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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 1–10 (part 1 of 11)

  1. Exercise 1

    For a loaded die, the probabilities of outcomes are given as under: P(1)=P(2)=0.2,P(3)=P(5)=P(6)=0.1\displaystyle \mathrm{P}(1)=\mathrm{P}(2)=0.2, \mathrm{P}(3)=\mathrm{P}(5)=\mathrm{P}(6)=0.1 and P(4)=0.3\displaystyle \mathrm{P}(4)=0.3. The die is thrown two times. Let A and B be the events, 'same number each time', and 'a total score is 10\displaystyle 10 or more', respectively. Determine whether or not A and B are independent.

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    NCERT’s answer
    Independent
    Probabilities multiply for the two throws; list the favourable pairs. \[P(A) = (0.2)^2 + (0.2)^2 + (0.1)^2 + (0.3)^2 + (0.1)^2 + (0.1)^2 = 0.20 \] \[B = \{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\} \] \[P(B) = 2(0.3)(0.1) + (0.1)^2 + 2(0.1)(0.1) + (0.1)^2 = 0.10 \] \[A \cap B = \{(5,5),(6,6)\} \] \[P(A \cap B) = (0.1)^2 + (0.1)^2 = 0.02 \] \[P(A)\,P(B) = 0.20 \times 0.10 = 0.02 = P(A \cap B) \] Answer: \(\displaystyle A\) and \(\displaystyle B\) are independent.

  2. Exercise 2

    Refer to Exercise 1\displaystyle 1 above. If the die were fair, determine whether or not the events A and B are independent.

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    NCERT’s answer
    not independent
    All $\displaystyle 36$ outcomes are equally likely. \[P(A) = \frac{6}{36} = \frac{1}{6} \] \[B = \{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\} \Rightarrow P(B) = \frac{6}{36} = \frac{1}{6} \] \[A \cap B = \{(5,5),(6,6)\} \Rightarrow P(A \cap B) = \frac{2}{36} = \frac{1}{18} \] \[P(A)\,P(B) = \frac{1}{36} \ne \frac{1}{18} = P(A \cap B) \] Answer: \(\displaystyle A\) and \(\displaystyle B\) are not independent.

  3. Exercise 3

    The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3\displaystyle 0.3, evaluate P(A‾)+P(B‾)\displaystyle \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}}).

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    NCERT’s answer
    1.$\displaystyle 1$
    \[P(A \cup B) = P(A) + P(B) - P(A \cap B) \] \[0.6 = P(A) + P(B) - 0.3 \Rightarrow P(A) + P(B) = 0.9 \] \[P(\bar{A}) + P(\bar{B}) = \bigl(1 - P(A)\bigr) + \bigl(1 - P(B)\bigr) = 2 - 0.9 = 1.1 \] Answer: \(\displaystyle P(\bar{A}) + P(\bar{B}) = 1.1\)

  4. Exercise 4

    A bag contains 5\displaystyle 5 red marbles and 3\displaystyle 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

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    After a red first draw, $\displaystyle 4$ red and $\displaystyle 3$ black marbles remain; two more are drawn. \[P(\text{no black in the next two}) = \frac{4}{7} \times \frac{3}{6} = \frac{2}{7} \] \[P(\text{at least one black}) = 1 - \frac{2}{7} = \frac{5}{7} \] Answer: \(\displaystyle \dfrac{5}{7}\)

  5. Exercise 5

    Two dice are thrown together and the total score is noted. The events E, F and G are 'a total of 4\displaystyle 4', 'a total of 9\displaystyle 9 or more', and 'a total divisible by 5\displaystyle 5', respectively. Calculate P(E),P(F)\displaystyle \mathrm{P}(\mathrm{E}), \mathrm{P}(\mathrm{F}) and P(G)\displaystyle \mathrm{P}(\mathrm{G}) and decide which pairs of events, if any, are independent.

