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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 11–20 (part 2 of 11)

  1. Exercise 11

    Prove that
    (i)
    P(A)=P(A∩B)+P(A∩Bˉ)\displaystyle \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \bar{B})
    (ii)
    P(A∪B)=P(A∩B)+P(A∩Bˉ)+P(Aˉ∩B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \bar{B})+\mathrm{P}(\bar{A} \cap \mathrm{B})

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    (i) \[A = A\cap(B\cup\bar B) = (A\cap B)\cup(A\cap\bar B) \] \[(A\cap B)\cap(A\cap\bar B) = A\cap(B\cap\bar B) = \varnothing \quad \text{(mutually exclusive)} \] \[P(A) = P(A\cap B) + P(A\cap\bar B) \](ii) \[A\cup B = A\cup(\bar A\cap B) \] \[A\cap(\bar A\cap B) = \varnothing \quad \text{(mutually exclusive)} \] \[P(A\cup B) = P(A) + P(\bar A\cap B) \] Put \(\displaystyle P(A)\) from (i): \[P(A\cup B) = P(A\cap B) + P(A\cap\bar B) + P(\bar A\cap B) \]

  2. Exercise 12

    If X is the number of tails in three tosses of a coin, determine the standard deviation of X.

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    NCERT’s answer
    \(\displaystyle \frac{\sqrt{3}}{2}\)
    \[X\in\{0,1,2,3\}, \qquad P(X=r) = \tbinom{3}{r}\tfrac18 \] \[P(0)=\tfrac18,\quad P(1)=\tfrac38,\quad P(2)=\tfrac38,\quad P(3)=\tfrac18 \] \[E(X) = 0 + \tfrac38 + \tfrac68 + \tfrac38 = \tfrac32 \] \[E(X^2) = 0 + \tfrac38 + \tfrac{12}{8} + \tfrac98 = 3 \] \[\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 3 - \tfrac94 = \tfrac34 \] \[\sigma = \sqrt{\tfrac34} = \tfrac{\sqrt3}{2} \]Answer: standard deviation \(\displaystyle =\dfrac{\sqrt3}{2}\approx 0.87\).

  3. Exercise 13

    In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5\displaystyle 5 if the die shows a 3\displaystyle 3, Rs 2\displaystyle 2 if the die shows a 1\displaystyle 1 or 6\displaystyle 6, and nothing otherwise. What is the player's expected profit per throw over a long series of throws?

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    NCERT’s answer
    Rs $\displaystyle 0.50$
    Net profit \(\displaystyle Y\) per throw = prize \(\displaystyle -\) stake of Re 1.\[\text{die shows } 3:\quad P=\tfrac16,\quad Y = 5-1 = 4 \] \[\text{die shows } 1 \text{ or } 6:\quad P=\tfrac26,\quad Y = 2-1 = 1 \] \[\text{otherwise}:\quad P=\tfrac36,\quad Y = 0-1 = -1 \] \[E(Y) = 4\cdot\tfrac16 + 1\cdot\tfrac26 + (-1)\cdot\tfrac36 = \tfrac{4+2-3}{6} = \tfrac12 \]Answer: expected profit \(\displaystyle =\) Re \(\displaystyle 0.50\) per throw.

  4. Exercise 14

    Three dice are thrown at the sametime. Find the probability of getting three two's, if it is known that the sum of the numbers on the dice was six.

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    NCERT’s answer
    \(\displaystyle \frac{1}{10}\)
    Let \(\displaystyle A\): three $\displaystyle 2$'s; \(\displaystyle B\): sum is 6.\[B:\ (1,1,4)\times 3,\quad (1,2,3)\times 6,\quad (2,2,2)\times 1 \] \[n(B) = 3+6+1 = 10 \] \[A\cap B = \{(2,2,2)\}, \qquad n(A\cap B) = 1 \] \[P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{1/216}{10/216} = \frac1{10} \]Answer: \(\displaystyle \dfrac1{10}\)

  5. Exercise 15

    Suppose 10,000\displaystyle 10{,}000 tickets are sold in a lottery each for Re 1. First prize is of Rs 3000\displaystyle 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500\displaystyle 500 each. If you buy one ticket, what is your expectation.

