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Probability Test - 10

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Probability Test - 10
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  • Question 1
    1 / -0

    Probability of selecting a bulb from a bag is 1/3, then, the probability of not selecting a bulb from bag is

    Solution

    Probability of selecting a bulb from a bag,P(E) = 1/3

    Probability of not selecting a bulb from bag,P(not E) = 1 - P(E) = 1 - 1/3 = 2/3

     

  • Question 2
    1 / -0

    A number is selected at random from the first 20 numbers, then the probability that the number is a multiple of 3 is :-

    Solution

    Total possibles outcomes = 20

    Number of favourable outcomes of getting a multiple of 3 = { 3,6,9,12,15,18} = 6

    Therefore, the the probability that the number is a multiple of 3 = 6/20= 3/10

     

  • Question 3
    1 / -0

    The probability that a boy will get married to his girlfriend is 2/7, then the probability that he will not get married to his girlfriend will be:-

    Solution

    The probability that a boy will get married to his girlfriend ,P(E) = 2/7

    The probability that he will not get married to his girlfriend, P(not E) = 1 - P(E) = 1 - 2/7 = 5/7

     

  • Question 4
    1 / -0

    Odds in favour of a target are 2 : 5, then the probability of success is

    Solution

    Odds in favour are 2:5

    This means the event can occur in 2 ways and there are 5 ways the event cannot occur

    This means that there are 7 total outcomes and 2 are favourable

    The probability of success 2 out of 7 = 2/7

     

  • Question 5
    1 / -0

    Probability of getting 53 thursdays in a non leap year is

    Solution

    Number of days in a non-leap year = 365 = 52 weeks and 1 day

    Thus, the probability is to be calculated for the remaining 1 day.

    That one day can be {Sun, Mon, Tues, Wed, Thurs, Fri, Sat}

    Total number of outcomes = 7

    Number of favourable outcomes = 1 (Since only one day is a Thursday)

    Thus, the required probability of getting 53 thursdays in a non leap year = 1/7

     

  • Question 6
    1 / -0

    Probability of getting two heads in a simultaneous throw of 3 coins is

    Solution

    Number of possible outcomes in in a simultaneous throw of 3 coins = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT } = 8

    Number of favourable outcomes of getting two heads {HHT, HTH, THH} = 3

    So,the probability of getting two heads in a simultaneous throw of 3 coins = 3/8

     

  • Question 7
    1 / -0

    Probability of getting a red king from a deck of   well shuffled 52 cards is

    Solution

    Total possible outcomes = 52

    Number of favourable outcomes of getting a red King =2{ King of Diamonds or King of Hearts}

    Probability of getting a red king from a deck of well shuffled 52 cards = 2/52 = 1/26

     

  • Question 8
    1 / -0

    The probability of throwing a sum of 9 with two dice is :

    Solution

    Number of possible outcomes when two dice are thrown = 36

    Number of favourable outcomes of throwing a sum of 9 = {(3,6),(4,5),(5,4),(6,3)} = 4

    The probability of throwing a sum of 9 = 4/36 = 1/9

     

  • Question 9
    1 / -0

    Two coins are tossed simultaneously. The probability of getting no head is :

    Solution

    When two different coins are tossed randomly, the sample space is given by

    S = {HH, HT, TH, TT}

    Therefore, n(S) = 4

    Let E = event of getting no head.

    Then, E = {TT} and, therefore, n(E) = 1. 
    Therefore, P(getting no head) = P(E) = n(E) / n(S) = 1/4

     

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