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NCERT Exemplar · Class 11 Mathematics Probability

43 questions · 43 still being checked

EXERCISE 16.3 11–20 (part 2 of 4)

  1. Exercise 11

    The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P(A∩B)=.07\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=.07). Determine
    (a)
    P(A)\displaystyle \mathrm{P}(\mathrm{A})
    (b)
    P(B∩C‾)\displaystyle \mathrm{P}(\mathrm{B} \cap \overline{\mathrm{C}})
    (c)
    P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})
    (d)
    P(A∩B‾)\displaystyle \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})
    (e)
    P(B∩C)\displaystyle \mathrm{P}(\mathrm{B} \cap \mathrm{C})
    (f)
    Probability of exactly one of the three occurs.
    NCERT_Question_Class11_Maths_Exemplar_Ch16_Ex16-3_Q11

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    NCERT’s answer
    (a)
    0.$\displaystyle 20$
    (b)
    0.$\displaystyle 17$
    (c)
    0.$\displaystyle 45$
    (d)
    0.$\displaystyle 13$
    (e)
    0.$\displaystyle 15$
    (f)
    0.$\displaystyle 51$
    From the diagram, A and C do not overlap.
    (a)
    \[P(A)=.13+.07=.20 \]
    (b)
    \[P(B\cap\bar C)=P(B)-P(B\cap C)=(.07+.10+.15)-.15=.17 \]
    (c)
    \[P(A\cup B)=.13+.07+.10+.15=.45 \]
    (d)
    \[P(A\cap\bar B)=.13 \]
    (e)
    \[P(B\cap C)=.15 \]
    (f)
    Exactly one occurs: only A, only B or only C.
    \[.13+.10+.28=.51 \]
    Answer: (a) .20; (b) .17; (c) .45; (d) .13; (e) .15; (f) .51.

  2. Exercise 12

    One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls (labelled W1 and W2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
    (a)
    Write the sample space showing all possible outcomes
    (b)
    What is the probability that two black balls are chosen?
    (c)
    What is the probability that two balls of opposite colour are chosen?

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    NCERT’s answer
    (a)
    \(\displaystyle S=\left\{B_1 B_2, B_1 W, B_2 B_1, B_2 W, W B_1, W B_2 B W_1, B W_2, W_1 B, W_1 W_2, W_2 B, W_2 W_1\right\}\)
    (b)
    \(\displaystyle \frac{1}{6}\)
    (c)
    \(\displaystyle \frac{2}{3}\)
    (a)
    Write \(\displaystyle U_1\) for the urn \(\displaystyle \{B_1,B_2,W\}\) and \(\displaystyle U_2\) for \(\displaystyle \{B,W_1,W_2\}\); the draws are ordered.
    \[S=\{U_1B_1B_2,\ U_1B_2B_1,\ U_1B_1W,\ U_1WB_1,\ U_1B_2W,\ U_1WB_2,\ U_2BW_1,\ U_2W_1B,\ U_2BW_2,\ U_2W_2B,\ U_2W_1W_2,\ U_2W_2W_1\} \]
    Each outcome has probability
    \[\tfrac12\cdot\tfrac13\cdot\tfrac12=\tfrac1{12} \]
    (b)
    Two black: \(\displaystyle U_1B_1B_2,\ U_1B_2B_1\).
    \[P=\tfrac2{12}=\tfrac16 \]
    (c)
    Opposite colours: $\displaystyle 4$ outcomes in \(\displaystyle U_1\) and $\displaystyle 4$ in \(\displaystyle U_2\).
    \[P=\tfrac8{12}=\tfrac23 \]
    Answer: (b) \(\displaystyle \tfrac16\); (c) \(\displaystyle \tfrac23\).

  3. Exercise 13

    A bag contains 8\displaystyle 8 red and 5\displaystyle 5 white balls. Three balls are drawn at random. Find the Probability that
    (a)
    All the three balls are white
    (b)
    All the three balls are red
    (c)
    One ball is red and two balls are white

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    NCERT’s answer
    (a)
    \(\displaystyle \frac{5}{143}\)
    (b)
    \(\displaystyle \frac{28}{143}\)
    (c)
    \(\displaystyle \frac{40}{143}\)
    Three balls from $\displaystyle 13$ can be drawn in
    \[n(S)=\binom{13}{3}=286 \]
    (a)
    All white:
    \[P=\frac{\binom53}{286}=\frac{10}{286}=\frac5{143} \]
    (b)
    All red:
    \[P=\frac{\binom83}{286}=\frac{56}{286}=\frac{28}{143} \]
    (c)
    One red, two white:
    \[P=\frac{\binom81\binom52}{286}=\frac{80}{286}=\frac{40}{143} \]
    Answer: (a) \(\displaystyle \tfrac5{143}\); (b) \(\displaystyle \tfrac{28}{143}\); (c) \(\displaystyle \tfrac{40}{143}\).

  4. Exercise 14

    If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
    (a)
    Four S's come consecutively in the word
    (b)
    Two I's and two N's come together
    (c)
    All A's are not coming together
    (d)
    No two A's are coming together.

