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Probability Test - 38

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Probability Test - 38
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  • Question 1
    1 / -0
    Two boys A and B find the jumble of n ropes lying on the floor. Each takes hold of one loose and randomly. If the probability that they are both holding the same rope is $$\dfrac{1}{101}$$ then the number of ropes is equal to
    Solution
    A and B have n ropes to choose. So, the number
    of ways in which they can choose ropes = $$n\times n$$
    Number of ways in which A and B
    choose the same rope = n
    Probability = $$\dfrac{n}{n\times n}= \dfrac{1}{n}= \dfrac{1}{101}$$
    $$\Rightarrow n= 101$$
    $$\therefore $$ Number of ropes = 101

  • Question 2
    1 / -0
    Counters numbered $$1,2,3$$ are placed in a bag and one is drawn at random and replaced. The operation is being repeated three times. The probabolity of obtaining a total of $$6$$ is ?
    Solution
    $$\quad To\quad get\quad a\quad sum\quad of\quad 6\quad ,the\quad three\quad numbers\quad could\quad be\quad either(1,2,3)or(2,2,2)$$
    $$\quad For(1,2,3),the\quad total\quad number\quad of\quad different\quad arrangements\quad _{ 3 }^{ 3 }{ p }=\frac { 3 }{ 0 } =6$$
    $$Total\quad \quad number\quad of\quad possibilities\quad that\quad \quad could\quad occur\quad \quad in\quad 3\quad operations=3*3*3=27$$
    $$Probability\quad ofobtaininga1in\quad one\quad operation=Probability\quad ofobtaining\quad a2in\quad one\quad operation=Probability\quad ofobtaining\quad a3\quad in\quad one\quad operation=\frac { 1 }{ 3 } $$
    $$Probability\quad ofobtaining1\quad and\quad 2\quad and3\quad in\quad any\quad order\quad 6\times \frac { 1 }{ 3 } .\frac { 1 }{ 3 } .\frac { 1 }{ 3 } =\frac { 6 }{ 27 } $$
    $$Probability\quad ofobtaining\quad \quad 2\quad in\quad each\quad of\quad the\quad 3\quad operations\frac { 1 }{ 3 } \frac { 1 }{ 3 } .\frac { 1 }{ 3 } =\frac { 1 }{ 27 } $$
    $$Chance\quad or\quad probability\quad of\quad obtaining\quad a\quad total\quad of\quad 6=\frac { 6 }{ 27 } +\frac { 1 }{ 27 } =\frac { 7 }{ 27 } $$

  • Question 3
    1 / -0
    Rajdhani Express stops at six intermediate stations between Kota and Mumbai.Five passengers board at Kota. Each passengers can get down at any station till Mumbai. The probability that all five passengers will get down at different stations, is  
    Solution
    Total number of intermediate Station = 6
    Total Number of passenger = 5

    Total Number of ways in which the passengers will get down at any station $$=6^5$$ 
    Number of ways in which all 5 passengers get down at different stations
    $$=^6P_5$$

    Hence, Required Probability $$=\dfrac{^6P_5}{6^5}$$
  • Question 4
    1 / -0
    If $$A$$ & $$B$$ are two events such that $$P(A)=0.6$$ & $$P(B)=0.8$$, then the greatest value that $$P(A/B)$$ can have is
    Solution
    First of all, $$P(A/B)=\dfrac{P(A⋂B)}{P(B)}$$

    Now, it is given that $$P(B)=0.8$$

    So for maximizing $$P(A/B)$$ , we have to maximize $$P(A⋂B)$$ .

    Notice that, if we draw the Venn Diagram of the above probability chart, if event B completely overlaps with event A, we get the maximum intersection area between the two. In that case, the entire probability region of A is the overlapping part. Thus for maximum intersection,

