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Probability Test 6

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Probability Test 6
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  • Question 1
    1 / -0
    Probability of getting a prime (or) composite is ________.
    Solution
    Since if prime occur composite can never occur and vice versa.
    So both events occurring of prime number and composite number are mutually exclusive.
    Option $$A$$ is correct.
  • Question 2
    1 / -0
    $$A, B, C$$ are three mutually independent with probabilities $$0.3, 0.2$$ and $$0.4$$ respectively.
    What is $$P(A \cap B\cap C)$$?
    Solution
    Given: $$A, B, C$$ are mutually independent event so this means
    $$P(A\cap B\cap C)=P(A)P(B)P(C)=0.3\times 0.2\times 0.4=0.024$$
  • Question 3
    1 / -0
    Two events A and B will be independent if
    Solution
    Two events A & B will be independent if
    $$P\left( A\cap { B }  \right) =P\left( A \right) \cap { P\left( B \right)  } $$
    If A & B are independent, then $${ A }^{ ' }\& \ { B }^{ ' }$$ are also independent
    $$\therefore P\left( { A }^{ ' }\cap { { B }^{ ' } }  \right) =P\left( { A }^{ ' } \right) P\left( { B }^{ ' } \right) $$
    $$=\left( 1-P(A) \right) \left( 1-P(B) \right) $$
  • Question 4
    1 / -0
    "The occurrence of one event excludes the occurrence of another event". In a random experiment of probability theory, it is called
    Solution
    "The occurrence of one event excludes the occurrence of the other event."
    In a random experiment of probability theory, this means that the occurence of one event does not affect the occurrence of another. Hence, they are called mutually exclusive events.
    Therefore, $$C$$ is the correct option.
  • Question 5
    1 / -0
    If three events $$A$$, $$B$$, $$C$$ are mutually exclusive, then which one of the following is correct?
    Solution
    Three events $$A.B.C$$  are mutually exclusive if they are disjoint or cannot be true at the same time. 
    Thus, the events are mutually exclusive if $$A\cap B\cap C=\phi$$
    Thus, $$P(A\cap B\cap C)=0$$
    Hence, C is correct.
  • Question 6
    1 / -0
    Which one of the following is correct?
    Solution
    An elementary event is an event which contains only a single element in the sample space. So, it will have only $$1$$ sample point.
    Hence, option B is true
  • Question 7
    1 / -0
    Consider the following statements:
    1. If $$A$$ and $$B$$ are exhaustive events, then their union is the sample space.
    2. If $$A$$ and $$B$$ are exhaustive events, then their intersection must be an empty event.
    Which of the above statements is/are correct?
    Solution
    Exhaustive events are those events, whose union covers the whole sample space. The probability of occurring at least one of them is $$1$$. So, their intersection may or may not be empty.
    Hence, A is correct.
  • Question 8
    1 / -0
    Identify and write the like terms in each of the following groups.
    (i) $$ a^2, b^2, -2a^2 , c^2 , 4a$$ 
    Solution
    In $$a^{2},b^{2},-2a^{2},c^{2},4a.$$
     $$a^{2}$$ and  $$-2a^{2}$$ are like terms because  $$-2a^{2}$$ is a factor of $$a^{2}$$ 
    $$B$$ is correct.
  • Question 9
    1 / -0
    Box I contains $$2$$ white and $$3$$ red balls and box II contains $$4$$ white and $$5$$ red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box II, is?
    Solution
    Probability that the ball drawn is red and from ! =$$P(R/A)$$
    $$P(R/A) = \cfrac{P(A/R)\times P(A)}{P(B/R)\times P(B) + P(A/R) \times P(A)}$$
    $$P(R/A) = \cfrac{3}{5}, P(R/B) = \cfrac{5}{9}$$
    $$P(A) = \cfrac{1}{2}, P(B) = \cfrac{1}{2}$$
    $$P(R/A) = \cfrac{\cfrac{3}{5}\times \cfrac{1}{2}}{\cfrac{5}{9}\times \cfrac{1}{2} + \cfrac{3}{5} \times \cfrac{1}{2}} = \cfrac{54}{104}$$
  • Question 10
    1 / -0
    If $$A = \left\{ {1,2,3} \right\},\,B = \left\{ {2,4,5,6} \right\}$$ and $$S = \left\{ {1,2,3,4,5,6} \right\}$$, then $$A$$ and $$B$$ is called as __________
    Solution
    Since all the elements of A and B and in S, Hence A and B is called Exhaustive event
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