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NCERT Exemplar · Class 11 Mathematics Probability

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EXERCISE 16.3 1–10 (part 1 of 4)

  1. Exercise 1

    If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

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    NCERT’s answer
    \(\displaystyle \frac{1}{72}\)
    The $\displaystyle 9$ letters are distinct; treat GOR as one unit, giving $\displaystyle 7$ units in all.\[n(S) = 9! \]\[n(E) = 7! \]\[P(E) = \frac{7!}{9!} = \frac{1}{9 \times 8} = \frac{1}{72} \]Answer: \(\displaystyle \dfrac{1}{72}\)

  2. Exercise 2

    Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks? [Hint: First find the probability that the couple has adjacent desks, and then subtract it from 1.]

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    NCERT’s answer
    \(\displaystyle \frac{2}{3}\)
    Adjacent desks: the couple takes one of $\displaystyle 5$ neighbouring pairs, in $\displaystyle 2$ orders; the other four employees fill the rest in \(\displaystyle 4!\) ways.\[n(S) = 6! = 720 \]\[n(\text{adjacent}) = 5 \times 2 \times 4! = 240 \]\[P(\text{adjacent}) = \frac{240}{720} = \frac{1}{3} \]\[P(\text{nonadjacent}) = 1 - \frac{1}{3} = \frac{2}{3} \]Answer: \(\displaystyle \dfrac{2}{3}\)

  3. Exercise 3

    Suppose an integer from 1\displaystyle 1 through 1000\displaystyle 1000 is chosen at random, find the probability that the integer is a multiple of 2\displaystyle 2 or a multiple of 9.

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    NCERT’s answer
    0.$\displaystyle 556$
    Let \(\displaystyle A\) = multiples of $\displaystyle 2$, \(\displaystyle B\) = multiples of 9. Common multiples of $\displaystyle 2$ and $\displaystyle 9$ are multiples of 18.\[n(S) = 1000 \]\[n(A) = \left\lfloor \frac{1000}{2} \right\rfloor = 500, \qquad n(B) = \left\lfloor \frac{1000}{9} \right\rfloor = 111 \]\[n(A \cap B) = \left\lfloor \frac{1000}{18} \right\rfloor = 55 \]\[n(A \cup B) = 500 + 111 - 55 = 556 \]\[P(A \cup B) = \frac{556}{1000} = \frac{139}{250} \]Answer: \(\displaystyle \dfrac{139}{250}\)

  4. Exercise 4

    An experiment consists of rolling a die until a 2\displaystyle 2 appears.
    (i)
    How many elements of the sample space correspond to the event that the 2\displaystyle 2 appears on the kth \displaystyle k^{\text {th }} roll of the die?
    (ii)
    How many elements of the sample space correspond to the event that the 2\displaystyle 2 appears not later than the kth \displaystyle k^{\text {th }} roll of the die?
    [Hint:(a) First (k−1)\displaystyle (k-1) rolls have 5\displaystyle 5 outcomes each and kth \displaystyle k^{\text {th }} rolls should result in 1\displaystyle 1 outcomes. (b) 1+5+52+…+5k−1.\displaystyle 1+5+5^2+\ldots+5^{k-1.}]

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    NCERT’s answer
    (a)
    \(\displaystyle 5^{\mathrm{k}-1}\) elements
    (b)
    \(\displaystyle \frac{5^k-1}{4}\)
    (i)
    The first \(\displaystyle k-1\) rolls avoid $\displaystyle 2$ ($\displaystyle 5$ outcomes each) and the \(\displaystyle k^{\text{th}}\) roll is $\displaystyle 2$ ($\displaystyle 1$ outcome).
    \[n = 5^{k-1} \times 1 = 5^{k-1} \]
    (ii)
    The $\displaystyle 2$ appears on roll \(\displaystyle 1, 2, \ldots, k\); add the cases from (i).
    \[1 + 5 + 5^2 + \cdots + 5^{k-1} = \frac{5^k - 1}{5 - 1} = \frac{5^k - 1}{4} \]
    Answer: (i) \(\displaystyle 5^{k-1}\); (ii) \(\displaystyle \dfrac{5^k - 1}{4}\)

  5. Exercise 5

    A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G)\displaystyle \mathrm{P}(\mathrm{G}), where G is the event that a number greater than 3\displaystyle 3 occurs on a single roll of the die.

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    NCERT’s answer
    \(\displaystyle \frac{4}{9}\)
    Let each even number have probability \(\displaystyle p\); each odd number then has \(\displaystyle 2p\).\[3(2p) + 3p = 1 \]\[9p = 1 \;\Rightarrow\; p = \frac{1}{9} \]\(\displaystyle G = \{4, 5, 6\}\): one odd number ($\displaystyle 5$) and two even numbers $\displaystyle (4, 6)$.\[P(G) = p + 2p + p = 4p = \frac{4}{9} \]Answer: \(\displaystyle \dfrac{4}{9}\)

  6. Exercise 6

    In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

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    NCERT’s answer
    0.$\displaystyle 93$
    Let \(\displaystyle C\) = colour set, \(\displaystyle B\) = black and white set.\[P(C) = 0.87, \quad P(B) = 0.36, \quad P(C \cap B) = 0.30 \]\[P(C \cup B) = P(C) + P(B) - P(C \cap B) \]\[P(C \cup B) = 0.87 + 0.36 - 0.30 = 0.93 \]Answer: \(\displaystyle 0.93\)

