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NCERT Exemplar · Class 10 Mathematics Statistics and Probability

96 questions · 96 still being checked

EXERCISE 13.3 21–30 (part 8 of 11)

  1. Exercise 21

    Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
    (i)
    6\displaystyle 6 (ii) 12\displaystyle 12
    (iii)
    7\displaystyle 7

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{9}\)
    (ii)
    \(\displaystyle \frac{1}{9}\)
    (iii)
    $\displaystyle 0$
    \[n(S)=36 \]
    (i)
    Product \(\displaystyle 6\): \(\displaystyle (1,6),(6,1),(2,3),(3,2)\), \(\displaystyle 4\) outcomes.
    \[P(6)=\frac{4}{36}=\frac{1}{9} \]
    (ii)
    Product \(\displaystyle 12\): \(\displaystyle (2,6),(6,2),(3,4),(4,3)\), \(\displaystyle 4\) outcomes.
    \[P(12)=\frac{4}{36}=\frac{1}{9} \]
    (iii)
    \(\displaystyle 7\) is prime and exceeds \(\displaystyle 6\), so no pair of faces from \(\displaystyle 1\)-\(\displaystyle 6\) has that product.
    \[P(7)=\frac{0}{36}=0 \]
    Answer: (i) \(\displaystyle \frac{1}{9}\) (ii) \(\displaystyle \frac{1}{9}\) (iii) \(\displaystyle 0\)

  2. Exercise 22

    Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

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    NCERT’s answer
    \(\displaystyle \frac{4}{9}\)
    Count favorable products by first die.\[n(S) = 6 \times 6 = 36 \]\[\begin{aligned} d_1=1:\ &1,2,3,4,5,6\quad(6)\\ d_1=2:\ &2,4,6,8\quad(4)\\ d_1=3:\ &3,6\quad(2)\\ d_1=4:\ &4,8\quad(2)\\ d_1=5:\ &5\quad(1)\\ d_1=6:\ &6\quad(1) \end{aligned} \]\[n(E) = 6+4+2+2+1+1 = 16 \]\[P(E) = \frac{16}{36} = \frac{4}{9} \]Answer: \(\displaystyle \dfrac{4}{9} \)

  3. Exercise 23

    Two dice are numbered 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4, 5\displaystyle 5, 6\displaystyle 6 and 1\displaystyle 1, 1\displaystyle 1, 2\displaystyle 2, 2\displaystyle 2, 3\displaystyle 3, 3\displaystyle 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2\displaystyle 2 to 9\displaystyle 9 separately.

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    NCERT’s answer
    \(\displaystyle \mathrm{P}(2)=\frac{1}{18}, \mathrm{P}(3)=\frac{1}{9}, \mathrm{P}(4)=\frac{1}{6}, \mathrm{P}(5)=\frac{1}{6}, \mathrm{P}(6)=\frac{1}{6}, \mathrm{P}(7)=\frac{1}{6}, \mathrm{P}(8)=\frac{1}{9} \quad \mathrm{P}(9)=\frac{1}{18}\)
    Second die's value \(\displaystyle v\) has two faces each for \(\displaystyle v=1,2,3\); first die shows \(\displaystyle a=1,\dots,6\). Sum \(\displaystyle s=a+v\); count ordered pairs, doubled for the repeated faces.\[n(S) = 36 \]\[\begin{aligned} P(2)&=\frac{2}{36}=\frac{1}{18}, & P(3)&=\frac{4}{36}=\frac{1}{9}\\ P(4)&=\frac{6}{36}=\frac{1}{6}, & P(5)&=\frac{6}{36}=\frac{1}{6}\\ P(6)&=\frac{6}{36}=\frac{1}{6}, & P(7)&=\frac{6}{36}=\frac{1}{6}\\ P(8)&=\frac{4}{36}=\frac{1}{9}, & P(9)&=\frac{2}{36}=\frac{1}{18} \end{aligned} \]Answer: \(\displaystyle P(2)=\tfrac{1}{18}\), \(\displaystyle P(3)=\tfrac19\), \(\displaystyle P(4)=P(5)=P(6)=P(7)=\tfrac16\), \(\displaystyle P(8)=\tfrac19\), \(\displaystyle P(9)=\tfrac{1}{18}\)

  4. Exercise 24

    A coin is tossed two times. Find the probability of getting at most one head.

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    NCERT’s answer
    \(\displaystyle \frac{3}{4}\)
    At most one head excludes only \(\displaystyle HH\).\[S = \{HH, HT, TH, TT\}, \quad n(S) = 4 \]\[E = \{HT, TH, TT\}, \quad n(E) = 3 \]\[P(E) = \frac{3}{4} \]Answer: \(\displaystyle \dfrac{3}{4} \)

  5. Exercise 25

    A coin is tossed 3\displaystyle 3 times. List the possible outcomes. Find the probability of getting
    (i)
    all heads
    (ii)
    at least 2\displaystyle 2 heads

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{8}\)
    (ii)
    \(\displaystyle \frac{1}{2}\)
    \[S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}, \quad n(S) = 8 \]
    (i)
    \[E_1 = \{HHH\}, \quad P(E_1) = \frac{1}{8} \]
    (ii)
    \[E_2 = \{HHH, HHT, HTH, THH\}, \quad P(E_2) = \frac{4}{8} = \frac{1}{2} \]
    Answer: (i) \(\displaystyle \dfrac{1}{8} \) (ii) \(\displaystyle \dfrac{1}{2} \)

  6. Exercise 26

    Two dice are thrown at the same time. Determine the probabiity that the difference of the numbers on the two dice is 2.

