SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Probability

62 questions · 62 still being checked

EXERCISE 13.3 11–14 (part 6 of 7)

  1. Exercise 11

    A manufacturer has three machine operators A, B and C. The first operator A produces 1\displaystyle 1% defective items, where as the other two operators B and C produce 5\displaystyle 5% and 7\displaystyle 7% defective items respectively. A is on the job for 50\displaystyle 50% of the time, B\displaystyle B is on the job for 30%\displaystyle 30 \% of the time and C\displaystyle C is on the job for 20%\displaystyle 20 \% of the time. A defective item is produced, what is the probability that it was produced by A?

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    NCERT’s answer
    \(\displaystyle \frac{5}{34}\)
    Let \(\displaystyle E_1, E_2, E_3\) be the events that the item was produced by operator A, B, C respectively, and \(\displaystyle A\) the event that the item is defective. The time each operator is on the job gives the prior probabilities: \[P(E_1)=50\%=0.5,\qquad P(E_2)=30\%=0.3,\qquad P(E_3)=20\%=0.2 .\]Their defective rates are \[P(A\mid E_1)=1\%=0.01,\qquad P(A\mid E_2)=5\%=0.05,\qquad P(A\mid E_3)=7\%=0.07 .\]By Bayes' theorem, \[P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}.\]Numerator \(\displaystyle =0.5\times 0.01=0.005\).Denominator \(\displaystyle =0.005+0.3\times 0.05+0.2\times 0.07=0.005+0.015+0.014=0.034\).\[P(E_1\mid A)=\frac{0.005}{0.034}=\frac{5}{34}.\]The probability that a defective item was produced by operator A is \(\displaystyle \dfrac{5}{34}\).

  2. Exercise 12

    A card from a pack of 52\displaystyle 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

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    NCERT’s answer
    \(\displaystyle \frac{11}{50}\)
    Let \(\displaystyle E_1\): the lost card is a diamond, \(\displaystyle E_2\): the lost card is not a diamond, and \(\displaystyle A\): the two cards drawn from the remaining $\displaystyle 51$ are both diamonds.Since the pack has $\displaystyle 13$ diamonds among $\displaystyle 52$ cards, \[P(E_1)=\frac{13}{52}=\frac14,\qquad P(E_2)=\frac{39}{52}=\frac34 .\]The step to get right: after one card is lost only $\displaystyle 51$ cards remain, and the number of diamonds among them depends on which case we are in.If the lost card is a diamond, $\displaystyle 12$ diamonds remain among $\displaystyle 51$ cards: \[P(A\mid E_1)=\frac{\binom{12}{2}}{\binom{51}{2}}=\frac{66}{1275}.\]If the lost card is not a diamond, all $\displaystyle 13$ diamonds remain among $\displaystyle 51$ cards: \[P(A\mid E_2)=\frac{\binom{13}{2}}{\binom{51}{2}}=\frac{78}{1275}.\]By Bayes' theorem, \[P(E_1\mid A)=\frac{\frac14\cdot\dfrac{66}{1275}}{\frac14\cdot\dfrac{66}{1275}+\frac34\cdot\dfrac{78}{1275}} =\frac{66}{66+3\times 78}=\frac{66}{66+234}=\frac{66}{300}.\] (The factors \(\displaystyle \frac14\) and \(\displaystyle \frac{1}{1275}\) cancel throughout.)The probability that the lost card was a diamond is \(\displaystyle \dfrac{11}{50}\).

  3. Exercise 13

    Probability that A speaks truth is 45\displaystyle \frac{4}{5}. A coin is tossed. A reports that a head appears. The probability that actually there was head is (A) 45\displaystyle \frac{4}{5} (B) 12\displaystyle \frac{1}{2} (C) 15\displaystyle \frac{1}{5} (D) 25\displaystyle \frac{2}{5}

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    NCERT’s answer
    A
    Let \(\displaystyle E\): a head actually appears on the toss, and \(\displaystyle A\): A reports that a head appears. For a fair coin, \[P(E)=\frac12,\qquad P(E')=\frac12 .\]A reports a head either when a head occurred and he speaks the truth, or when a tail occurred and he lies: \[P(A\mid E)=P(\text{A speaks truth})=\frac45,\qquad P(A\mid E')=P(\text{A lies})=1-\frac45=\frac15 .\]By Bayes' theorem, \[P(E\mid A)=\frac{P(E)P(A\mid E)}{P(E)P(A\mid E)+P(E')P(A\mid E')} =\frac{\frac12\cdot\frac45}{\frac12\cdot\frac45+\frac12\cdot\frac15} =\frac{\dfrac{4}{10}}{\dfrac{4}{10}+\dfrac{1}{10}}=\frac{4}{5}.\](The equal priors \(\displaystyle \frac12\) cancel, so the posterior coincides numerically with the truth-telling probability.)Hence the correct option is (A) \(\displaystyle \dfrac{4}{5}\).

  4. Exercise 14

    If A and B are two events such that A ∈B\displaystyle \in \mathrm{B} and P(B)≠0\displaystyle \mathrm{P}(\mathrm{B}) \neq 0, then which of the following is correct? (A) P(A∣\displaystyle \mathrm{P}(\mathrm{A} \mid B)=P(B)P(A)\displaystyle \mathrm{B})=\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{A})} (B) P(A∣B)<P(A)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})<\mathrm{P}(\mathrm{A}) (C) P(A∣B)≥P(A)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A}) (D) None of these

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    NCERT’s answer
    C
    Here the symbol is read as \(\displaystyle A\subset B\), and \(\displaystyle P(B)\neq 0\).If \(\displaystyle A\subset B\), then every outcome of \(\displaystyle A\) lies in \(\displaystyle B\), so \[A\cap B=A\quad\Longrightarrow\quad P(A\cap B)=P(A).\]By the definition of conditional probability, \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}.\]This already rules out (A), which has the ratio inverted.Now \(\displaystyle 0<P(B)\leq 1\), so dividing by \(\displaystyle P(B)\) cannot decrease the value: \[P(A\mid B)=\frac{P(A)}{P(B)}\geq P(A),\] with equality when \(\displaystyle P(B)=1\). Hence (B), which asserts \(\displaystyle P(A\mid B)<P(A)\), is false, and (C) holds in every case.The correct option is (C) \(\displaystyle P(A\mid B)\geq P(A)\).