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ICSE-VIII-Mathematics

23: Probability Class 8 Maths

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  • #
    Exercise 23
  • Qstn #1
    A die is thrown, find the probability of getting:
    Ans : A die has six numbers : 1, 2, 3, 4, 5, 6
    ∴ Number of possible outcomes = 6
  • #1-i
    (a) prime number
    Ans : A prime number
    Number of favourable outcomes = a prime number = 1, 3, 5 which are in numbers
    P(E) = (Numbers of favourable outcome)/(Numbers of all outcome)
    = 3/6 = ½
  • #1-ii
    (a) number greater than 4
    Ans : Numbers of favourable outcome = greater than four i.e two number 5 and 6
    ∴ P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)
    = 2/6 = 1/3
  • #1-iii
    (a) number not greater than 4.
    Ans : Number of favourable outcome = not graeter than 4 or numbers will be 1, 2, 3, 4 which are 4 in numbers
    ∴ P(E) = (Numbers of favourable outcome)/(Number of all outcome)
    = 4/6
    = 2/3
  • Qstn #2
    A coin is tossed. What is the probability of getting:
    Ans : On tossing a coin once,
    Number of possible outcome = 2
  • #2-i
    (a) tail?(ii) ahead?
    Ans : Favourable outcome getting a tail = 1
    ⇒ number of favorable outcome = 2
    ∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
    = ½
    (ii) a head
    Similarly, favourable outcome getting a head = 1
    But number of possible outcome = 2
    ∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
    = ½
  • Qstn #3
    A coin is tossed twice. Find the probability of getting:
  • #3-i
    exactly one head (ii) exactly one tail
    (iii) two tails (iv) two heads
    Ans : Exactly one head
    Possible number of favourable outcomes = 2
    (i.e. TH and HT)
    Total number of possible outcomes = 4
    ∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
    = 2/4
    = ½
    (ii) Exactly one tail
    Possible number of favourable outcomes = 2
    (i.e. TH and HT)
    Total number of possible outcomes = 4
    P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
    = 2/4
    = ½
    (iii) Two tails
    Possible number of favourable outcomes = 1
    (i.e., TT)
    Total number of possible outcome s = 4
    ∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
    = ¼
    (iv) Two heads
    Possible number of favourable outcomes = 1
    (i.e. HH)
    Total number of possible outcomes = 4
    ∴ P(E) = ¼
  • Qstn #4
    A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
    Ans : Total no. of letters in the word ‘PENCIL = 6
    Total Number of Consonant = ‘PNCL’ i.e. 4
    P(E) = (Total No. of consonants)/(Total No. of letters in the word PENCIL)
    = 4/6
    = 2/3
  • Qstn #5
    A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
  • #5-i
    (a) red ball
    Ans : Total number of possible outcomes = 3
    ∴ P(E) = 1/3
  • #5-ii
    not a red ball
    Ans : Not a red ball
    Number of favourable outcomes = Green ball + Black Ball
    = 1 + 1 + 2
    ∴ P(E) = 2/3
  • #5-iii
    (a) white ball.
    Ans : A white ball
    Number of favourable outcome = 0
    ∴ P(E) = 0/3
    = 0
  • Qstn #6
    In a single throw of a die, find the probability of getting a number
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