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NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.3 31–42 (part 9 of 11)

  1. Exercise 31

    An integer is chosen between 0\displaystyle 0 and 100. What is the probability that it is
    (i)
    divisible by 7\displaystyle 7?
    (ii)
    not divisible by 7\displaystyle 7?

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{14}{99}\)
    (ii)
    \(\displaystyle \frac{85}{99}\)
    Integers strictly between $\displaystyle 0$ and $\displaystyle 100$:
    \[n(S) = 99 \]
    Multiples of $\displaystyle 7$ up to $\displaystyle 99$:
    \[n(\text{div by }7) = \left\lfloor \frac{99}{7} \right\rfloor = 14 \]
    (i)
    \[P(\text{div by }7) = \frac{14}{99} \]
    (ii)
    \[P(\text{not div by }7) = 1 - \frac{14}{99} = \frac{85}{99} \]
    Answer: \(\displaystyle \dfrac{14}{99} \), \(\displaystyle \dfrac{85}{99} \)

  2. Exercise 32

    Cards with numbers 2\displaystyle 2 to 101\displaystyle 101 are placed in a box. A card is selected at random. Find the probability that the card has
    (i)
    an even number
    (ii)
    a square number

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{2}\)
    (ii)
    \(\displaystyle \frac{9}{100}\)
    \[n(S) = 101 - 2 + 1 = 100 \]
    (i)
    even numbers $\displaystyle 2$ to $\displaystyle 100$:
    \[n(\text{even}) = 50 \]
    \[P(\text{even}) = \frac{50}{100} = \frac{1}{2} \]
    (ii)
    squares in range: \(\displaystyle 4,9,16,25,36,49,64,81,100\)
    \[n(\text{square}) = 9 \]
    \[P(\text{square}) = \frac{9}{100} \]
    Answer: \(\displaystyle P(\text{even}) = \dfrac{1}{2} \), \(\displaystyle P(\text{square}) = \dfrac{9}{100} \)

  3. Exercise 33

    A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.

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    NCERT’s answer
    \(\displaystyle \frac{21}{26}\)
    \[n(S) = 26 \] Vowels: \(\displaystyle a, e, i, o, u\) \[n(\text{consonant}) = 26 - 5 = 21 \] \[P(\text{consonant}) = \frac{21}{26} \]Answer: \(\displaystyle \dfrac{21}{26} \)

  4. Exercise 34

    There are 1000\displaystyle 1000 sealed envelopes in a box, 10\displaystyle 10 of them contain a cash prize of Rs 100\displaystyle 100 each, 100\displaystyle 100 of them contain a cash prize of Rs 50\displaystyle 50 each and 200\displaystyle 200 of them contain a cash prize of Rs 10\displaystyle 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?

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    NCERT’s answer
    0.$\displaystyle 69$
    \[n(S) = 1000 \] \[n(\text{no prize}) = 1000 - (10 + 100 + 200) = 690 \] \[P(\text{no prize}) = \frac{690}{1000} = \frac{69}{100} \]Answer: \(\displaystyle \dfrac{69}{100} \)

  5. Exercise 35

    Box A contains 25\displaystyle 25 slips of which 19\displaystyle 19 are marked Re 1\displaystyle 1 and other are marked Rs 5\displaystyle 5 each. Box B contains 50\displaystyle 50 slips of which 45\displaystyle 45 are marked Re 1\displaystyle 1 each and others are marked Rs 13\displaystyle 13 each. Slips of both boxes are poured into a third box and resuffled. A slip is drawn at random. What is the probability that it is marked other than Re 1\displaystyle 1?

