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Probability Test - 43

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Probability Test - 43
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  • Question 1
    1 / -0
    Given two mutually exclusive events $$A$$ and $$B$$ such that $$P(A)=\dfrac 12$$ and $$P(B)=\dfrac 13$$, find $$P(A\ or\ B)$$
    Solution
    Given mutually exclusive events
    $$\Rightarrow \ A\cap B=\phi$$
    $$\therefore \ P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
    $$\Rightarrow \ P(A\cup B)=\dfrac {1}{2}+\dfrac {1}{3}-0$$
    $$\Rightarrow \ P(A\cup B)=\dfrac {3+2}{6}$$
    $$\Rightarrow \ P(A\ or\ B)=P(A\cup B)=\dfrac {5}{6}$$
  • Question 2
    1 / -0
    Choose the most appropriate option.
    $$12$$ defective pens are accidentally mixed with $$132$$ good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
    Solution
    Total number of pens $$=$$ Total defective pens $$+$$ Total good pens
                                         $$=12+132$$
                                         $$=144$$
    Number of good pens $$=132$$
    Probability of getting good pen $$=\dfrac{Number\,of\,good\,pens}{Total\,number\, of\, pens}$$

                                                          $$=\dfrac{132}{144}$$

                                                          $$=\dfrac{11}{12}$$ 
  • Question 3
    1 / -0
    A card is picked at random from a pack of $$52$$ playing cards. given that the picked card is a queen, the probability of this card to be a card of spade is
    Solution
    No. of spade cards $$= 13$$

    No. of queen in spade set $$=1$$

    $$\therefore n$$(spade queen) $$= 13_{c_1}$$

    $$\therefore n$$ (Total no. of cases) $$= 52_{c_1}$$

    $$\therefore P$$ (spade queen) $$= \dfrac{13_{c_1}}{52_{c_1}}$$

                                      $$= \dfrac{13}{52} = \dfrac{1}{4}$$

    option C
  • Question 4
    1 / -0
    A quadratic equation $$ax^2 + bx + c = 0$$, with distinct coefficients is formed. It a, b, c are chosen from the numbers $$2, 3, 5$$ then the probability that the equation has real roots is
    Solution
    Total number of ways of assigning values $$2,3,5$$ to $$a,b,c,=3!=6$$
    Now, for quadratic equation $$ax^2+bx+c=0$$ to have real roots $$b^2-4ac \ge 0$$. This is possible only when $$a=2,b=5,c=3$$ or $$a=3,b=5,c=2$$
    $$\Rightarrow$$ Required probability $$=\dfrac{2}{6}=\dfrac{1}{3}$$
  • Question 5
    1 / -0
    Two dice of different colours are thrown at a time. The probability that the sum is either $$7$$ or $$11$$ is 
    Solution
    event $$A=$$ sum is either $$7$$ or $$11$$

    Probability of sum of $$7$$
    $$=(6,1),(5,2),(4,3),(3,4),(2,5),(1,6)$$ and probability of sum of $$11$$
    $$=(6,5),(5,6)$$
    $$P(A)=\dfrac{6}{36}+\dfrac{2}{36}$$
    $$=\dfrac{8}{36}=\dfrac{2}{9}$$
  • Question 6
    1 / -0
    A coin is tossed $$5$$ times. What is the probability that tail appears an odd number of times?
    Solution

  • Question 7
    1 / -0
    If $$A$$ and$$B$$ are mutually exclusive events such that $$P(A)=0.4,P(B)=x$$ and $$P(A\cup B)=0.5$$, then $$x=$$?
    Solution

  • Question 8
    1 / -0
    In a class, $$60$$% of the students read mathematics, $$25$$% biology and $$15$$% both mathematics and biology. One student is selected at random. What is the probability that he reads mathematics, if it is known that he reads biology?
    Solution

  • Question 9
    1 / -0
    A fair coin is tossed $$6$$ times. What is the probability of getting at least $$3$$ heads?
    Solution

  • Question 10
    1 / -0
    A die is thrown twice and the sum of the numbers appearing is observed to be $$7$$. What is the conditional probability that the number $$2$$ has appeared at least once?
    Solution
    $${\textbf{Step -1: Find required events.}}$$

                      $${\text{Let S be the sample space of a die is thrown twice.}}$$

                      $$\Rightarrow n(S) = 36.$$

                      $${\text{Let X be the event getting the sum is 7 and Y be the event getting that 2 appears at least once.}}$$

                      $$\Rightarrow P(X)= \dfrac{6}{36}$$

                      $$\Rightarrow P(Y)= \dfrac{11}{36}$$

                      $${\text{So,}}$$ $$X \cap Y =$$ $${\text{{(2, 5), (5, 2)}}}$$

    $${\textbf{Step -2: Find required probability.}}$$

                      $$\Rightarrow P(X \cap Y) = \dfrac{2}{36}$$

                      $${\text{Required probability(P(Y|X))}}$$ $$= \dfrac{P(X \cap Y)}{P(X)}= \dfrac{2}{6}= \dfrac{1}{3}.$$

    $${\textbf{Hence, the option B is correct.}}$$
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