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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 51–60 (part 6 of 11)

  1. Exercise 51

    The probability distribution of a random variable x\displaystyle x is given as under: P(X=x)={kx2 for x=1,2,32kx for x=4,5,60 otherwise \mathrm{P}(\mathrm{X}=x)=\left\{\begin{array}{l} k x^2 \text { for } x=1,2,3 \\ 2 k x \text { for } x=4,5,6 \\ 0 \quad \text { otherwise } \end{array}\right. where k\displaystyle k is a constant. Calculate
    (i)
    E(X)
    (ii)
    E(3X2)\displaystyle \mathrm{E}\left(3 \mathrm{X}^2\right)
    (iii)
    P(X≥4)\displaystyle \mathrm{P}(\mathrm{X} \geq 4)

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    NCERT’s answer
    (i)
    4.$\displaystyle 32$, (ii) $\displaystyle 61.9$, (iii) \(\displaystyle \frac{15}{22}\)
    \[\begin{array}{c|cccccc} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & k & 4k & 9k & 8k & 10k & 12k \end{array} \] \[\sum P(x) = 44k = 1 \Rightarrow k = \tfrac{1}{44} \] (i) \[E(X) = k\,(1+8+27+32+50+72) = 190k = \tfrac{95}{22} \] (ii) \[E(X^2) = k\,(1+16+81+128+250+432) = 908k = \tfrac{227}{11} \] \[E(3X^2) = 3E(X^2) = \tfrac{681}{11} \] (iii) \[P(X \ge 4) = P(4)+P(5)+P(6) = (8+10+12)k = 30k = \tfrac{15}{22} \] Answer: \(\displaystyle E(X) = \tfrac{95}{22}\), \(\displaystyle E(3X^2) = \tfrac{681}{11}\), \(\displaystyle P(X \ge 4) = \tfrac{15}{22}\).

  2. Exercise 52

    A bag contains (2n+1)\displaystyle (2 n+1) coins. It is known that n\displaystyle n of these coins have a head on both sides where as the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is 3142\displaystyle \frac{31}{42}, determine the value of n\displaystyle n.

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    NCERT’s answer
    $\displaystyle 10$
    Let \(\displaystyle D\) = a double-headed coin is picked, \(\displaystyle F\) = a fair coin is picked. \[P(D) = \tfrac{n}{2n+1}, \qquad P(F) = \tfrac{n+1}{2n+1} \] \[P(H) = P(D)\cdot 1 + P(F)\cdot\tfrac12 \] \[\tfrac{n}{2n+1} + \tfrac{n+1}{2(2n+1)} = \tfrac{31}{42} \] \[\tfrac{3n+1}{2(2n+1)} = \tfrac{31}{42} \] \[21(3n+1) = 31(2n+1) \] \[63n + 21 = 62n + 31 \] \[n = 10 \] Answer: \(\displaystyle n = 10\).

  3. Exercise 53

    Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.

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    \(\displaystyle X = 0, 1, 2\), with \(\displaystyle \binom{52}{2} = 1326\) equally likely pairs. \[P(X=0) = \tfrac{\binom{48}{2}}{\binom{52}{2}} = \tfrac{188}{221} \] \[P(X=1) = \tfrac{4\cdot 48}{\binom{52}{2}} = \tfrac{32}{221} \] \[P(X=2) = \tfrac{\binom{4}{2}}{\binom{52}{2}} = \tfrac{1}{221} \] \[\text{Mean} = E(X) = 0 + \tfrac{32}{221} + 2\cdot\tfrac{1}{221} = \tfrac{34}{221} = \tfrac{2}{13} \] \[E(X^2) = \tfrac{32}{221} + 4\cdot\tfrac{1}{221} = \tfrac{36}{221} \] \[\operatorname{Var}(X) = \tfrac{36}{221} - \left(\tfrac{2}{13}\right)^2 = \tfrac{468 - 68}{2873} = \tfrac{400}{2873} \] \[\sigma = \tfrac{20}{\sqrt{2873}} = \tfrac{20}{13\sqrt{17}} = \tfrac{20\sqrt{17}}{221} \approx 0.373 \] Answer: Mean \(\displaystyle = \tfrac{2}{13}\); standard deviation \(\displaystyle = \tfrac{20\sqrt{17}}{221} \approx 0.373\).

  4. Exercise 54

    A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of the number of successes.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\)
    Each toss: an even number is one of $\displaystyle 2$, $\displaystyle 4$, 6. \[p = \tfrac36 = \tfrac12, \qquad q = \tfrac12, \qquad n = 2 \] \(\displaystyle X\), the number of successes, is binomial: \(\displaystyle X \sim B\!\left(2, \tfrac12\right)\). \[\operatorname{Var}(X) = npq = 2\cdot\tfrac12\cdot\tfrac12 = \tfrac12 \] Answer: \(\displaystyle \operatorname{Var}(X) = \tfrac12\).

  5. Exercise 55

    There are 5\displaystyle 5 cards numbered 1\displaystyle 1 to 5\displaystyle 5, one number on one card. Two cards are drawn at random without replacement. Let X\displaystyle X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.

