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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 41–50 (part 5 of 11)

  1. Exercise 41

    Three bags contain a number of red and white balls as follows:
    Bag 1\displaystyle 1 : 3\displaystyle 3 red balls, Bag 2\displaystyle 2 : 2\displaystyle 2 red balls and 1\displaystyle 1 white ball
    Bag 3\displaystyle 3 : 3\displaystyle 3 white balls.
    The probability that bag i\displaystyle i will be chosen and a ball is selected from it is i6\displaystyle \frac{i}{6}, i=1,2,3\displaystyle i=1,2,3. What is the probability that
    (i)
    a red ball will be selected?
    (ii)
    a white ball is selected?

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{7}{18}\), (ii) \(\displaystyle \frac{11}{18}\)
    Let \(\displaystyle B_i\) = bag \(\displaystyle i\) is chosen, \(\displaystyle R\) = red ball, \(\displaystyle W\) = white ball.
    \[P(B_1)=\frac16,\quad P(B_2)=\frac26,\quad P(B_3)=\frac36 \]
    \[P(R\mid B_1)=1,\quad P(R\mid B_2)=\frac23,\quad P(R\mid B_3)=0 \]
    (i)
    Total probability:
    \[P(R)=\sum_{i=1}^{3}P(B_i)\,P(R\mid B_i) \]
    \[P(R)=\frac16\cdot 1+\frac26\cdot\frac23+\frac36\cdot 0 \]
    \[P(R)=\frac16+\frac29=\frac{3+4}{18}=\frac{7}{18} \]
    (ii)
    Every ball is red or white:
    \[P(W)=1-P(R)=1-\frac{7}{18}=\frac{11}{18} \]
    Answer: (i) \(\displaystyle \dfrac{7}{18}\) (ii) \(\displaystyle \dfrac{11}{18}\)

  2. Exercise 42

    Refer to Question 41\displaystyle 41 above. If a white ball is selected, what is the probability that it came from
    (i)
    Bag 2\displaystyle 2
    (ii)
    Bag 3\displaystyle 3

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{2}{11}\), (ii) \(\displaystyle \frac{9}{11}\)
    Bag \(\displaystyle i\) is chosen with \(\displaystyle P(B_i)=\dfrac i6\); \(\displaystyle W\) = white ball.
    \[P(W\mid B_1)=0,\quad P(W\mid B_2)=\frac13,\quad P(W\mid B_3)=1 \]
    \[P(W)=\frac16\cdot 0+\frac26\cdot\frac13+\frac36\cdot 1=\frac19+\frac12=\frac{11}{18} \]
    Bayes' theorem:
    \[P(B_i\mid W)=\frac{P(B_i)\,P(W\mid B_i)}{P(W)} \]
    (i)
    \[P(B_2\mid W)=\frac{\frac26\cdot\frac13}{\dfrac{11}{18}}=\frac{\frac19}{\dfrac{11}{18}}=\frac{2}{11} \]
    (ii)
    \[P(B_3\mid W)=\frac{\frac36\cdot 1}{\dfrac{11}{18}}=\frac{\frac12}{\dfrac{11}{18}}=\frac{9}{11} \]
    Answer: (i) \(\displaystyle \dfrac{2}{11}\) (ii) \(\displaystyle \dfrac{9}{11}\)

  3. Exercise 43

    A shopkeeper sells three types of flower seeds A1, A2\displaystyle \mathrm{A}_1, \mathrm{~A}_2 and A3\displaystyle \mathrm{A}_3. They are sold as a mixture where the proportions are 4\displaystyle 4:4\displaystyle 4:2\displaystyle 2 respectively. The germination rates of the three types of seeds are 45\displaystyle 45%, 60\displaystyle 60% and 35\displaystyle 35%. Calculate the probability
    (i)
    of a randomly chosen seed to germinate
    (ii)
    that it will not germinate given that the seed is of type A3\displaystyle \mathrm{A}_3,
    (iii)
    that it is of the type A2\displaystyle \mathrm{A}_2 given that a randomly chosen seed does not germinate.

