SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Probability

38 questions · 38 still being checked

Miscellaneous Exercise 1–10 (part 4 of 4)

  1. Exercise 1

    A box contains 10\displaystyle 10 red marbles, 20\displaystyle 20 blue marbles and 30\displaystyle 30 green marbles. 5\displaystyle 5 marbles are drawn from the box, what is the probability that
    (i)
    all will be blue?
    (ii)
    atleast one will be green?

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    NCERT’s answer
    (i)
    $\displaystyle \frac{{ }^{20} \mathrm{C}_{5}}{{ }^{60} \mathrm{C}_{5}}$ (ii) $\displaystyle 1-\frac{{ }^{30} \mathrm{C}_{5}}{{ }^{60} \mathrm{C}_{5}}$
    Combinations, equally likely selections. The box holds \(\displaystyle 10+20+30=60\) marbles, and $\displaystyle 5$ are drawn together, so the total number of equally likely draws is \(\displaystyle {}^{60}C_5\).(i) All five blue. The $\displaystyle 5$ marbles must come from the $\displaystyle 20$ blue ones, which can be done in \(\displaystyle {}^{20}C_5\) ways: \[P(\text{all blue})=\frac{{}^{20}C_5}{{}^{60}C_5}=\frac{15504}{5461512}=\frac{34}{11977}\approx 0.0028 .\](ii) At least one green. Counting "at least one" directly means four separate cases, so use the complement: the opposite of "at least one green" is "no green", i.e. all $\displaystyle 5$ drawn from the \(\displaystyle 60-30=30\) non-green marbles. \[P(\text{no green})=\frac{{}^{30}C_5}{{}^{60}C_5}=\frac{142506}{5461512}=\frac{117}{4484},\] \[P(\text{at least one green})=1-\frac{117}{4484}=\frac{4367}{4484}\approx 0.974 .\]Answers: (i) \(\displaystyle \dfrac{{}^{20}C_5}{{}^{60}C_5}=\dfrac{34}{11977}\) (ii) \(\displaystyle 1-\dfrac{{}^{30}C_5}{{}^{60}C_5}=\dfrac{4367}{4484}\).

  2. Exercise 2

    4\displaystyle 4 cards are drawn from a well - shuffled deck of 52\displaystyle 52 cards. What is the probability of obtaining 3\displaystyle 3 diamonds and one spade?

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    NCERT’s answer
    $\displaystyle \frac{{ }^{13} \mathrm{C}_{3} \cdot{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}$
    Combinations. Four cards are drawn together from $\displaystyle 52$, so the number of equally likely selections is \[{}^{52}C_4=270725 .\]A deck has $\displaystyle 13$ diamonds and $\displaystyle 13$ spades. Choose $\displaystyle 3$ of the diamonds and $\displaystyle 1$ of the spades independently of each other: \[{}^{13}C_3\times{}^{13}C_1=286\times 13=3718 \text{ favourable selections.}\]Hence \[P=\frac{{}^{13}C_3\times{}^{13}C_1}{{}^{52}C_4}=\frac{3718}{270725}=\frac{286}{20825}\approx 0.0137 .\]The probability is \(\displaystyle \dfrac{286}{20825}\).

  3. Exercise 3

    A die has two faces each with number '1\displaystyle 1', three faces each with number '2\displaystyle 2' and one face with number '3\displaystyle 3'. If die is rolled once, determine
    (i)
    P(2\displaystyle 2)
    (ii)
    P(1\displaystyle 1 or 3\displaystyle 3)
    (iii)
    P(not 3\displaystyle 3)

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    NCERT’s answer
    (i)
    $\displaystyle \frac{1}{2}$ (ii) $\displaystyle \frac{1}{2}$ (iii) $\displaystyle \frac{5}{6}$
    Equally likely faces, unequal labels. The die still has $\displaystyle 6$ equally likely faces; what changes is how many faces carry each number. Two faces show $\displaystyle 1$, three show $\displaystyle 2$, one shows $\displaystyle 3$, so the sample space is \[S=\{1,1,2,2,2,3\},\qquad n(S)=6 .\](i) Three faces show $\displaystyle 2$: \[P(2)=\frac{3}{6}=\frac12 .\](ii) The events "$\displaystyle 1$" and "$\displaystyle 3$" are mutually exclusive (one roll cannot show both), so \[P(1\text{ or }3)=P(1)+P(3)=\frac26+\frac16=\frac36=\frac12 .\](iii) By the complement rule, \[P(\text{not }3)=1-P(3)=1-\frac16=\frac56 .\]Answers: (i) \(\displaystyle \dfrac12\) (ii) \(\displaystyle \dfrac12\) (iii) \(\displaystyle \dfrac56\).