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    NCERT’s answer
    (a)
    \(\displaystyle \frac{2}{143}\)
    (b)
    \(\displaystyle \frac{2}{143}\)
    (c)
    \(\displaystyle \frac{25}{26}\)
    (d)
    \(\displaystyle \frac{15}{26}\)
    ASSASSINATION has $\displaystyle 13$ letters: \(\displaystyle A^3,\ S^4,\ I^2,\ N^2,\ T,\ O\).
    \[n(S)=\frac{13!}{3!\,4!\,2!\,2!} \]
    (a)
    Treat SSSS as one block ($\displaystyle 10$ objects).
    \[P=\frac{10!/(3!\,2!\,2!)}{n(S)}=\frac{10!\cdot4!}{13!}=\frac{2}{143} \]
    (b)
    Treat IINN as one block ($\displaystyle 10$ objects: \(\displaystyle A^3,\ S^4,\ T,\ O\), block); inside it the letters can be ordered in \(\displaystyle \frac{4!}{2!\,2!}=6\) ways.
    \[P=\frac{\dfrac{10!}{3!\,4!}\cdot6}{n(S)}=\frac{10!\cdot4!}{13!}=\frac{2}{143} \]
    (c)
    All A's together: AAA as a block ($\displaystyle 11$ objects).
    \[P(\text{together})=\frac{11!/(4!\,2!\,2!)}{n(S)}=\frac{11!\cdot3!}{13!}=\frac1{26} \]
    \[P(\text{not all together})=1-\tfrac1{26}=\tfrac{25}{26} \]
    (d)
    Arrange the $\displaystyle 10$ other letters, then put the A's in $\displaystyle 3$ of the $\displaystyle 11$ gaps.
    \[P=\frac{\binom{11}{3}\cdot10!/(4!\,2!\,2!)}{n(S)}=\frac{165\cdot10!\cdot3!}{13!}=\frac{15}{26} \]
    Answer: (a) \(\displaystyle \tfrac2{143}\); (b) \(\displaystyle \tfrac2{143}\); (c) \(\displaystyle \tfrac{25}{26}\); (d) \(\displaystyle \tfrac{15}{26}\).

  5. Exercise 15

    A card is drawn from a deck of 52\displaystyle 52 cards. Find the probability of getting a king or a heart or a red card.

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    NCERT’s answer
    \(\displaystyle \frac{7}{13}\)
    Every heart is a red card, so \(\displaystyle K\cup H\cup R=K\cup R\). The red kings are the only cards in both \(\displaystyle K\) and \(\displaystyle R\).\[P(K)=\tfrac4{52},\quad P(R)=\tfrac{26}{52},\quad P(K\cap R)=\tfrac2{52} \]\[P(K\cup H\cup R)=\tfrac4{52}+\tfrac{26}{52}-\tfrac2{52}=\tfrac{28}{52}=\tfrac7{13} \]Answer: \(\displaystyle \tfrac7{13}\).

  6. Exercise 16

    A sample space consists of 9\displaystyle 9 elementary outcomes e1,e2,…,e9\displaystyle e_1, e_2, \ldots, e_9 whose probabilities are
    P(e1)=P(e2)=.08,P(e3)=P(e4)=P(e5)=.1P(e6)=P(e7)=.2,P(e8)=P(e9)=.07\begin{aligned} & \mathrm{P}\left(e_1\right)=\mathrm{P}\left(e_2\right)=.08, \mathrm{P}\left(e_3\right)=\mathrm{P}\left(e_4\right)=\mathrm{P}\left(e_5\right)=.1 \\ & \mathrm{P}\left(e_6\right)=\mathrm{P}\left(e_7\right)=.2, \mathrm{P}\left(e_8\right)=\mathrm{P}\left(e_9\right)=.07 \end{aligned}
    Suppose A={e1,e5,e8},B={e2,e5,e8,e9}\displaystyle \mathrm{A}=\left\{e_1, e_5, e_8\right\}, \mathrm{B}=\left\{e_2, e_5, e_8, e_9\right\}
    (a)
    Calculate P(A),P(B)\displaystyle \mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B}), and P(A∩B)\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})
    (b)
    Using the addition law of probability, calculate P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})
    (c)
    List the composition of the event A∪B\displaystyle \mathrm{A} \cup \mathrm{B}, and calculate P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B}) by adding the probabilities of the elementary outcomes.
    (d)
    Calculate P(B‾)\displaystyle \mathrm{P}(\overline{\mathrm{B}}) from P(B)\displaystyle \mathrm{P}(\mathrm{B}), also calculate P(B‾)\displaystyle \mathrm{P}(\overline{\mathrm{B}}) directly from the elementary outcomes of B‾\displaystyle \overline{\mathrm{B}}