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    NCERT’s answer
    \(\displaystyle \mathrm{P}(\mathrm{E})=\frac{1}{12}, \mathrm{P}(\mathrm{F}): \frac{5}{18}, \mathrm{P}(\mathrm{G})=\frac{7}{36}\), no pair is independent
    Sample space: $\displaystyle 36$ equally likely pairs. \[E = \{(1,3),(2,2),(3,1)\} \Rightarrow P(E) = \frac{3}{36} = \frac{1}{12} \] \[F:\ \text{totals } 9, 10, 11, 12 \Rightarrow P(F) = \frac{4+3+2+1}{36} = \frac{5}{18} \] \[G:\ \text{totals } 5, 10 \Rightarrow P(G) = \frac{4+3}{36} = \frac{7}{36} \] \[E \cap F = \varnothing: \quad P(E \cap F) = 0 \ne \frac{1}{12} \cdot \frac{5}{18} = P(E)P(F) \] \[E \cap G = \varnothing: \quad P(E \cap G) = 0 \ne \frac{1}{12} \cdot \frac{7}{36} = P(E)P(G) \] \[F \cap G = \{\text{total } 10\}: \quad P(F \cap G) = \frac{3}{36} = \frac{54}{648} \ne \frac{35}{648} = \frac{5}{18} \cdot \frac{7}{36} = P(F)P(G) \] Answer: \(\displaystyle P(E)=\frac{1}{12}\), \(\displaystyle P(F)=\frac{5}{18}\), \(\displaystyle P(G)=\frac{7}{36}\); no pair of events is independent.

  6. Exercise 6

    Explain why the experiment of tossing a coin three times is said to have binomial distribution.

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    Let \(\displaystyle X\) be the number of heads in three tosses. There are a fixed number of tosses, each with only two outcomes (head or tail), the tosses are independent, and a head has the same probability on each. \[n = 3, \qquad p = \tfrac{1}{2}, \qquad q = \tfrac{1}{2} \] \[P(X = r) = \binom{3}{r} \left(\tfrac{1}{2}\right)^{r} \left(\tfrac{1}{2}\right)^{3-r} = \binom{3}{r}\cdot\tfrac{1}{8}, \quad r = 0, 1, 2, 3 \] \[P(X = 0, 1, 2, 3) = \tfrac{1}{8},\ \tfrac{3}{8},\ \tfrac{3}{8},\ \tfrac{1}{8} \] Answer: \(\displaystyle X\) follows the binomial distribution \(\displaystyle B\left(3, \tfrac{1}{2}\right)\).

  7. Exercise 7

    A and B are two events such that P(A)=12,P(B)=13\displaystyle \mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3} and P(A∩B)=14\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}.
    Find :
    (i)
    P(A∣B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B})
    (ii)
    P(B∣A)\displaystyle \mathrm{P}(\mathrm{B} \mid \mathrm{A})
    (iii)
    P(A′∣B)\displaystyle \mathrm{P}(\mathrm{A}^{\prime} \mid \mathrm{B})
    (iv)
    P(A′∣B′)\displaystyle \mathrm{P}(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime})

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{3}{4}\), (ii) \(\displaystyle \frac{1}{2}\), (iii) \(\displaystyle \frac{1}{4}\), (iv) \(\displaystyle \frac{5}{8}\)
    (i) \[P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4} \] (ii) \[P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{1/2} = \frac{1}{2} \] (iii) \[P(A' \mid B) = 1 - P(A \mid B) = 1 - \frac{3}{4} = \frac{1}{4} \] (iv) \[P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12} \] \[P(A' \cap B') = 1 - P(A \cup B) = \frac{5}{12}, \qquad P(B') = 1 - \frac{1}{3} = \frac{2}{3} \] \[P(A' \mid B') = \frac{5/12}{2/3} = \frac{5}{8} \] Answer: (i) \(\displaystyle \frac{3}{4}\); (ii) \(\displaystyle \frac{1}{2}\); (iii) \(\displaystyle \frac{1}{4}\); (iv) \(\displaystyle \frac{5}{8}\)

  8. Exercise 8

    Three events A,B\displaystyle \mathrm{A}, \mathrm{B} and C have probabilities 25,13\displaystyle \frac{2}{5}, \frac{1}{3} and 12\displaystyle \frac{1}{2}, respectively. Given that P(A∩C)=15\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{5} and P(B∩C)=14\displaystyle \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{4}, find the values of P(C∣B)\displaystyle \mathrm{P}(\mathrm{C} \mid \mathrm{B}) and P(A′∩C′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{C}^{\prime}\right).