  4. Exercise 4

    In a certain lottery 10,000\displaystyle 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy
    (a)
    one ticket
    (b)
    two tickets
    (c)
    10\displaystyle 10 tickets.

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    NCERT’s answer
    (a)
    $\displaystyle \frac{999}{1000}$ (b) $\displaystyle \frac{{ }^{9990} \mathrm{C}_{2}}{{ }^{10000} \mathrm{C}_{2}}$ (c) $\displaystyle \frac{{ }^{9990} \mathrm{C}_{10}}{{ }^{10000} \mathrm{C}_{10}}$
    Complement with combinations. Of the $\displaystyle 10,000$ tickets exactly $\displaystyle 10$ win a prize, so \(\displaystyle 10000-10=9990\) tickets are blanks. Buying \(\displaystyle k\) tickets means choosing \(\displaystyle k\) of the $\displaystyle 10,000$, which can happen in \(\displaystyle {}^{10000}C_k\) equally likely ways; you get no prize exactly when all \(\displaystyle k\) come from the $\displaystyle 9990$ blanks. \[P(\text{no prize with }k\text{ tickets})=\frac{{}^{9990}C_k}{{}^{10000}C_k}.\](a) One ticket. \[P=\frac{{}^{9990}C_1}{{}^{10000}C_1}=\frac{9990}{10000}=\frac{999}{1000}=0.999 .\](b) Two tickets. \[P=\frac{{}^{9990}C_2}{{}^{10000}C_2}=\frac{9990\times 9989}{10000\times 9999}=\frac{1108779}{1111000}\approx 0.998 .\](c) Ten tickets. \[P=\frac{{}^{9990}C_{10}}{{}^{10000}C_{10}}\approx 0.990 .\]Answers: (a) \(\displaystyle \dfrac{9990}{10000}\) (b) \(\displaystyle \dfrac{{}^{9990}C_2}{{}^{10000}C_2}\) (c) \(\displaystyle \dfrac{{}^{9990}C_{10}}{{}^{10000}C_{10}}\).

  5. Exercise 5

    Out of 100\displaystyle 100 students, two sections of 40\displaystyle 40 and 60\displaystyle 60 are formed. If you and your friend are among the 100\displaystyle 100 students, what is the probability that
    (a)
    you both enter the same section?
    (b)
    you both enter the different sections?

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    NCERT’s answer
    (a)
    $\displaystyle \frac{17}{33}$ (b) $\displaystyle \frac{16}{33}$
    Counting the two sections. The $\displaystyle 100$ students are split into a section of $\displaystyle 40$ and a section of 60. Think of the places being filled at random and track just two people — you and your friend.(a) Both in the same section. Two mutually exclusive cases.Both in the section of $\displaystyle 40$: your chance of landing there is \(\displaystyle \dfrac{40}{100}\), and then $\displaystyle 39$ of the remaining $\displaystyle 99$ places in that section are left for your friend: \[\frac{40}{100}\times\frac{39}{99}=\frac{1560}{9900}.\]Both in the section of $\displaystyle 60$: \[\frac{60}{100}\times\frac{59}{99}=\frac{3540}{9900}.\]Adding the two exclusive cases, \[P(\text{same section})=\frac{1560+3540}{9900}=\frac{5100}{9900}=\frac{17}{33}.\](b) Different sections. This is the complement: \[P(\text{different sections})=1-\frac{17}{33}=\frac{16}{33}.\]Answers: (a) \(\displaystyle \dfrac{17}{33}\) (b) \(\displaystyle \dfrac{16}{33}\).