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    NCERT’s answer
    (a)
    \(\displaystyle p(\mathrm{A})=.25, p(\mathrm{B})=.32, p(\mathrm{A} \cap \mathrm{B})=.17\)
    (b)
    \(\displaystyle p(\mathrm{A} \cup \mathrm{B})=.40\)
    (c)
    .40
    (d)
    .68
    (a)
    \[P(A)=P(e_1)+P(e_5)+P(e_8)=.08+.1+.07=.25 \]
    \[P(B)=P(e_2)+P(e_5)+P(e_8)+P(e_9)=.08+.1+.07+.07=.32 \]
    \[A\cap B=\{e_5,e_8\},\quad P(A\cap B)=.1+.07=.17 \]
    (b)
    \[P(A\cup B)=P(A)+P(B)-P(A\cap B)=.25+.32-.17=.40 \]
    (c)
    \[A\cup B=\{e_1,e_2,e_5,e_8,e_9\} \]
    \[P(A\cup B)=.08+.08+.1+.07+.07=.40 \]
    (d)
    \[P(\bar B)=1-P(B)=1-.32=.68 \]
    \[\bar B=\{e_1,e_3,e_4,e_6,e_7\},\quad P(\bar B)=.08+.1+.1+.2+.2=.68 \]
    Answer: (a) .25, .32, .17; (b) .40; (c) .40; (d) .68.

  7. Exercise 17

    Determine the probability p\displaystyle p, for each of the following events.
    (a)
    An odd number appears in a single toss of a fair die.
    (b)
    At least one head appears in two tosses of a fair coin.
    (c)
    A king, 9\displaystyle 9 of hearts, or 3\displaystyle 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52\displaystyle 52 cards.
    (d)
    The sum of 6\displaystyle 6 appears in a single toss of a pair of fair dice.

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    NCERT’s answer
    (a)
    \(\displaystyle \frac{1}{2}\)
    (b)
    \(\displaystyle \frac{3}{4}\)
    (c)
    \(\displaystyle \frac{3}{26}\)
    (d)
    \(\displaystyle \frac{5}{36}\)
    (a)
    Odd number on a die:
    \[n(S)=6,\quad E=\{1,3,5\} \]
    \[p=\frac{3}{6}=\frac12 \]
    (b)
    At least one head in two tosses:
    \[S=\{HH,HT,TH,TT\},\quad E=\{HH,HT,TH\} \]
    \[p=\frac34 \]
    (c)
    King, $\displaystyle 9$ of hearts or $\displaystyle 3$ of spades:
    \[n(E)=4+1+1=6 \]
    \[p=\frac{6}{52}=\frac{3}{26} \]
    (d)
    Sum $\displaystyle 6$ with two dice:
    \[n(S)=36,\quad E=\{(1,5),(2,4),(3,3),(4,2),(5,1)\} \]
    \[p=\frac{5}{36} \]
    Answer: (a) \(\displaystyle \frac12\); (b) \(\displaystyle \frac34\); (c) \(\displaystyle \frac{3}{26}\); (d) \(\displaystyle \frac{5}{36}\)
  8. Choose the correct answer out of four given options in each of the Exercises $\displaystyle 18$ to $\displaystyle 29$ (M.C.Q.).

    Exercise 18

    In a non-leap year, the probability of having 53\displaystyle 53 tuesdays or 53\displaystyle 53 wednesdays is
    (A)
    17\displaystyle \frac{1}{7}
    (B)
    27\displaystyle \frac{2}{7}
    (C)
    37\displaystyle \frac{3}{7}
    (D)
    none of these

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    (B) \(\displaystyle \frac{2}{7}\)A non-leap year has one day beyond $\displaystyle 52$ full weeks: \[365=52\times 7+1 \] The extra day is equally likely to be any weekday, and $\displaystyle 53$ Tuesdays and $\displaystyle 53$ Wednesdays cannot both occur: \[P=\frac17+\frac17=\frac27 \]

  9. Exercise 19

    Three numbers are chosen from 1\displaystyle 1 to 20. Find the probability that they are not consecutive
    (A)
    186190\displaystyle \frac{186}{190}
    (B)
    187190\displaystyle \frac{187}{190}
    (C)
    188190\displaystyle \frac{188}{190}
    (D)
    1820C3\displaystyle \frac{18}{{ }^{20} \mathrm{C}_3}

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac{187}{190}\)\[n(S)={}^{20}C_3=1140 \] The consecutive triples are \(\displaystyle (1,2,3),(2,3,4),\dots,(18,19,20)\): \[n(E)=18 \] \[P(\text{consecutive})=\frac{18}{1140}=\frac{3}{190} \] \[P(\text{not consecutive})=1-\frac{3}{190}=\frac{187}{190} \]

  10. Exercise 20

    While shuffling a pack of 52\displaystyle 52 playing cards, 2\displaystyle 2 are accidentally dropped. Find the probability that the missing cards to be of different colours
    (A)
    2952\displaystyle \frac{29}{52}
    (B)
    12\displaystyle \frac{1}{2}
    (C)
    2651\displaystyle \frac{26}{51}
    (D)
    2751\displaystyle \frac{27}{51}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{26}{51}\)One red and one black card are dropped: \[n(S)={}^{52}C_2=1326 \] \[n(E)={}^{26}C_1\cdot{}^{26}C_1=676 \] \[P=\frac{676}{1326}=\frac{26}{51} \]