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Probability Test 19

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Probability Test 19
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  • Question 1
    1 / -0
    Given that $$A \subset B$$, then identify the correct statement 
    Solution
    $$A \subset B$$
    $$P (A/B) = \dfrac{P(A \cap B)}{P (B)} = \dfrac{P (A)}{P(B)} \neq P(A)$$ (always)
    $$P(A/B) = \dfrac{P(A)}{P(B)} \ge P(A)$$
  • Question 2
    1 / -0
    The number of committees formed by taking $$5men$$ and $$5women$$ from $$6women$$ and $$7men$$ are
    Solution

  • Question 3
    1 / -0
    Probability of hitting a target independently of $$4$$ persons are $$\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{8}$$. Then the probability that target is hit, is?
    Solution
    $$P(H)=1-P$$(Not Hitting)
    $$=1-\dfrac{1}{2}\cdot \dfrac{2}{3}\cdot \dfrac{3}{4}\cdot \dfrac{7}{8}=\dfrac{25}{32}$$.
  • Question 4
    1 / -0
    The probability of happening of an event A is $$0.5$$ and that of B is $$0.3$$. If A and B are mutually exclusive events, then the probability of neither A nor B is?
    Solution
    $$P(A)=0.5$$; $$P(B)=0.3$$

    $$\therefore P(\bar{A}\cap\bar{B})=1-P(A\cup B)=1-[P(A)+P(B)]$$      $$[\because\ P(A\cap B)=\phi $$ as they are mutually exclusive events $$]$$ 

                                                         $$=1-[0.5+0.3]$$  

                                                         $$=0.2$$  
  • Question 5
    1 / -0
    If $$6P(A)=8P(B)=14P(A\cap B)=1$$, then $$P\left( \cfrac { A' }{ B }  \right) =........$$
    Solution
    we know that
    $$P(A)= \dfrac16$$ $$P(B)=\dfrac18$$  $$P(A\cap B)=\dfrac1{14}$$
    by using conditional probability
    $$P\left( \cfrac { \overline { A }  }{ B }  \right) =\cfrac { P(\overline { A } \cap B) }{ P(B) }$$
    $$ =\cfrac { P(B)-P( { A } \cap B) }{ P(B) }$$
    $$ =1-\cfrac { \cfrac { 1 }{ 14 }  }{ \cfrac { 1 }{ 8 }  }$$
    $$ =\cfrac { 3 }{ 7 } $$
  • Question 6
    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 7
    1 / -0
    From a well shuffled deck of $$52$$ cards, $$4$$ cards are drawn at random. What is the probability that all the drawn cards are of the same colour.
    Solution

  • Question 8
    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 9
    1 / -0
    Let $$S$$ and $$T$$ are two events defined on a sample space with probabilities $$P(S)=0.5,P(T)=0.69,P(S/T)=0.5$$ Event $$S$$ and $$T$$ are 
    Solution
    Given, $$S$$ and $$T$$ are two events defined on a sample space with probabilities 
    $$P(S)=0.5,P(T)=0.69,P(S/T)=0.5$$
    $$P(S/T)=\dfrac{P(S \cap T)}{P(T)}$$
    $$\Rightarrow 0.5=\dfrac{P(S \cap T)}{0.69}$$
    $$\Rightarrow P(S \cap T)=0.5 \times 0.69=P(S)P(T)$$
    Therefore, $$S$$ and $$T$$ are independent.

    $$\therefore P(S \cap T)=P(S)P(T)$$
    $$=0.69\times 0.5=0.345$$
    As intersection is not zero $$S$$ and $$ T$$ are not mutually exclusive.
    Option B is correct.
  • Question 10
    1 / -0
    In a random experiment,it the occurrence of one event prevents the occurrence  of other event,is 
    Solution
    In a random experiment if the occurrence of one event prevents the occurrence of other event it is mutually exclusive.

    Explanation:

    As we know

    The two events both of which can not occur at the same time i.e one event prevent the occurrence of the other i.e their intersection is null are called mutually exclusive events.

    Hence, In a random experiment if the occurrence of one event prevents the occurrence of other event it is mutually exclusive.

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