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    NCERT’s answer
    (i)
    0.$\displaystyle 49$, (ii) $\displaystyle 0.65$, (iii) .314
    Let \(\displaystyle A_i\) = seed is of type \(\displaystyle A_i\), \(\displaystyle G\) = seed germinates.
    \[P(A_1)=\frac4{10},\quad P(A_2)=\frac4{10},\quad P(A_3)=\frac2{10} \]
    \[P(G\mid A_1)=0.45,\quad P(G\mid A_2)=0.60,\quad P(G\mid A_3)=0.35 \]
    (i)
    \[P(G)=0.4(0.45)+0.4(0.60)+0.2(0.35) \]
    \[P(G)=0.18+0.24+0.07=0.49 \]
    (ii)
    \[P(\overline G\mid A_3)=1-P(G\mid A_3)=1-0.35=0.65 \]
    (iii)
    \[P(\overline G)=1-0.49=0.51 \]
    \[P(A_2\mid\overline G)=\frac{P(A_2)\,P(\overline G\mid A_2)}{P(\overline G)}=\frac{0.4\times 0.40}{0.51} \]
    \[P(A_2\mid\overline G)=\frac{0.16}{0.51}=\frac{16}{51} \]
    Answer: (i) \(\displaystyle 0.49\) (ii) \(\displaystyle 0.65\) (iii) \(\displaystyle \dfrac{16}{51}\)

  4. Exercise 44

    A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from TATA NAGAR.

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    NCERT’s answer
    \(\displaystyle \frac{7}{11}\)
    Let \(\displaystyle T\) = letter from TATA NAGAR, \(\displaystyle C\) = from CALCUTTA, \(\displaystyle E\) = "TA" is visible.\[P(T)=P(C)=\frac12 \]Pairs of consecutive letters:\[\text{TATANAGAR: } TA,\ AT,\ TA,\ AN,\ NA,\ AG,\ GA,\ AR \quad (8\text{ pairs, }2\text{ are }TA) \]\[\text{CALCUTTA: } CA,\ AL,\ LC,\ CU,\ UT,\ TT,\ TA \quad (7\text{ pairs, }1\text{ is }TA) \]\[P(E\mid T)=\frac28=\frac14,\qquad P(E\mid C)=\frac17 \]Bayes' theorem:\[P(T\mid E)=\frac{\frac12\cdot\frac14}{\frac12\cdot\frac14+\frac12\cdot\frac17} \]\[P(T\mid E)=\frac{\frac14}{\dfrac{11}{28}}=\frac{7}{11} \]Answer: \(\displaystyle \dfrac{7}{11}\)

  5. Exercise 45

    There are two bags, one of which contains 3\displaystyle 3 black and 4\displaystyle 4 white balls while the other contains 4\displaystyle 4 black and 3\displaystyle 3 white balls. A die is thrown. If it shows up 1\displaystyle 1 or 3\displaystyle 3, a ball is taken from the Ist bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

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    NCERT’s answer
    \(\displaystyle \frac{11}{21}\)
    Let \(\displaystyle E_1\) = die shows $\displaystyle 1$ or $\displaystyle 3$, \(\displaystyle E_2\) = any other number, \(\displaystyle B\) = black ball.\[P(E_1)=\frac26=\frac13,\qquad P(E_2)=\frac46=\frac23 \]\[P(B\mid E_1)=\frac37,\qquad P(B\mid E_2)=\frac47 \]Total probability:\[P(B)=P(E_1)P(B\mid E_1)+P(E_2)P(B\mid E_2) \]\[P(B)=\frac13\cdot\frac37+\frac23\cdot\frac47 \]\[P(B)=\frac{3}{21}+\frac{8}{21}=\frac{11}{21} \]Answer: \(\displaystyle \dfrac{11}{21}\)

  6. Exercise 46

    There are three urns containing 2\displaystyle 2 white and 3\displaystyle 3 black balls, 3\displaystyle 3 white and 2\displaystyle 2 black balls, and 4\displaystyle 4 white and 1\displaystyle 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}\)
    Let \(\displaystyle U_i\) = urn \(\displaystyle i\) is chosen, \(\displaystyle W\) = ball drawn is white.\[P(U_1)=P(U_2)=P(U_3)=\frac13 \]\[P(W\mid U_1)=\frac25,\quad P(W\mid U_2)=\frac35,\quad P(W\mid U_3)=\frac45 \]Bayes' theorem:\[P(U_2\mid W)=\frac{P(U_2)\,P(W\mid U_2)}{\sum_{i=1}^{3}P(U_i)\,P(W\mid U_i)} \]\[P(U_2\mid W)=\frac{\frac13\cdot\frac35}{\frac13\left(\frac25+\frac35+\frac45\right)} \]\[P(U_2\mid W)=\frac{3/5}{9/5}=\frac13 \]Answer: \(\displaystyle \dfrac13\)

  7. Exercise 47

    By examining the chest X ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1\displaystyle 1 in 1000\displaystyle 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

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    NCERT’s answer
    \(\displaystyle \frac{110}{221}\)
    Let \(\displaystyle T\) = person has TB, \(\displaystyle D\) = diagnosed to have TB.\[P(T)=\frac{1}{1000}=0.001,\qquad P(\overline T)=0.999 \]\[P(D\mid T)=0.99,\qquad P(D\mid\overline T)=0.001 \]Bayes' theorem:\[P(T\mid D)=\frac{P(T)\,P(D\mid T)}{P(T)\,P(D\mid T)+P(\overline T)\,P(D\mid\overline T)} \]\[P(T\mid D)=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.001} \]\[P(T\mid D)=\frac{0.99}{1.989}=\frac{990}{1989}=\frac{110}{221}\approx 0.498 \]Answer: \(\displaystyle \dfrac{110}{221}\)