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    NCERT’s answer
    \(\displaystyle \frac{2}{9}\)
    Pairs with \(\displaystyle |d_1-d_2|=2\).\[n(S) = 36 \]\[E = \{(1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4)\}, \quad n(E) = 8 \]\[P(E) = \frac{8}{36} = \frac{2}{9} \]Answer: \(\displaystyle \dfrac{2}{9} \)

  7. Exercise 27

    A bag contains 10\displaystyle 10 red, 5\displaystyle 5 blue and 7\displaystyle 7 green balls. A ball is drawn at random. Find the probability of this ball being a
    (i)
    red ball
    (ii)
    green ball
    (iii)
    not a blue ball

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{5}{11}\)
    (ii)
    \(\displaystyle \frac{7}{22}\)
    (iii)
    \(\displaystyle \frac{17}{22}\)
    \[n(S) = 10+5+7 = 22 \]
    (i)
    \[P(\text{red}) = \frac{10}{22} = \frac{5}{11} \]
    (ii)
    \[P(\text{green}) = \frac{7}{22} \]
    (iii)
    \[P(\text{not blue}) = 1-\frac{5}{22} = \frac{17}{22} \]
    Answer: (i) \(\displaystyle \dfrac{5}{11} \) (ii) \(\displaystyle \dfrac{7}{22} \) (iii) \(\displaystyle \dfrac{17}{22} \)

  8. Exercise 28

    The king, queen and jack of clubs are removed from a deck of 52\displaystyle 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
    (i)
    a heart
    (ii)
    a king

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{13}{49}\)
    (ii)
    \(\displaystyle \frac{3}{49}\)
    Removing clubs' K, Q, J leaves hearts untouched.
    \[n(S) = 52-3 = 49 \]
    (i)
    \[P(\text{heart}) = \frac{13}{49} \]
    (ii)
    \[P(\text{king}) = \frac{4-1}{49} = \frac{3}{49} \]
    Answer: (i) \(\displaystyle \dfrac{13}{49} \) (ii) \(\displaystyle \dfrac{3}{49} \)

  9. Exercise 29

    Refer to Q.28. What is the probability that the card is
    (i)
    a club
    (ii)
    10\displaystyle 10 of hearts

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{10}{49}\)
    (ii)
    \(\displaystyle \frac{1}{49}\)
    \[n(S) = 52 - 3 = 49 \]
    (i)
    Clubs remaining after removing K, Q, J of clubs:
    \[n(\text{club}) = 13 - 3 = 10 \]
    \[P(\text{club}) = \frac{10}{49} \]
    (ii)
    The $\displaystyle 10$ of hearts was not removed:
    \[P(10\text{ of hearts}) = \frac{1}{49} \]
    Answer: \(\displaystyle P(\text{club}) = \dfrac{10}{49} \), \(\displaystyle P(10\text{ of hearts}) = \dfrac{1}{49} \)

  10. Exercise 30

    All the jacks, queens and kings are removed from a deck of 52\displaystyle 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1\displaystyle 1 similar value for other cards, find the probability that the card has a value
    (i)
    7\displaystyle 7 (ii) greater than 7\displaystyle 7
    (iii)
    less than 7\displaystyle 7

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    NCERT’s answer
    (i)
    \(\displaystyle \frac{1}{10}\)
    (ii)
    \(\displaystyle \frac{3}{10}\)
    (iii)
    \(\displaystyle \frac{3}{5}\)
    \[n(S) = 52 - 12 = 40 \]
    Each value $\displaystyle 1$ to $\displaystyle 10$ occurs in $\displaystyle 4$ suits.
    (i)
    \[n(7) = 4 \]
    \[P(7) = \frac{4}{40} = \frac{1}{10} \]
    (ii)
    values $\displaystyle 8$, $\displaystyle 9$, $\displaystyle 10$:
    \[n(>7) = 3 \times 4 = 12 \]
    \[P(>7) = \frac{12}{40} = \frac{3}{10} \]
    (iii)
    values $\displaystyle 1$ to $\displaystyle 6$:
    \[n(<7) = 6 \times 4 = 24 \]
    \[P(<7) = \frac{24}{40} = \frac{3}{5} \]
    Answer: \(\displaystyle P(7) = \dfrac{1}{10} \), \(\displaystyle P(>7) = \dfrac{3}{10} \), \(\displaystyle P(<7) = \dfrac{3}{5} \)