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Probability Test - 57

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Probability Test - 57
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  • Question 1
    1 / -0
    A throws a coin $$3$$ times. If he get a head all three times, he is to get a reward of Rs.$$200$$, on the other hand if he does not get $$3$$ heads he is to loose Rs.40. He is expected to win Rs. ...............
    Solution
    A throw of $$3$$ coins, total probability $$=8$$
    $$\Rightarrow$$ Probability of getting head on all three throws $$=\dfrac{1}{8}$$
    $$\Rightarrow$$ Probability of not getting head on all three throws $$=\dfrac{7}{8}$$
    $$\therefore \dfrac{7}{8}\times40=35$$
    $$\Rightarrow 35-(40+5)=Rs.(-10)$$
    Hence, the answer is $$Rs.(-10).$$

  • Question 2
    1 / -0
    If $$\displaystyle P(A) = \frac{2}{3}$$, $$\displaystyle P(B) = \frac{4}{9}$$ and $$\displaystyle P(A \cap B) = \frac{14}{45}$$ then $$\displaystyle P(A' \cap B') $$ is greater than or equal to
    Solution

  • Question 3
    1 / -0
    The percentage of marks obtained by a student in the monthly unit tests are given.
    Unit test$$1$$$$2$$$$3$$$$4$$$$5$$
    Percentage of marks obtained$$60$$$$75$$$$55$$$$72$$$$85$$
    Find the probability that the student gets Marks between $$80\%$$ and $$90\%$$.
    Solution

  • Question 4
    1 / -0
    Let $${E}^{e}$$ denotes the complement of an event $$E$$. If $$E,F,G$$ are pairwise independent events with $$P(G)> 0$$ and $$P(E\cap F\cap G)=0$$. Then, $$P({ E }^{ e }\cap { F }^{ e }|G)$$ equals
    Solution
    Given: $$P(E^e)=1-P(E)$$ and E, F, G are pairwise independent events, P(G)>0, $$P(E\cap F\cap G)=0$$
    To find: $$P(E^e\cap F^e|G)=?$$
    Sol:
    $$P(E^e\cap F^e|G)=\dfrac{P\left(E^e\cap F^e\cap G\right)}{P\left(G\right)}=\dfrac{P(E^e)P(F^e)P(G)}{P\left(G\right)}= P(E^e)P(F^e)\\\implies =P(E^e)[1-P(F)]=P(E^e)-P(E^e)P(F)\\\implies = P(E^e)-[1-P(E)]P(F)\\\implies = P(E^e)-P(F)+P(E)P(F)$$
    E, F, G are independent and $$P(E\cap F\cap G)=0$$
    $$\implies P(E)P(F)P(G)=0$$ but $$P(G)\ne 0$$
    $$\implies P(E)P(F)=0$$
    Hence $$P(E^e\cap F^e|G)= P(E^e)-P(F)$$
  • Question 5
    1 / -0
    The probability that $$A$$ can solve a problem is $$\dfrac {2 }{ 3}$$ and that $$B$$ can solve is $$\dfrac {3 }{4}$$. If both of them attempt the problem. What is probability that the problem act solved?
    Solution
    The problem won't be solved if both A and B.
    dont solve which has probability of $$\left( 1-\cfrac { 2 }{ 3 }  \right) \left( 1-\cfrac { 3 }{ 4 }  \right) $$ $$=\quad \cfrac { 1 }{ 12 } $$
    So, the probability that the problem is solve
    = 1  -  $$\ \cfrac { 1 }{ 12 } $$
    $$=\ \cfrac { 11 }{ 12 } $$
  • Question 6
    1 / -0
    If the probability for A to fail in an examination is 0.8 and that for B is 0.7 , then the probability that atleast one of A or B fails, is 
    Solution

  • Question 7
    1 / -0
    Let $$A,\ B,\ C$$ be pairwise independent even such that $$P(C)> 0$$ and $$P(A\cap B\cap C)=0$$, then $$P(A'\cap B'\cap C)$$ is equal to  
    Solution

  • Question 8
    1 / -0
    Two numbers $$b$$ and $$c$$ are chosen at random with replacement from the numbers $$1,\,2,\,3,\,4,\,5,\,6,\,7,\,8$$ and $$9$$.The probability that $${x}^{2}+bx+c>0$$ for all $$x\in R$$ is
    Solution

  • Question 9
    1 / -0
    If $$A$$ and $$B$$ are two independent events such that $$P(A)=\dfrac{1}{3}$$ and $$P(B)=\dfrac{3}{4}$$, then $$p\left\{\dfrac{B}{(A\cup B)}\right\}$$
    Solution

  • Question 10
    1 / -0
    If A and B are two events such that P (A)=$$\frac{1}{2}$$ and P (B) = $$\frac{2}{3}$$ , then which of the
    Solution

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