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    NCERT’s answer
    Expectation = Rs $\displaystyle 0.65$
    Let \(\displaystyle X\) be the prize on one ticket.\[P(X=3000)=\tfrac1{10000},\quad P(X=2000)=\tfrac1{10000},\quad P(X=500)=\tfrac3{10000} \] \[E(X) = \frac{3000 + 2000 + 3\cdot 500}{10000} = \frac{6500}{10000} = 0.65 \] Subtract the price of the ticket: \[0.65 - 1 = -0.35 \]Answer: expected prize Rs \(\displaystyle 0.65\); net of the Re $\displaystyle 1$ ticket, the expectation is \(\displaystyle -\) Rs \(\displaystyle 0.35\).

  6. Exercise 16

    A bag contains 4\displaystyle 4 white and 5\displaystyle 5 black balls. Another bag contains 9\displaystyle 9 white and 7\displaystyle 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

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    NCERT’s answer
    \(\displaystyle \frac{85}{153}\)
    Let \(\displaystyle E_1\): white ball transferred, \(\displaystyle E_2\): black ball transferred, \(\displaystyle W\): white ball drawn from the second bag.\[P(E_1)=\tfrac49, \qquad P(E_2)=\tfrac59 \] \[E_1:\ \text{second bag } 10\text{W},7\text{B} \;\Rightarrow\; P(W\mid E_1)=\tfrac{10}{17} \] \[E_2:\ \text{second bag } 9\text{W},8\text{B} \;\Rightarrow\; P(W\mid E_2)=\tfrac{9}{17} \] \[P(W) = P(E_1)\,P(W\mid E_1) + P(E_2)\,P(W\mid E_2) \] \[= \tfrac49\cdot\tfrac{10}{17} + \tfrac59\cdot\tfrac{9}{17} = \tfrac{40+45}{153} = \tfrac{85}{153} = \tfrac59 \]Answer: \(\displaystyle \dfrac59\)

  7. Exercise 17

    Bag I contains 3\displaystyle 3 black and 2\displaystyle 2 white balls, Bag II contains 2\displaystyle 2 black and 4\displaystyle 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.

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    NCERT’s answer
    \(\displaystyle \frac{7}{15}\)
    Choose the bag, then the ball (total probability).\[P(\text{I})=P(\text{II})=\frac12 \]\[P(B\mid \text{I})=\frac35,\qquad P(B\mid \text{II})=\frac26=\frac13 \]\[P(B)=P(\text{I})\,P(B\mid \text{I})+P(\text{II})\,P(B\mid \text{II}) \]\[P(B)=\frac12\cdot\frac35+\frac12\cdot\frac13=\frac{3}{10}+\frac16=\frac{14}{30} \]Answer: \(\displaystyle \dfrac{7}{15}\)

  8. Exercise 18

    A box has 5\displaystyle 5 blue and 4\displaystyle 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?

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    NCERT’s answer
    \(\displaystyle \frac{5}{9}\)
    The unseen first ball is blue or red; split on it.\[P(\text{2nd blue})=P(B_1)\,P(B_2\mid B_1)+P(R_1)\,P(B_2\mid R_1) \]\[P(B_1)=\frac59,\quad P(B_2\mid B_1)=\frac48,\quad P(R_1)=\frac49,\quad P(B_2\mid R_1)=\frac58 \]\[P(\text{2nd blue})=\frac59\cdot\frac48+\frac49\cdot\frac58=\frac{20}{72}+\frac{20}{72}=\frac{40}{72} \]Answer: \(\displaystyle \dfrac{5}{9}\)

  9. Exercise 19

    Four cards are successively drawn without replacement from a deck of 52\displaystyle 52 playing cards. What is the probability that all the four cards are kings?

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    NCERT’s answer
    \(\displaystyle \frac{1}{270725}\)
    Each king drawn leaves one fewer king and one fewer card.\[P=\frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}\cdot\frac{1}{49} \]\[P=\frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{25}\cdot\frac{1}{49}=\frac{1}{270725} \]Answer: \(\displaystyle \dfrac{1}{270725}\)

  10. Exercise 20

    A die is thrown 5\displaystyle 5 times. Find the probability that an odd number will come up exactly three times.

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    NCERT’s answer
    \(\displaystyle \frac{5}{16}\)
    Five throws are independent trials; an odd number has probability \(\displaystyle \frac12\) each time.\[n=5,\qquad p=\frac12,\qquad q=\frac12 \]\[P(X=r)={}^{n}C_r\,p^{\,r}q^{\,n-r} \]\[P(X=3)={}^{5}C_3\left(\frac12\right)^3\left(\frac12\right)^2=10\cdot\frac{1}{32} \]Answer: \(\displaystyle \dfrac{5}{16}\)