  6. Exercise 6

    Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

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    NCERT’s answer
    $\displaystyle \frac{2}{3}$
    Complement over all arrangements. The three letters can be put into the three addressed envelopes in \[3!=6\] equally likely ways. Label the letters \(\displaystyle L_1,L_2,L_3\) and the matching envelopes \(\displaystyle E_1,E_2,E_3\); an arrangement is a permutation saying which letter goes in which envelope."At least one letter in its proper envelope" is awkward to count directly, so count its complement: arrangements with no letter in the right envelope. Listing the six permutations of \(\displaystyle (L_1,L_2,L_3)\) into \(\displaystyle (E_1,E_2,E_3)\): \[(1,2,3),\ (1,3,2),\ (2,1,3),\ (2,3,1),\ (3,1,2),\ (3,2,1),\] only \(\displaystyle (2,3,1)\) and \(\displaystyle (3,1,2)\) have no letter in its own envelope — $\displaystyle 2$ arrangements. \[P(\text{no letter correct})=\frac26=\frac13 .\]Therefore \[P(\text{at least one correct})=1-\frac13=\frac23 .\]The probability is \(\displaystyle \dfrac{2}{3}\).

  7. Exercise 7

    A and B are two events such that P(A)=0.54,P(B)=0.69\displaystyle \mathrm{P}(\mathrm{A})=0.54, \mathrm{P}(\mathrm{B})=0.69 and P(A∩B)=0.35\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.35. Find
    (i)
    P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})
    (ii)
    P(A′∩B′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)
    (iii)
    P(A∩B′)\displaystyle \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)
    (iv)
    P(B∩A′)\displaystyle \mathrm{P}\left(\mathrm{B} \cap \mathrm{A}^{\prime}\right)

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    NCERT’s answer
    (i)
    0.$\displaystyle 88$ (ii) $\displaystyle 0.12$ (iii) $\displaystyle 0.19$ (iv) $\displaystyle 0.34$
    Addition rule, De Morgan and the "only" pieces. Given \(\displaystyle P(A)=0.54,\ P(B)=0.69,\ P(A\cap B)=0.35\).(i) \[P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.54+0.69-0.35=0.88 .\](ii) By De Morgan's law \(\displaystyle A'\cap B'=(A\cup B)'\), so \[P(A'\cap B')=1-P(A\cup B)=1-0.88=0.12 .\](iii) \(\displaystyle A\) splits into the part inside \(\displaystyle B\) and the part outside it, so \(\displaystyle P(A\cap B')=P(A)-P(A\cap B)\): \[P(A\cap B')=0.54-0.35=0.19 .\](iv) Similarly \(\displaystyle P(B\cap A')=P(B)-P(A\cap B)\): \[P(B\cap A')=0.69-0.35=0.34 .\]Check: the four disjoint pieces \(\displaystyle 0.19+0.35+0.34+0.12=1.00\). A Venn diagram carrying these four numbers — \(\displaystyle A\) only, the overlap, \(\displaystyle B\) only, and outside both — makes parts (ii)–(iv) immediate.Answers: (i) \(\displaystyle 0.88\) (ii) \(\displaystyle 0.12\) (iii) \(\displaystyle 0.19\) (iv) \(\displaystyle 0.34\).

  8. Exercise 8

    From the employees of a company, 5\displaystyle 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:
    S. No.NameSexAge in years
    1.HarishM30\displaystyle 30
    2.RohanM33\displaystyle 33
    3.SheetalF46\displaystyle 46
    4.AlisF28\displaystyle 28
    5.SalimM41\displaystyle 41
    A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35\displaystyle 35 years?

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    NCERT’s answer
    $\displaystyle \frac{4}{5}$
    Addition rule on a $\displaystyle 5$-element sample space. One of the five listed people is chosen at random, so \[S=\{\text{Harish},\ \text{Rohan},\ \text{Sheetal},\ \text{Alis},\ \text{Salim}\},\qquad n(S)=5,\] each outcome equally likely.Let \(\displaystyle M\) be the event "the spokesperson is male" and \(\displaystyle E\) the event "the spokesperson is over $\displaystyle 35$ years".
    Males: Harish ($\displaystyle 30$), Rohan ($\displaystyle 33$), Salim ($\displaystyle 41$) \(\displaystyle \Rightarrow n(M)=3\).
    Over $\displaystyle 35$: Sheetal ($\displaystyle 46$), Salim ($\displaystyle 41$) \(\displaystyle \Rightarrow n(E)=2\).
    Both male and over $\displaystyle 35$: Salim only \(\displaystyle \Rightarrow n(M\cap E)=1\).
    So \(\displaystyle P(M)=\dfrac35,\ P(E)=\dfrac25,\ P(M\cap E)=\dfrac15\), and by the addition rule \[P(M\cup E)=\frac35+\frac25-\frac15=\frac45 .\](Directly: the favourable people are Harish, Rohan, Salim and Sheetal — $\displaystyle 4$ out of 5.)The probability is \(\displaystyle \dfrac{4}{5}=0.8\).