  7. Exercise 7

    If A and B are mutually exclusive events, P(A)=0.35\displaystyle \mathrm{P}(\mathrm{A})=0.35 and P(B)=0.45\displaystyle \mathrm{P}(\mathrm{B})=0.45, find
    (a)
    P(A′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime}\right)
    (b)
    P(B′)\displaystyle \mathrm{P}\left(\mathrm{B}^{\prime}\right)
    (c)
    P(A∪B)\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})
    (d)
    P(A∩B)\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B})
    (e)
    P(A∩B′)\displaystyle \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)
    (f)
    P(A′∩B′)\displaystyle \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)

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    NCERT’s answer
    (a)
    0.$\displaystyle 65$
    (b)
    0.$\displaystyle 55$
    (c)
    0.$\displaystyle 8$
    (d)
    $\displaystyle 0$ (e) $\displaystyle 0.35$
    (f)
    0.$\displaystyle 2$
    Mutually exclusive: \(\displaystyle A \cap B = \varnothing\).
    (a)
    \[P(A') = 1 - 0.35 = 0.65 \]
    (b)
    \[P(B') = 1 - 0.45 = 0.55 \]
    (c)
    \[P(A \cup B) = P(A) + P(B) = 0.35 + 0.45 = 0.80 \]
    (d)
    \[P(A \cap B) = 0 \]
    (e)
    Since \(\displaystyle A \subseteq B'\),
    \[P(A \cap B') = P(A) = 0.35 \]
    (f)
    \[P(A' \cap B') = P\big((A \cup B)'\big) = 1 - 0.80 = 0.20 \]
    Answer: (a) $\displaystyle 0.65$ (b) $\displaystyle 0.55$ (c) $\displaystyle 0.80$ (d) $\displaystyle 0$ (e) $\displaystyle 0.35$ (f) $\displaystyle 0.20$

  8. Exercise 8

    A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15\displaystyle 0.15, 0.20\displaystyle 0.20, 0.31\displaystyle 0.31, 0.26\displaystyle 0.26, .08. Find the probabilities that a particular surgery will be rated
    (a)
    complex or very complex;
    (b)
    neither very complex nor very simple;
    (c)
    routine or complex
    (d)
    routine or simple

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    NCERT’s answer
    (a)
    0.$\displaystyle 35$
    (b)
    0.$\displaystyle 77$
    (c)
    0.$\displaystyle 51$
    (d)
    0.$\displaystyle 57$
    The five ratings are mutually exclusive, so probabilities add.
    (a)
    \[P = 0.15 + 0.20 = 0.35 \]
    (b)
    \[P = 1 - (0.15 + 0.08) = 0.77 \]
    (c)
    \[P = 0.31 + 0.20 = 0.51 \]
    (d)
    \[P = 0.31 + 0.26 = 0.57 \]
    Answer: (a) $\displaystyle 0.35$ (b) $\displaystyle 0.77$ (c) $\displaystyle 0.51$ (d) $\displaystyle 0.57$

  9. Exercise 9

    Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that
    (a)
    C will be selected?
    (b)
    A will not be selected?

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    NCERT’s answer
    (a)
    \(\displaystyle \frac{2}{9}\)
    (b)
    \(\displaystyle \frac{5}{9}\)
    Exactly one candidate is selected, so the four probabilities sum to 1. Let \(\displaystyle P(D)=p\).
    \[P(C)=2p,\quad P(B)=P(C)=2p,\quad P(A)=2P(B)=4p \]
    \[p+2p+2p+4p=1 \Rightarrow p=\tfrac19 \]
    (a)
    \[P(C)=2p=\tfrac29 \]
    (b)
    \[P(\bar A)=1-P(A)=1-4p=1-\tfrac49=\tfrac59 \]
    Answer: (a) \(\displaystyle \tfrac29\); (b) \(\displaystyle \tfrac59\).

  10. Exercise 10

    One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes
    S={\displaystyle \mathrm{S}=\{ John promoted, Rita promoted, Aslam promoted, Gurpreet promoted }\displaystyle \}
    You are told that the chances of John's promotion is same as that of Gurpreet, Rita's chances of promotion are twice as likely as Johns. Aslam's chances are four times that of John.
    (a)
    Determine P (John promoted)
    P (Rita promoted)
    P (Aslam promoted)
    P (Gurpreet promoted)
    (b)
    If A={\displaystyle \mathrm{A}=\{ John promoted or Gurpreet promoted }\displaystyle \}, find P(A)\displaystyle \mathrm{P}(\mathrm{A}).

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    NCERT’s answer
    (a)
    \(\displaystyle p(\text{John promoted})=\frac{1}{8}\), \(\displaystyle p(\text{Rita promoted})=\frac{1}{4}\), \(\displaystyle p(\text{Aslam promoted})=\frac{1}{2}\), \(\displaystyle p(\text{Gurpreet promoted})=\frac{1}{8}\)
    (b)
    \(\displaystyle \mathrm{P}(\mathrm{A})=\frac{1}{4}\)
    Let \(\displaystyle P(\text{John})=p\).
    \[P(\text{Gurpreet})=p,\quad P(\text{Rita})=2p,\quad P(\text{Aslam})=4p \]
    \[p+2p+4p+p=1 \Rightarrow p=\tfrac18 \]
    (a)
    \[P(\text{John})=\tfrac18,\quad P(\text{Rita})=\tfrac28=\tfrac14,\quad P(\text{Aslam})=\tfrac48=\tfrac12,\quad P(\text{Gurpreet})=\tfrac18 \]
    (b)
    \[P(A)=P(\text{John})+P(\text{Gurpreet})=\tfrac18+\tfrac18=\tfrac14 \]
    Answer: (a) \(\displaystyle \tfrac18,\ \tfrac14,\ \tfrac12,\ \tfrac18\); (b) \(\displaystyle P(A)=\tfrac14\).