  8. Exercise 48

    An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50%\displaystyle 50 \% are manufactured on A, 30\displaystyle 30% on B and 20\displaystyle 20% on C. 2\displaystyle 2% of the items produced on A and 2\displaystyle 2% of items produced on B are defective, and 3\displaystyle 3% of these produced on C are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

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    NCERT’s answer
    \(\displaystyle \frac{5}{11}\)
    Let \(\displaystyle E_A,E_B,E_C\) = item made on A, B, C; \(\displaystyle D\) = item is defective.\[P(E_A)=0.5,\quad P(E_B)=0.3,\quad P(E_C)=0.2 \]\[P(D\mid E_A)=0.02,\quad P(D\mid E_B)=0.02,\quad P(D\mid E_C)=0.03 \]Total probability:\[P(D)=0.5(0.02)+0.3(0.02)+0.2(0.03) \]\[P(D)=0.010+0.006+0.006=0.022 \]Bayes' theorem:\[P(E_A\mid D)=\frac{P(E_A)\,P(D\mid E_A)}{P(D)}=\frac{0.010}{0.022} \]\[P(E_A\mid D)=\frac{5}{11} \]Answer: \(\displaystyle \dfrac{5}{11}\)

  9. Exercise 49

    Let X be a discrete random variable whose probability distribution is defined as follows: P(X=x)={k(x+1) for x=1,2,3,42kx for x=5,6,70 otherwise P(\mathrm{X}=x)= \begin{cases}k(x+1) & \text { for } x=1,2,3,4 \\ 2 k x & \text { for } x=5,6,7 \\ 0 & \text { otherwise }\end{cases} where k\displaystyle k is a constant. Calculate
    (i)
    the value of k\displaystyle k
    (ii)
    E (X)
    (iii)
    Standard deviation of X.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{50}\), (ii) $\displaystyle 5.2$, (iii) $\displaystyle 1.7$ (approx.)
    (i) \[\begin{array}{c|ccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & 2k & 3k & 4k & 5k & 10k & 12k & 14k \end{array} \] \[\sum P(x) = 50k = 1 \] \[k = \tfrac{1}{50} \] (ii) \[E(X) = \sum xP(x) = k\,(2+6+12+20+50+72+98) = 260k = \tfrac{26}{5} \] (iii) \[E(X^2) = \sum x^2P(x) = k\,(2+12+36+80+250+432+686) = 1498k = \tfrac{749}{25} \] \[\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \tfrac{749}{25} - \tfrac{676}{25} = \tfrac{73}{25} \] \[\sigma = \sqrt{\operatorname{Var}(X)} = \tfrac{\sqrt{73}}{5} \approx 1.71 \] Answer: \(\displaystyle k = \tfrac{1}{50}\), \(\displaystyle E(X) = \tfrac{26}{5}\), standard deviation \(\displaystyle = \tfrac{\sqrt{73}}{5}\).

  10. Exercise 50

    The probability distribution of a discrete random variable X is given as under:
    X1\displaystyle 12\displaystyle 24\displaystyle 42A3A5A
    P(X)12\displaystyle \frac{1}{2}15\displaystyle \frac{1}{5}325\displaystyle \frac{3}{25}110\displaystyle \frac{1}{10}125\displaystyle \frac{1}{25}125\displaystyle \frac{1}{25}
    Calculate :
    (i)
    The value of A if E(X)=2.94\displaystyle \mathrm{E}(\mathrm{X})=2.94
    (ii)
    Variance of X.

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    (i) \[E(X) = 1\cdot\tfrac12 + 2\cdot\tfrac15 + 4\cdot\tfrac{3}{25} + 2A\cdot\tfrac1{10} + 3A\cdot\tfrac1{25} + 5A\cdot\tfrac1{25} \] \[2.94 = 1.38 + 0.52A \] \[A = \tfrac{1.56}{0.52} = 3 \] (ii) With \(\displaystyle A = 3\), \(\displaystyle X\) takes the values \(\displaystyle 1, 2, 4, 6, 9, 15\). \[E(X^2) = 1(0.5) + 4(0.2) + 16(0.12) + 36(0.1) + 81(0.04) + 225(0.04) = 19.06 \] \[\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 19.06 - (2.94)^2 = 19.06 - 8.6436 = 10.4164 \] Answer: \(\displaystyle A = 3\); \(\displaystyle \operatorname{Var}(X) = 10.4164\).