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Probability Test 13

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Probability Test 13
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  • Question 1
    1 / -0
    We draw two cards from a deck of shuffled cards without replacement. Find the probability of getting the second card a queen.
    Solution
    There are two cases here:
    Case 1: First card chosen is a queen
    $$\dfrac 4{52}\times \dfrac 3{51}=\dfrac {1}{221}$$
    Case 2: First card chosen is not a queen.
    $$\dfrac {48}{52}\times \dfrac 4{51}=\dfrac {16}{221}$$
    Adding both the cases, we get the probability of getting the second card a queen. $$\dfrac {17}{221} =\dfrac  4{52} = \dfrac 1{13}$$
  • Question 2
    1 / -0
    Tossing a coin is an example of .........
    Solution
    A coin is tossed.
    The possible outcomes are "$$\text{head,tail}$$"
    Thus $$ S=\{H,T\}$$
    We get a finite sample space.
    Thus tossing a coin is an example of finite sample space.
  • Question 3
    1 / -0
    For a random experiment, all possible outcomes are called
    Solution
    We know that,
    An outcome is a result of a random experiment
    The set of all possible outcomes is called the sample space.
    The subset of the sample space is called event space.
    Thus, for a random experiment, all possible outcomes are called sample space and event space.
    Thus option $$(D)$$ is correct.
  • Question 4
    1 / -0
    A man is know to speak the truth $$3$$ out if $$4$$ times. He throws a die and reports that it is a six. The probability that it is actually a six is:
  • Question 5
    1 / -0
    Sample space for experiment in which a dice is rolled is
    Solution
    A dice is rolled.
    Thus $$S=\{1,2,3,4,5,6\}$$
    $$\Rightarrow n(S)=6$$.
    That is, sample space for experiment in which a dice is rolled is $$6$$.
    Since $$6$$ is not listed in the given options we choose D.
  • Question 6
    1 / -0
    Let A, B, C be three events such that A and B are independent and $$P(C) = 0$$, then events A, B, C are
    Solution
    Given three events $$A,B,C.$$
    A and B are independent $$\xi$$ $$P(C)=0.$$
    A term 'independent event' means probability of events. They have a common outcomes and each event is non-empty.
    As A and B are independent
    i.e, $$P(A)\neq 0$$ and $$P(B)\neq 0.$$
    For $$P(C)$$ to be independent,
    $$\Rightarrow P(A\cap B)=P(A).P(B)$$            $$P(A\cap B\cap C)=P(A).P(B).P(C)$$
    $$\Rightarrow P(A\cap C)=P(A).P(C)$$
    $$\Rightarrow P(B\cap C)=P(B).P(C)$$
    If all the given events are true, then A,B,C are independent.
    Hence, the answer is independent.
  • Question 7
    1 / -0
    Sample space for experiment in which two coins are tossed is
    Solution
    Two coins are tossed.
    Number of outcomes in sample space when two coins are tossed is given by $$2^2=4$$.
    Hence, sample space for experiment in which two coins are tossed is $$4$$.
  • Question 8
    1 / -0
    Choosing a birthdate is an example of .........
    Solution
    Let us see the sample space for choosing a birthdate.
    A person can be born in any date of a month and a month has maximum of $$30$$ or $$31$$ days.
    Therefore, $$S=\{1,2,3,4,5,6,7,8,9,10,11,12,13,...,30,31\}$$ which is a finite set.
    Thus choosing a birthdate is an example of finite sample space.
  • Question 9
    1 / -0
    One bag contains 3 white balls, 7 red balls and 15 black balls. Another bag contains 10 white balls, 6 red balls and 9 black balls. One ball is taken from each bag. What is the probability that both the balls will be of the same colour?
    Solution
    $$\textbf{Step-1: Find the number of balls in each bag.}$$

                     $$\text{Bag}$$ $$I$$ $$\text{ contains 3 white, 7 red and 15 black balls.}$$

                     $$\text{Bag}$$ $$II$$ $$\text{contains 10white, 6 red and 9 black balls.}$$

                     $$\therefore$$ $$\text{Each bag contains total of}$$ $$ 25$$ $$\text{balls.}$$

    $$\textbf{Step-2: Find the number of cases for the selection of a ball.}$$

                     $$\text{There are three cases for selection of a particular ball:-}$$

                     $$\text{Case 1:}$$ $$\text{Probability of getting one white ball each from Bag}$$ $$I$$ $$\text{and Bag}$$ $$II=\dfrac{3}{25}\times\dfrac{10}{25}$$

                     $$\text{Case 2:}$$ $$\text{Probability of getting one red ball each from Bag}$$ $$I$$ $$\text{and Bag}$$ $$II=\dfrac{7}{25}\times\dfrac{6}{25}$$
     
                     $$\text{Case 3:}$$ $$\text{Probability of getting one black ball each from Bag}$$ $$I$$ $$\text{and Bag}$$ $$II=\dfrac{15}{25}\times\dfrac{9}{25}$$

    $$\textbf{Step-3: Add the above cases to get the required probability.}$$

                    $$\therefore$$ $$\text{Total probability that both the ball will be of same colour}$$ $$=\dfrac{30}                                    {625}+\dfrac{42}{625}+\dfrac{135}{25}=\dfrac{207}{625}.$$

    $$\textbf{Hence, the answer is}$$ $$\boldsymbol{\dfrac{207}{625}}.$$
  • Question 10
    1 / -0
    An experiment can result in only $$3$$ mutually exclusive events $$A, B$$ and $$C$$. If $$P(A) = 2P(B) = 3P(C)$$, then $$P(A) =$$
    Solution
    Given that $$P(A)=2P(B)=3P(C)$$
    Let $$P(C)=x$$ , we get $$P(B)=\cfrac{3}{2}x$$ and $$P(A)=3x$$
    Given that there are only these three events.
    So we have $$P(A)+P(B)+P(C)=1$$
    $$\Rightarrow 3x+\cfrac{3x}{2}+x=1$$
    $$\Rightarrow x=\cfrac{2}{11}$$
    Therefore the value of $$P(A)=3x=\cfrac{6}{11}$$
    So the correct option is $$A$$.
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