F=\text{'the two numbers are different'},\qquad E=\text{'the sum of the numbers is }4\text{'}\] The pairs with equal numbers are the six doublets \(\displaystyle (1,1),(2,2),\dots,(6,6)\), so \[n(F)=36-6=30,\qquad P(F)=\frac{30}{36}\] The sum is $\displaystyle 4$ for \(\displaystyle (1,3),(2,2),(3,1)\); of these \(\displaystyle (2,2)\) is a doublet and so is excluded, giving \(\displaystyle E\cap F=\{(1,3),(3,1)\}\) and \(\displaystyle P(E\cap F)=\frac{2}{36}\). \[P(E\mid F)=\frac{P(E\cap F)}{P(F)}=\frac{2/36}{30/36}=\frac{2}{30}=\frac{1}{15}\] Hence the required probability is \(\displaystyle \dfrac{1}{15}\). (Dropping the doublet \(\displaystyle (2,2)\) is the step to watch: it is a sum of $\displaystyle 4$ but is not allowed by the given condition.)

  5. Exercise 15

    Consider the experiment of throwing a die, if a multiple of 3\displaystyle 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 3\displaystyle 3'.

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    NCERT’s answer
    $\displaystyle 0$
    The multiples of $\displaystyle 3$ on a die are $\displaystyle 3$ and 6. So the experiment produces either a pair of die scores (when the first throw is $\displaystyle 3$ or $\displaystyle 6$) or a die score with a coin face (otherwise): \[S=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),\] \[(1,H),(1,T),(2,H),(2,T),(4,H),(4,T),(5,H),(5,T)\}\] Let \[A=\text{'the coin shows a tail'}=\{(1,T),(2,T),(4,T),(5,T)\}\] \[B=\text{'at least one die shows a }3\text{'}=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}\] The coin is tossed only when the first throw is not a multiple of $\displaystyle 3$, so every outcome of \(\displaystyle A\) has first score \(\displaystyle 1,2,4\) or \(\displaystyle 5\) and no second die at all. Hence no outcome lies in both events: \[A\cap B=\phi\quad\Rightarrow\quad P(A\cap B)=0\] Since \(\displaystyle P(B)\neq 0\) (the outcome \(\displaystyle (3,3)\), for instance, has positive probability), the conditional probability is defined and \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0}{P(B)}=0\] Hence the required conditional probability is \(\displaystyle 0\).
  6. In each of the Exercises $\displaystyle 16$ and $\displaystyle 17$ choose the correct answer:

    Exercise 16

    If P(A)=12,P(B)=0\displaystyle \mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=0, then P(A∣B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B}) is (A) 0\displaystyle 0 (B) 12\displaystyle \frac{1}{2} (C) not defined (D) 1\displaystyle 1

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    NCERT’s answer
    C
    Conditional probability is defined by \[P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad\text{provided } P(B)\neq 0\] Here \(\displaystyle P(B)=0\), so the quotient has zero denominator and the definition does not apply; \(\displaystyle P(A\mid B)\) simply does not exist. (It is not \(\displaystyle 0\): \(\displaystyle P(A\cap B)=0\) too, and \(\displaystyle \frac{0}{0}\) is meaningless, not zero.)Hence the correct answer is (C) not defined.

  7. Exercise 17

    If A and B are events such that P(A∣B)=P(B∣A)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A}), then (A) A⊂B\displaystyle \mathrm{A} \subset \mathrm{B} but A≠B\displaystyle \mathrm{A} \neq \mathrm{B} (B) A=B\displaystyle \mathrm{A}=\mathrm{B} (C) A∩B=ϕ\displaystyle \mathrm{A} \cap \mathrm{B}=\phi (D) P(A)=P(B)\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})

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    NCERT’s answer
    D
    For \(\displaystyle P(A\mid B)\) and \(\displaystyle P(B\mid A)\) to exist we need \(\displaystyle P(A)\neq 0\) and \(\displaystyle P(B)\neq 0\). Using the definition and \(\displaystyle A\cap B=B\cap A\), \[P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B\mid A)=\frac{P(A\cap B)}{P(A)}\] Equating them and cross-multiplying, \[P(A\cap B)\,P(A)=P(A\cap B)\,P(B)\quad\Rightarrow\quad P(A\cap B)\big[P(A)-P(B)\big]=0\] So, apart from the degenerate case \(\displaystyle P(A\cap B)=0\) (in which both conditional probabilities are \(\displaystyle 0\) and nothing is being compared), the equality forces \[P(A)=P(B)\] The other options are not forced: \(\displaystyle A\subset B\) with \(\displaystyle A\neq B\) gives \(\displaystyle P(B\mid A)=1\) and \(\displaystyle P(A\mid B)=\frac{P(A)}{P(B)}\), which agree only if the probabilities are equal; \(\displaystyle A=B\) is sufficient but far stronger than necessary; and \(\displaystyle A\cap B=\phi\) is only the degenerate case above.Hence the correct answer is (D) \(\displaystyle P(A)=P(B)\).