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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 21–30 (part 3 of 11)

  1. Exercise 21

    Ten coins are tossed. What is the probability of getting at least 8\displaystyle 8 heads?

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    NCERT’s answer
    \(\displaystyle \frac{7}{128}\)
    Ten independent tosses, head with probability \(\displaystyle \frac12\).\[n=10,\qquad p=q=\frac12 \]\[P(X\ge 8)=P(8)+P(9)+P(10) \]\[P(X\ge 8)=\left({}^{10}C_8+{}^{10}C_9+{}^{10}C_{10}\right)\left(\frac12\right)^{10} \]\[P(X\ge 8)=\frac{45+10+1}{1024}=\frac{56}{1024} \]Answer: \(\displaystyle \dfrac{7}{128}\)

  2. Exercise 22

    The probability of a man hitting a target is 0.25. He shoots 7\displaystyle 7 times. What is the probability of his hitting at least twice?

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    NCERT’s answer
    \(\displaystyle \frac{4547}{8192}\)
    Seven independent shots; use the complement of "$\displaystyle 0$ or $\displaystyle 1$ hit".\[n=7,\qquad p=\frac14,\qquad q=\frac34 \]\[P(X\ge 2)=1-P(0)-P(1) \]\[P(0)=\left(\frac34\right)^7=\frac{2187}{16384} \]\[P(1)={}^{7}C_1\cdot\frac14\left(\frac34\right)^6=\frac{7\cdot 729}{16384}=\frac{5103}{16384} \]\[P(X\ge 2)=1-\frac{2187+5103}{16384}=1-\frac{7290}{16384}=\frac{9094}{16384} \]Answer: \(\displaystyle \dfrac{4547}{8192}\approx 0.555\)

  3. Exercise 23

    A lot of 100\displaystyle 100 watches is known to have 10\displaystyle 10 defective watches. If 8\displaystyle 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?

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    NCERT’s answer
    \(\displaystyle 1-\left(\frac{9}{10}\right)^8\)
    Draws are with replacement, so each is an independent trial with a defective watch having probability \(\displaystyle \frac{1}{10}\).\[n=8,\qquad p=\frac{10}{100}=\frac1{10},\qquad q=\frac9{10} \]\[P(X\ge 1)=1-P(X=0) \]\[P(X=0)={}^{8}C_0\left(\frac{9}{10}\right)^8=\left(\frac{9}{10}\right)^8 \]\[P(X\ge 1)=1-\left(\frac{9}{10}\right)^8 \approx 1-0.4305 \]Answer: \(\displaystyle 1-\left(\dfrac{9}{10}\right)^8\approx 0.570\)

  4. Exercise 24

    Consider the probability distribution of a random variable X:
    X0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 34\displaystyle 4
    P(X)0.1\displaystyle 0.10.25\displaystyle 0.250.3\displaystyle 0.30.2\displaystyle 0.20.15\displaystyle 0.15
    Calculate
    (i)
    V(X2)\displaystyle \mathrm{V}\left(\frac{\mathrm{X}}{2}\right)
    (ii)
    Variance of X.

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    Find \(\displaystyle V(X)\) first; then scale.\[E(X)=\sum xP(x)=0(0.1)+1(0.25)+2(0.3)+3(0.2)+4(0.15)=2.05 \]\[E(X^2)=\sum x^2P(x)=0+0.25+1.2+1.8+2.4=5.65 \]\[V(X)=E(X^2)-\left[E(X)\right]^2=5.65-4.2025=1.4475 \]\[V\!\left(\frac X2\right)=\frac14\,V(X)=\frac{1.4475}{4}=0.361875 \]Answer: (i) \(\displaystyle V\!\left(\dfrac X2\right)=0.361875\); (ii) \(\displaystyle V(X)=1.4475\)

  5. Exercise 25

    The probability distribution of a random variable X is given below:
    X0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 3
    P(X)k\displaystyle kk2\displaystyle \frac{k}{2}k4\displaystyle \frac{k}{4}k8\displaystyle \frac{k}{8}
    (i)
    Determine the value of k\displaystyle k.
    (ii)
    Determine P(X≤2)\displaystyle \mathrm{P}(\mathrm{X} \leq 2) and P(X>2)\displaystyle \mathrm{P}(\mathrm{X}>2)
    (iii)
    Find P(X≤2)+P(X>2)\displaystyle \mathrm{P}(\mathrm{X} \leq 2)+\mathrm{P}(\mathrm{X}>2).

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{8}{15}\), (ii) \(\displaystyle \frac{14}{15}, \frac{1}{15}\), (iii) $\displaystyle 1$
    Probabilities sum to 1.
    (i)
    \[k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1 \]
    \[\frac{15k}{8} = 1 \;\Rightarrow\; k = \frac{8}{15} \]
    (ii)
    \[P(X \le 2) = k\left(1 + \frac12 + \frac14\right) = \frac{7}{4}\cdot\frac{8}{15} = \frac{14}{15} \]
    \[P(X > 2) = P(X = 3) = \frac{k}{8} = \frac{1}{15} \]
    (iii)
    \[P(X \le 2) + P(X > 2) = \frac{14}{15} + \frac{1}{15} = 1 \]
    Answer: (i) \(\displaystyle k = \frac{8}{15}\); (ii) \(\displaystyle P(X \le 2) = \frac{14}{15}\), \(\displaystyle P(X > 2) = \frac{1}{15}\); (iii) \(\displaystyle 1\).