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    NCERT’s answer
    \(\displaystyle \frac{3}{4}, \frac{3}{10}\)
    \[P(C \mid B) = \frac{P(B \cap C)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4} \] By De Morgan, \(\displaystyle A' \cap C' = (A \cup C)'\). \[P(A \cup C) = P(A) + P(C) - P(A \cap C) = \frac{2}{5} + \frac{1}{2} - \frac{1}{5} = \frac{7}{10} \] \[P(A' \cap C') = 1 - P(A \cup C) = 1 - \frac{7}{10} = \frac{3}{10} \] Answer: \(\displaystyle P(C \mid B) = \frac{3}{4}\), \(\displaystyle P(A' \cap C') = \frac{3}{10}\)

  9. Exercise 9

    Let E1\displaystyle \mathrm{E}_1 and E2\displaystyle \mathrm{E}_2 be two independent events such that p(E1)=p1\displaystyle p\left(\mathrm{E}_1\right)=p_1 and P(E2)=p2\displaystyle \mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{p}_2. Describe in words of the events whose probabilities are:
    (i)
    p1p2\displaystyle p_1 p_2
    (ii)
    (1−p1)p2\displaystyle \left(1-p_1\right) p_2
    (iii)
    1−(1−p1)(1−p2)\displaystyle 1-\left(1-p_1\right)\left(1-p_2\right)
    (iv)
    p1+p2−2p1p2\displaystyle p_1+p_2-2 p_1 p_2

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    NCERT’s answer
    (i)
    \(\displaystyle \mathrm{E}_1\) and \(\displaystyle \mathrm{E}_2\) occur
    (ii)
    \(\displaystyle \mathrm{E}_1\) does not occur, but \(\displaystyle \mathrm{E}_2\) occurs
    (iii)
    Either \(\displaystyle \mathrm{E}_1\) or \(\displaystyle \mathrm{E}_2\), or both \(\displaystyle \mathrm{E}_1\) and \(\displaystyle \mathrm{E}_2\) occurs
    (iv)
    Either \(\displaystyle \mathrm{E}_1\) or \(\displaystyle \mathrm{E}_2\) occurs, but not both
    (i) \[p_1p_2 = P(E_1)\,P(E_2) = P(E_1\cap E_2) \quad \text{(independent)} \] Both \(\displaystyle E_1\) and \(\displaystyle E_2\) occur.(ii) \[(1-p_1)\,p_2 = P(\bar E_1)\,P(E_2) = P(\bar E_1\cap E_2) \quad \text{(complement of an independent event is independent)} \] \(\displaystyle E_1\) does not occur, but \(\displaystyle E_2\) occurs.(iii) \[(1-p_1)(1-p_2) = P(\bar E_1\cap\bar E_2) = 1-P(E_1\cup E_2) \] \[1-(1-p_1)(1-p_2) = P(E_1\cup E_2) \] At least one of \(\displaystyle E_1, E_2\) occurs.(iv) \[p_1+p_2-2p_1p_2 = p_1(1-p_2)+(1-p_1)\,p_2 \] \[= P(E_1\cap\bar E_2)+P(\bar E_1\cap E_2) \] Exactly one of \(\displaystyle E_1, E_2\) occurs.Answer: (i) both occur; (ii) \(\displaystyle E_1\) does not occur but \(\displaystyle E_2\) occurs; (iii) at least one occurs; (iv) exactly one occurs.

  10. Exercise 10

    A discrete random variable X has the probability distribution given as below:
    X0.5\displaystyle 0.51\displaystyle 11.5\displaystyle 1.52\displaystyle 2
    P(X)k\displaystyle kk2\displaystyle k^22k2\displaystyle 2 k^2k\displaystyle k
    (i)
    Find the value of k\displaystyle k
    (ii)
    Determine the mean of the distribution.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{3}\), (ii) \(\displaystyle \frac{23}{18}\)
    (i) The probabilities add to 1. \[k + k^2 + 2k^2 + k = 1 \] \[3k^2 + 2k - 1 = 0 \] \[(3k-1)(k+1) = 0 \] \[P(X)\ge 0 \;\Rightarrow\; k = \tfrac13 \] \[P(X) = \tfrac13,\ \tfrac19,\ \tfrac29,\ \tfrac13 \](ii) \[\mu = \sum x\,P(x) = 0.5\cdot\tfrac13 + 1\cdot\tfrac19 + 1.5\cdot\tfrac29 + 2\cdot\tfrac13 \] \[= \tfrac16 + \tfrac19 + \tfrac13 + \tfrac23 = \tfrac{3+2+6+12}{18} = \tfrac{23}{18} \]Answer: \(\displaystyle k=\tfrac13\); mean \(\displaystyle =\tfrac{23}{18}\).