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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 81–90 (part 9 of 11)

  1. Choose the correct answer from the given four options in each of the exercises from $\displaystyle 56$ to 82.

    Exercise 81

    Eight coins are tossed together. The probability of getting exactly 3\displaystyle 3 heads is
    (A)
    1256\displaystyle \frac{1}{256}
    (B)
    732\displaystyle \frac{7}{32}
    (C)
    532\displaystyle \frac{5}{32}
    (D)
    332\displaystyle \frac{3}{32}

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac{7}{32}\)Binomial with \(\displaystyle n=8,\ p=\tfrac12\), exactly $\displaystyle 3$ heads.\[P(X=3) = {}^8\mathrm{C}_3\left(\tfrac12\right)^3\left(\tfrac12\right)^5 \] \[= \frac{56}{256} \] \[= \frac{7}{32} \]

  2. Exercise 82

    Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6\displaystyle 6, the probability of getting a sum 3\displaystyle 3, is
    (A)
    118\displaystyle \frac{1}{18}
    (B)
    518\displaystyle \frac{5}{18}
    (C)
    15\displaystyle \frac{1}{5}
    (D)
    25\displaystyle \frac{2}{5}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{1}{5}\)Reduced sample space: sums $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, 5.\[n(\text{sum}<6) = 1+2+3+4 = 10 \] \[n(\text{sum}=3) = 2 \quad \{(1,2),(2,1)\} \] \[P(\text{sum}=3 \mid \text{sum}<6) = \frac{2}{10} = \frac{1}{5} \]

  3. Exercise 83

    Which one is not a requirement of a binomial distribution?
    (A)
    There are 2\displaystyle 2 outcomes for each trial
    (B)
    There is a fixed number of trials
    (C)
    The outcomes must be dependent on each other
    (D)
    The probability of success must be the same for all the trials

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    NCERT’s answer
    C
    (C) The outcomes must be dependent on each otherBinomial trials are independent, with two outcomes, a fixed number of trials and constant \(\displaystyle p\). Dependence is not a requirement.

  4. Exercise 84

    Two cards are drawn from a well shuffled deck of 52\displaystyle 52 playing cards with replacement. The probability, that both cards are queens, is
    (A)
    113×113\displaystyle \frac{1}{13} \times \frac{1}{13}
    (B)
    113+113\displaystyle \frac{1}{13}+\frac{1}{13}
    (C)
    113×117\displaystyle \frac{1}{13} \times \frac{1}{17}
    (D)
    113×451\displaystyle \frac{1}{13} \times \frac{4}{51}

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    NCERT’s answer
    A
    (A) \(\displaystyle \frac{1}{13}\times\frac{1}{13}\)With replacement, the draws are independent.\[P(\text{queen}) = \frac{4}{52} = \frac{1}{13} \] \[P(\text{both queens}) = \frac{1}{13}\times\frac{1}{13} \]

  5. Exercise 85

    The probability of guessing correctly at least 8\displaystyle 8 out of 10\displaystyle 10 answers on a true-false type examination is
    (A)
    764\displaystyle \frac{7}{64}
    (B)
    7128\displaystyle \frac{7}{128}
    (C)
    451024\displaystyle \frac{45}{1024}
    (D)
    741\displaystyle \frac{7}{41}

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac{7}{128}\)Binomial with \(\displaystyle n=10,\ p=\tfrac12\); each term has \(\displaystyle \left(\tfrac12\right)^{10}\).\[P(X\ge 8) = P(8)+P(9)+P(10) \] \[= \left({}^{10}\mathrm{C}_8+{}^{10}\mathrm{C}_9+{}^{10}\mathrm{C}_{10}\right)\frac{1}{1024} \] \[= \frac{45+10+1}{1024} = \frac{56}{1024} = \frac{7}{128} \]

  6. Exercise 86

    The probability that a person is not a swimmer is 0.3. The probability that out of 5\displaystyle 5 persons 4\displaystyle 4 are swimmers is
    (A)
    5C4(0.7)4(0.3)\displaystyle { }^5 \mathrm{C}_4(0.7)^4(0.3)
    (B)
    5C1(0.7)(0.3)4\displaystyle { }^5 \mathrm{C}_1(0.7)(0.3)^4
    (C)
    5C4(0.7)(0.3)4\displaystyle { }^5 \mathrm{C}_4(0.7)(0.3)^4
    (D)
    (0.7)4(0.3)\displaystyle (0.7)^4(0.3)