  6. Exercise 26

    For the following probability distribution determine standard deviation of the random variable X.
    X2\displaystyle 23\displaystyle 34\displaystyle 4
    P(X)0.2\displaystyle 0.20.5\displaystyle 0.50.3\displaystyle 0.3

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    NCERT’s answer
    0.$\displaystyle 7$ (approx.)
    \[E(X) = 2(0.2) + 3(0.5) + 4(0.3) = 3.1 \] \[E(X^2) = 4(0.2) + 9(0.5) + 16(0.3) = 10.1 \] \[\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 10.1 - 9.61 = 0.49 \] \[\sigma = \sqrt{0.49} = 0.7 \] Answer: standard deviation \(\displaystyle = 0.7\).

  7. Exercise 27

    A biased die is such that P(4)=110\displaystyle \mathrm{P}(4)=\frac{1}{10} and other scores being equally likely. The die is tossed twice. If X is the 'number of fours seen', find the variance of the random variable X.

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    NCERT’s answer
    0.$\displaystyle 18$
    A four is a success: \(\displaystyle p = \frac{1}{10}\), two tosses, so \(\displaystyle X\) is binomial. \[X \sim B\left(2, \tfrac{1}{10}\right), \qquad q = 1 - \tfrac{1}{10} = \tfrac{9}{10} \] \[\operatorname{Var}(X) = npq = 2\cdot\frac{1}{10}\cdot\frac{9}{10} = \frac{9}{50} \] Answer: \(\displaystyle \operatorname{Var}(X) = \frac{9}{50} = 0.18\).

  8. Exercise 28

    A die is thrown three times. Let X be 'the number of twos seen'. Find the expectation of X.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\)
    A two is a success, \(\displaystyle p = \frac16\), three throws, so \(\displaystyle X \sim B\left(3, \frac16\right)\). \[P(X=0) = \left(\frac56\right)^3 = \frac{125}{216}, \qquad P(X=1) = 3\cdot\frac16\left(\frac56\right)^2 = \frac{75}{216} \] \[P(X=2) = 3\left(\frac16\right)^2\frac56 = \frac{15}{216}, \qquad P(X=3) = \left(\frac16\right)^3 = \frac{1}{216} \] \[E(X) = \frac{0\cdot125 + 1\cdot75 + 2\cdot15 + 3\cdot1}{216} = \frac{108}{216} = \frac12 \] \[\text{Check: } np = 3\cdot\frac16 = \frac12 \] Answer: \(\displaystyle E(X) = \frac12\).

  9. Exercise 29

    Two biased dice are thrown together. For the first die P(6)=12\displaystyle \mathrm{P}(6)=\frac{1}{2}, the other scores being equally likely while for the second die, P(1)=25\displaystyle \mathrm{P}(1)=\frac{2}{5} and the other scores are equally likely. Find the probability distribution of 'the number of ones seen'.

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    NCERT’s answer
    X$\displaystyle 0$$\displaystyle 1$$\displaystyle 2$
    P (X).54.42.04
    First die: the other five scores share the remaining probability equally. \[P(1) = \frac15\left(1 - \frac12\right) = \frac{1}{10} \] Second die: \(\displaystyle P(1) = \frac25\). Let \(\displaystyle X\) be the number of ones. \[P(X=0) = \frac{9}{10}\cdot\frac{3}{5} = \frac{27}{50} \] \[P(X=1) = \frac{1}{10}\cdot\frac{3}{5} + \frac{9}{10}\cdot\frac{2}{5} = \frac{21}{50} \] \[P(X=2) = \frac{1}{10}\cdot\frac{2}{5} = \frac{1}{25} \] \[\frac{27}{50} + \frac{21}{50} + \frac{2}{50} = 1 \] Answer: \(\displaystyle P(X=0) = \frac{27}{50}\), \(\displaystyle P(X=1) = \frac{21}{50}\), \(\displaystyle P(X=2) = \frac{1}{25}\).

  10. Exercise 30

    Two probability distributions of the discrete random variable X and Y are given below.
    X0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 3
    P(X)15\displaystyle \frac{1}{5}25\displaystyle \frac{2}{5}15\displaystyle \frac{1}{5}15\displaystyle \frac{1}{5}
    Y0\displaystyle 01\displaystyle 12\displaystyle 23\displaystyle 3
    P(Y)15\displaystyle \frac{1}{5}310\displaystyle \frac{3}{10}25\displaystyle \frac{2}{5}110\displaystyle \frac{1}{10}
    Prove that E(Y2)=2E(X)\displaystyle E\left(Y^2\right)=2 E(X).

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    \[E(X) = 0\cdot\frac15 + 1\cdot\frac25 + 2\cdot\frac15 + 3\cdot\frac15 = \frac75 \] \[E(Y^2) = 0\cdot\frac15 + 1\cdot\frac{3}{10} + 4\cdot\frac25 + 9\cdot\frac{1}{10} = \frac{3 + 16 + 9}{10} = \frac{14}{5} \] \[2E(X) = 2\cdot\frac75 = \frac{14}{5} = E(Y^2) \] Answer: \(\displaystyle E(Y^2) = \frac{14}{5} = 2E(X)\).