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    NCERT’s answer
    Mean \(\displaystyle =6\), Variance \(\displaystyle =3\)
    \[\binom52 = 10 \text{ equally likely pairs} \] \[\begin{array}{c|ccccccc} X & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline P(X) & \tfrac1{10} & \tfrac1{10} & \tfrac2{10} & \tfrac2{10} & \tfrac2{10} & \tfrac1{10} & \tfrac1{10} \end{array} \] \[E(X) = \tfrac{1}{10}\,(3+4+10+12+14+8+9) = \tfrac{60}{10} = 6 \] \[E(X^2) = \tfrac{1}{10}\,(9+16+50+72+98+64+81) = \tfrac{390}{10} = 39 \] \[\operatorname{Var}(X) = 39 - 6^2 = 3 \] Answer: Mean \(\displaystyle = 6\); variance \(\displaystyle = 3\).
  6. Choose the correct answer from the given four options in each of the exercises from $\displaystyle 56$ to 82.

    Exercise 56

    If P(A)=45\displaystyle \mathrm{P}(\mathrm{A})=\frac{4}{5}, and P(A∩B)=710\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{7}{10}, then P(B∣A)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A}) is equal to
    (A)
    110\displaystyle \frac{1}{10}
    (B)
    18\displaystyle \frac{1}{8}
    (C)
    78\displaystyle \frac{7}{8}
    (D)
    1720\displaystyle \frac{17}{20}

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    NCERT’s answer
    C
    (C) \(\displaystyle \tfrac78\) \[P(B\mid A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{7/10}{4/5} = \tfrac{7}{8} \]

  7. Exercise 57

    If P(A∩B)=710\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{7}{10} and P(B)=1720\displaystyle \mathrm{P}(\mathrm{B})=\frac{17}{20}, then P(A∣B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B}) equals
    (A)
    1417\displaystyle \frac{14}{17}
    (B)
    1720\displaystyle \frac{17}{20}
    (C)
    78\displaystyle \frac{7}{8}
    (D)
    18\displaystyle \frac{1}{8}

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{14}{17}\)\[P(A\mid B)=\frac{P(A\cap B)}{P(B)} \] \[=\frac{7/10}{17/20} \] \[=\frac{7}{10}\cdot\frac{20}{17}=\frac{14}{17} \]

  8. Exercise 58

    If P(A)=310,P(B)=25\displaystyle \mathrm{P}(\mathrm{A})=\frac{3}{10}, \mathrm{P}(\mathrm{B})=\frac{2}{5} and P(A∪B)=35\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{5}, then P(B∣A)+P(A∣B)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})+\mathrm{P}(\mathrm{A} \mid \mathrm{B}) equals
    (A)
    14\displaystyle \frac{1}{4}
    (B)
    13\displaystyle \frac{1}{3}
    (C)
    512\displaystyle \frac{5}{12}
    (D)
    72\displaystyle \frac{7}{2}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{7}{12}\)\[P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac{3}{10}+\frac{2}{5}-\frac{3}{5}=\frac{1}{10} \] \[P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{1/10}{3/10}=\frac{1}{3} \] \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/10}{2/5}=\frac{1}{4} \] \[P(B\mid A)+P(A\mid B)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12} \]Option (D) is printed as \(\displaystyle \frac{7}{2}\), a misprint of \(\displaystyle \frac{7}{12}\).

  9. Exercise 59

    If P(A)=25,P(B)=310\displaystyle \mathrm{P}(\mathrm{A})=\frac{2}{5}, \mathrm{P}(\mathrm{B})=\frac{3}{10} and P(A∩B)=15\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{5}, then P(A′∣B′)⋅P(B′∣A′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime} \mid \mathrm{A}^{\prime}\right) is equal to
    (A)
    56\displaystyle \frac{5}{6}
    (B)
    57\displaystyle \frac{5}{7}
    (C)
    2542\displaystyle \frac{25}{42}

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{25}{42}\)\[P(A\cup B)=\frac{2}{5}+\frac{3}{10}-\frac{1}{5}=\frac{1}{2} \] \[P(A'\cap B')=1-P(A\cup B)=\frac{1}{2} \] \[P(A'\mid B')=\frac{P(A'\cap B')}{P(B')}=\frac{1/2}{7/10}=\frac{5}{7} \] \[P(B'\mid A')=\frac{P(A'\cap B')}{P(A')}=\frac{1/2}{3/5}=\frac{5}{6} \] \[P(A'\mid B')\cdot P(B'\mid A')=\frac{5}{7}\cdot\frac{5}{6}=\frac{25}{42} \]

  10. Exercise 60

    If A and B are two events such that P(A)=12,P(B)=13,P(A/B)=14\displaystyle \mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{1}{4}, then P(A′∩B′)\displaystyle P\left(A^{\prime} \cap B^{\prime}\right) equals
    (A)
    112\displaystyle \frac{1}{12}
    (B)
    34\displaystyle \frac{3}{4}
    (C)
    14\displaystyle \frac{1}{4}
    (D)
    316\displaystyle \frac{3}{16}

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{1}{4}\)\[P(A\cap B)=P(A\mid B)\,P(B)=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12} \] \[P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{12}=\frac{3}{4} \] \[P(A'\cap B')=1-P(A\cup B)=1-\frac{3}{4}=\frac{1}{4} \]