    $$P(A⋂B)_{maximum}=P(A)=0.6$$

    $$P(A/B)_{maximum}=\dfrac{P(A)}{P(B)}=\dfrac{0.6}{0.8}=0.75$$
  • Question 5
    1 / -0
    Of the students in a college, it is known that $$60\%$$ reside in hostel and $$40\%$$ are day scholars (not residing in hostel). Previous year results report that $$30\%$$ of all students who reside in hostel attain A grade and $$20\%$$ of day scholar attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostile?
    Solution
    Let $$A$$ denote $$A-grade$$.
    $$H$$ denote from hostel.
    $$D$$ denote day scholar.
    $$P(A/H) = 0.3$$ and $$P(A/D) = 0.2$$
    $$P(H) = 0.6$$ and $$P(D) = 0.4$$
    $$P(H/A) = \cfrac{P(A/H) \times P(H)}{P(A/H) \times P(H) + P(A/D) \times P(D)}$$
    $$P(H/A) = \cfrac{0.3 \times 0.6}{0.3 \times 0.6 + 0.2 \times 0.4}$$
    $$P(H/A) = \cfrac{0.18}{0.18+0.08}$$
    $$P(H/A) = \cfrac{18}{26} = \cfrac{9}{13}$$
  • Question 6
    1 / -0
    A card is drawn from a deck of cards. What is the probability that it is either a spade or an ace or both 
    Solution
    There are 4 ace including spade ace
    13 spade includes its ace
    So there are remaining 3 ace
    Total number of favourable choice=13+3=16
    Probability=16/52=4/13
  • Question 7
    1 / -0
    If two events A and B are such that $$P(A)=0.75\   P(B|A)=0.8\ P(B|A')=0.6$$ Find P(B)
    Solution
    $$P(B|A)=\frac { P(B\cap A) }{ P(A) } \\ 0.8=\frac { P(B\cap A) }{ 0.75 } \\ P(B\cap A)=0.6\\ \\ P(B|A')=\frac { P(B\cap A') }{ P(A') } \\ 0.6(1-0.75)=P(B\cap A')\\ P(B\cap A')=0.15\\ P(Only\quad B)=0.15\\ P(B)=P(only\quad B)+P(B\cap A)=0.75$$
  • Question 8
    1 / -0
    A card is lost from a pack of $$52$$ playing cards. From the remainder of the pack, two cards are drawn and are found to be spade. The probability that the missing card is a spade is :
    Solution

  • Question 9
    1 / -0
    In a class 5% of boys and 10% of girls have an I.Q of more than 150.In this class 60% of students are boys. If a student is selected at random and found to have an I.Q. of more than 150. Find the probability that the student is a boy.
    Solution
    Let us consider the problem
    $$E_1$$ : Event that boys are selected
    $$E_2$$ : Event that girls are selected
    $$A$$ : event that have IQ $$150$$
    Implies that,
    \begin{array}{l} P\left( { A/{ E_{ 1 } } } \right) =\dfrac { 5 }{ { 100 } }  \\ P\left( { A/{ E_{ 2 } } } \right) =\dfrac { { 10 } }{ { 100 } }  \\ P\left( { { E_{ 1 } } } \right) =\dfrac { { 60 } }{ { 100 } } ,P\left( { { E_{ 2 } } } \right) =\dfrac { { 40 } }{ { 100 } }  \\ P\left( { A|{ E_{ 1 } } } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right)  } }{ { P\left( { { E_{ 1 } } } \right) P\left( { A|{ E_{ 1 } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { A|{ E_{ 2 } } } \right)  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } }  } }{ { \dfrac { { 60 } }{ { 100 } } \times \dfrac { 5 }{ { 100 } } +\dfrac { { 40 } }{ { 100 } } \times \dfrac { { 10 } }{ { 100 } }  } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 60\times 5 } }{ { 60\times 5+40\times 10 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { { 300 } }{ { 300+400 } }  \\ P\left( { { E_{ 1 } }|A } \right) =\dfrac { 3 }{ 7 }  \end{array}

    Hence, the probability is $$\dfrac {3}{7}$$
  • Question 10
    1 / -0
    If $$P(A/B)=P(B/A)$$. $$A$$ and $$B$$ are two non-mutually exclusive events then
    Solution
    Given $$p\left(\dfrac{A}{B}\right)=P\left(\dfrac{B}{A}\right)\Rightarrow \dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A\cap B)}{P(A)}$$

    $$P(A)=P(B)$$     [$$P(A\cap B)\neq P(A).P(B)$$ as they not independent]

    $$\therefore$$ Option $$B$$ is correct.

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