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    NCERT’s answer
    \(\displaystyle \frac{11}{75}\)
    \[n(S) = 25 + 50 = 75 \] \[n(\text{Re }1) = 19 + 45 = 64 \] \[n(\text{other}) = 75 - 64 = 11 \] \[P(\text{other than Re }1) = \frac{11}{75} \]Answer: \(\displaystyle \dfrac{11}{75} \)

  6. Exercise 36

    A carton of 24\displaystyle 24 bulbs contain 6\displaystyle 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

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    NCERT’s answer
    \(\displaystyle \mathrm{P}(\) not defective \(\displaystyle )=\frac{3}{4}, \mathrm{P}(2 \mathrm{nd}\) bulb defective \(\displaystyle )=\frac{5}{23}\)
    Bulbs not defective \(\displaystyle = 24-6=18\). \[P(\text{not defective}) = \frac{18}{24} = \frac{3}{4} \] One defective bulb removed, not replaced: \(\displaystyle 23\) bulbs remain, \(\displaystyle 5\) defective. \[P(\text{second bulb defective}) = \frac{5}{23} \] Answer: \(\displaystyle P(\text{not defective})=\frac{3}{4}\); \(\displaystyle P(\text{second defective})=\frac{5}{23}\)

  7. Exercise 37

    A child's game has 8\displaystyle 8 triangles of which 3\displaystyle 3 are blue and rest are red, and 10\displaystyle 10 squares of which 6\displaystyle 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a
    (i)
    triangle
    (ii)
    square
    (iii)
    square of blue colour
    (iv)
    triangle of red colour

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{4}{9}\)
    (ii)
    \(\displaystyle \frac{5}{9}\)
    (iii)
    \(\displaystyle \frac{1}{3}\)
    (iv)
    \(\displaystyle \frac{5}{18}\)
    \[\text{Total pieces} = 8+10 = 18 \]
    (i)
    \[P(\text{triangle}) = \frac{8}{18} = \frac{4}{9} \]
    (ii)
    \[P(\text{square}) = \frac{10}{18} = \frac{5}{9} \]
    (iii)
    \[P(\text{blue square}) = \frac{6}{18} = \frac{1}{3} \]
    (iv)
    \[P(\text{red triangle}) = \frac{8-3}{18} = \frac{5}{18} \]
    Answer: (i) \(\displaystyle \frac{4}{9}\) (ii) \(\displaystyle \frac{5}{9}\) (iii) \(\displaystyle \frac{1}{3}\) (iv) \(\displaystyle \frac{5}{18}\)

  8. Exercise 38

    In a game, the entry fee is Rs 5. The game consists of a tossing a coin 3\displaystyle 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3\displaystyle 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she
    (i)
    loses the entry fee.
    (ii)
    gets double entry fee.
    (iii)
    just gets her entry fee.

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{8}\)
    (ii)
    \(\displaystyle \frac{1}{8}\)
    (iii)
    \(\displaystyle \frac{3}{4}\)
    \[S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}, \quad n(S)=8 \]
    (i)
    Loses the fee only on \(\displaystyle 0\) heads.
    \[P(\text{loses fee}) = \frac{1}{8} \]
    (ii)
    Double fee only on \(\displaystyle 3\) heads.
    \[P(\text{double fee}) = \frac{1}{8} \]
    (iii)
    Fee returned on \(\displaystyle 1\) or \(\displaystyle 2\) heads, \(\displaystyle 6\) outcomes.
    \[P(\text{fee returned}) = \frac{6}{8} = \frac{3}{4} \]
    Answer: (i) \(\displaystyle \frac{1}{8}\) (ii) \(\displaystyle \frac{1}{8}\) (iii) \(\displaystyle \frac{3}{4}\)

  9. Exercise 39

    A die has its six faces marked 0\displaystyle 0, 1\displaystyle 1, 1\displaystyle 1, 1\displaystyle 1, 6\displaystyle 6, 6. Two such dice are thrown together and the total score is recorded.
    (i)
    How many different scores are possible?
    (ii)
    What is the probability of getting a total of 7\displaystyle 7?