  9. Exercise 9

    If 4\displaystyle 4-digit numbers greater than 5,000\displaystyle 5,000 are randomly formed from the digits 0,1,3,5\displaystyle 0,1,3,5, and 7\displaystyle 7 , what is the probability of forming a number divisible by 5\displaystyle 5 when,
    (i)
    the digits are repeated?
    (ii)
    the repetition of digits is not allowed?

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    NCERT’s answer
    (i)
    $\displaystyle \frac{33}{83}$ (ii) $\displaystyle \frac{3}{8}$
    Classical probability by counting.Two conditions control the count. A $\displaystyle 4$-digit number built from \(\displaystyle 0,1,3,5,7\) is greater than \(\displaystyle 5000\) only if its thousands digit is \(\displaystyle 5\) or \(\displaystyle 7\) — a leading \(\displaystyle 0\) is not allowed, and \(\displaystyle 1\) or \(\displaystyle 3\) gives a number below \(\displaystyle 5000\). And a number is divisible by \(\displaystyle 5\) exactly when its units digit is \(\displaystyle 0\) or \(\displaystyle 5\).(i) Digits repeated.Sample space: thousands digit \(\displaystyle 2\) ways, and each of the other three places \(\displaystyle 5\) ways, \[2\times 5\times 5\times 5 = 250.\] One of these strings is \(\displaystyle 5000\) itself, which is not greater than \(\displaystyle 5000\), so \[n(S) = 250 - 1 = 249.\]Favourable: thousands digit \(\displaystyle 2\) ways, units digit \(\displaystyle 2\) ways (\(\displaystyle 0\) or \(\displaystyle 5\)), the two middle places \(\displaystyle 5\) ways each, \[2\times 5\times 5\times 2 = 100,\] and again \(\displaystyle 5000\) must be dropped (it is divisible by \(\displaystyle 5\), so it was counted here): \[n(E) = 100 - 1 = 99.\]\[P(E)=\frac{99}{249}=\frac{33}{83}.\](ii) Repetition not allowed.Sample space: thousands digit \(\displaystyle 2\) ways, then the remaining three places filled from the \(\displaystyle 4\) unused digits, \[n(S) = 2\times 4\times 3\times 2 = 48\] (no exclusion is needed now — \(\displaystyle 5000\) repeats a digit, so it never arises).Favourable, split on the thousands digit:
    Thousands digit \(\displaystyle 5\): the \(\displaystyle 5\) is used up, so the units digit must be \(\displaystyle 0\); the two middle places take the remaining digits \(\displaystyle 1,3,7\) in \(\displaystyle 3\times 2 = 6\) ways.
    Thousands digit \(\displaystyle 7\): the units digit may be \(\displaystyle 0\) or \(\displaystyle 5\) (\(\displaystyle 2\) ways), and the two middle places take \(\displaystyle 3\times 2 = 6\) ways, giving \(\displaystyle 2\times 6 = 12\).
    \[n(E) = 6 + 12 = 18,\qquad P(E)=\frac{18}{48}=\frac{3}{8}.\]Answer: (i) \(\displaystyle \dfrac{33}{83}\) (ii) \(\displaystyle \dfrac{3}{8}\).

  10. Exercise 10

    The number lock of a suitcase has 4\displaystyle 4 wheels, each labelled with ten digits i.e., from 0\displaystyle 0 to 9\displaystyle 9 . The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

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    NCERT’s answer
    $\displaystyle \frac{1}{5040}$
    Classical probability with equally likely arrangements.Rule: the number of ordered arrangements of \(\displaystyle r\) objects chosen from \(\displaystyle n\) distinct objects is \[{}^{n}P_{r}=\frac{n!}{(n-r)!}.\]A setting of the lock is an ordered choice of \(\displaystyle 4\) digits from \(\displaystyle 0,1,2,\dots,9\) with no digit repeated, so the number of possible settings is \[{}^{10}P_{4}=10\times 9\times 8\times 7 = 5040.\]All \(\displaystyle 5040\) sequences are equally likely for a person guessing, and exactly one of them opens the suitcase, so \(\displaystyle n(E)=1\).\[P(\text{opens}) = \frac{1}{5040}.\]Answer: \(\displaystyle \dfrac{1}{5040}\).