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    NCERT’s answer
    A
    (A) \(\displaystyle {}^5\mathrm{C}_4(0.7)^4(0.3)\)\[p = P(\text{swimmer}) = 1-0.3 = 0.7,\quad q = 0.3 \] \[P(X=4) = {}^5\mathrm{C}_4\,p^4q^{1} \] \[= {}^5\mathrm{C}_4(0.7)^4(0.3) \]

  7. Exercise 87

    The probability distribution of a discrete random variable X is given below:
    X2\displaystyle 23\displaystyle 34\displaystyle 45\displaystyle 5
    P(X)5k\displaystyle \frac{5}{k}7k\displaystyle \frac{7}{k}9k\displaystyle \frac{9}{k}11k\displaystyle \frac{11}{k}
    The value of k\displaystyle k is
    (A)
    8\displaystyle 8 (B) 16\displaystyle 16
    (C)
    32\displaystyle 32
    (D)
    48\displaystyle 48

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    NCERT’s answer
    C
    (C) \(\displaystyle 32\)Probabilities of a distribution add to 1.\[\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k} = 1 \] \[\frac{32}{k} = 1 \] \[k = 32 \]

  8. Exercise 88

    For the following probability distribution:
    X−4\displaystyle -4−3\displaystyle -3−2\displaystyle -2−1\displaystyle -10\displaystyle 0
    P(X)0.1\displaystyle 0.10.2\displaystyle 0.20.3\displaystyle 0.30.2\displaystyle 0.20.2\displaystyle 0.2
    E(X)\displaystyle \mathrm{E}(\mathrm{X}) is equal to :
    (A)
    0\displaystyle 0 (B) −1\displaystyle -1 (C) −2\displaystyle -2 (D) −1.8\displaystyle -1.8

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    NCERT’s answer
    D
    (D) \(\displaystyle -1.8\)\[E(X) = \sum x_i\,p_i \] \[= (-4)(0.1)+(-3)(0.2)+(-2)(0.3)+(-1)(0.2)+0(0.2) \] \[= -0.4-0.6-0.6-0.2+0 \] \[= -1.8 \]

  9. Exercise 89

    For the following probability distribution
    X1\displaystyle 12\displaystyle 23\displaystyle 34\displaystyle 4
    P (X)110\displaystyle \frac{1}{10}15\displaystyle \frac{1}{5}310\displaystyle \frac{3}{10}25\displaystyle \frac{2}{5}
    E(X2)\displaystyle \mathrm{E}\left(\mathrm{X}^2\right) is equal to
    (A)
    3\displaystyle 3 (B) 5\displaystyle 5 (C) 7\displaystyle 7 (D) 10\displaystyle 10

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    NCERT’s answer
    D
    (D) \(\displaystyle 10\)\[E(X^2)=\sum x^2\,P(x) \] \[E(X^2)=1^2\cdot\tfrac{1}{10}+2^2\cdot\tfrac{1}{5}+3^2\cdot\tfrac{3}{10}+4^2\cdot\tfrac{2}{5} \] \[=\tfrac{1}{10}+\tfrac{8}{10}+\tfrac{27}{10}+\tfrac{64}{10} \] \[=\tfrac{100}{10}=10 \]

  10. Exercise 90

    Suppose a random variable X follows the binomial distribution with parameters n\displaystyle n and p\displaystyle p, where 0<p<1\displaystyle 0<p<1. If P(x=r)/P(x=n−r)\displaystyle \mathrm{P}(x=r) / \mathrm{P}(x=n-r) is independent of n\displaystyle n and r\displaystyle r, then p\displaystyle p equals
    (A)
    12\displaystyle \frac{1}{2}
    (B)
    13\displaystyle \frac{1}{3}
    (C)
    15\displaystyle \frac{1}{5}
    (D)
    17\displaystyle \frac{1}{7}

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    NCERT’s answer
    A
    (A) \(\displaystyle \tfrac12\)With \(\displaystyle q=1-p\): \[P(x=r)=\binom{n}{r}p^{r}q^{\,n-r},\qquad P(x=n-r)=\binom{n}{n-r}p^{\,n-r}q^{r} \] \[\binom{n}{r}=\binom{n}{n-r}\ \Rightarrow\ \frac{P(x=r)}{P(x=n-r)}=\frac{p^{r}q^{\,n-r}}{p^{\,n-r}q^{r}}=\left(\frac{p}{q}\right)^{2r-n} \] The ratio is free of \(\displaystyle n\) and \(\displaystyle r\) only when its base is $\displaystyle 1$: \[\frac{p}{q}=1\ \Rightarrow\ p=1-p\ \Rightarrow\ p=\tfrac12 \]