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Each die: faces \(\displaystyle 0,1,1,1,6,6\); \(\displaystyle 36\) equally likely pairs.
    \[\text{Sums} = \{0+0,\;0+1,\;0+6,\;1+1,\;1+6,\;6+6\} = \{0,1,2,6,7,12\} \]
    (i)
    \[\text{Number of different scores} = 6 \]
    (ii)
    Sum \(\displaystyle 7\) comes only from a \(\displaystyle 1\) with a \(\displaystyle 6\).
    \[P(1,6)=\frac{3}{6}\times\frac{2}{6}=\frac{6}{36}, \quad P(6,1)=\frac{2}{6}\times\frac{3}{6}=\frac{6}{36} \]
    \[P(\text{sum}=7) = \frac{6}{36}+\frac{6}{36} = \frac{12}{36} = \frac{1}{3} \]
    Answer: (i) \(\displaystyle 6\) (ii) \(\displaystyle \frac{1}{3}\)

  10. Exercise 40

    A lot consists of 48\displaystyle 48 mobile phones of which 42\displaystyle 42 are good, 3\displaystyle 3 have only minor defects and 3\displaystyle 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
    (i)
    acceptable to Varnika?
    (ii)
    acceptable to the trader?

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{7}{8}\)
    (ii)
    \(\displaystyle \frac{15}{16}\)
    (i)
    Varnika buys only a good phone.
    \[P(\text{acceptable to Varnika}) = \frac{42}{48} = \frac{7}{8} \]
    (ii)
    Trader refuses only a major defect, so good \(\displaystyle +\) minor defect qualify.
    \[P(\text{acceptable to trader}) = \frac{42+3}{48} = \frac{45}{48} = \frac{15}{16} \]
    Answer: (i) \(\displaystyle \frac{7}{8}\) (ii) \(\displaystyle \frac{15}{16}\)

  11. Exercise 41

    A bag contains 24\displaystyle 24 balls of which x\displaystyle x are red, 2x\displaystyle 2 x are white and 3x\displaystyle 3 x are blue. A ball is selected at random. What is the probability that it is
    (i)
    not red?
    (ii)
    white?

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{5}{6}\)
    (ii)
    \(\displaystyle \frac{1}{3}\)
    \[x+2x+3x=24 \implies 6x=24 \implies x=4 \]
    Red \(\displaystyle =4\), white \(\displaystyle =8\), blue \(\displaystyle =12\).
    (i)
    \[P(\text{not red}) = \frac{2x+3x}{24} = \frac{20}{24} = \frac{5}{6} \]
    (ii)
    \[P(\text{white}) = \frac{2x}{24} = \frac{8}{24} = \frac{1}{3} \]
    Answer: (i) \(\displaystyle \frac{5}{6}\) (ii) \(\displaystyle \frac{1}{3}\)

  12. Exercise 42

    At a fete, cards bearing numbers 1\displaystyle 1 to 1000\displaystyle 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500\displaystyle 500, the player wins a prize. What is the probability that
    (i)
    the first player wins a prize?
    (ii)
    the second player wins a prize, if the first has won?

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    NCERT’s answer
    (i)
    0.$\displaystyle 009$
    (ii)
    \(\displaystyle \frac{8}{999}\)
    [Hint : (ii) After first player has won the prize the number of perfect squares greater than $\displaystyle 500$ will be reduced by $\displaystyle 1$]
    \[22^2=484<500, \quad 23^2=529>500 \]
    \[31^2=961<1000, \quad 32^2=1024>1000 \]
    \[\text{Favourable} = \{23^2,24^2,\dots,31^2\} \implies 9 \text{ cards} \]
    (i)
    \[P(\text{first player wins}) = \frac{9}{1000} \]
    (ii)
    Winning card not replaced.
    \[P(\text{second player wins}) = \frac{8}{999} \]
    Answer: (i) \(\displaystyle \frac{9}{1000}=0.009\) (ii) \(\displaystyle \frac{8}{999}\)