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Probability Test 18

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Probability Test 18
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  • Question 1
    1 / -0
    A card is drawn from a well shuffled pack of 52 cards. Events A and B are defined as follows :
    A : Getting a card of spade
    B : Getting an ace, then A and B are .....
    Solution
    A: Getting a card of spade

    B : Getting an ace

    i) Events A & B are not mutually exclusive

    since the occurrence of one doesn't rule out the occurrence of others. For instance, getting an ace of spade

    ii) Events A & B are independent as occurrence

    of one is not dependent on other.

    also P(AB)=152P(A\cap B)=\dfrac {1}{52}  since ther is only 11 ace of spades

    & P(A)P(B)=1352×452=152P(A)P(B)=\dfrac{13}{52}\times \dfrac{4}{52}= \dfrac{1}{52}

    P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B)\Rightarrow Independent events

    \Rightarrow A & B are not mutually exclusive

    but independent.
  • Question 2
    1 / -0
    If A and B are two events, such that P(A or B)=P(A), then 
    Solution
    P(AorB)=P(AB)=P(A)P\left( AorB \right) =P\left( A\cup B \right) =P\left( A \right)
    In venn diagram it is shown that BB is a part of AA Hence probability of occurrance of event AA or event BB will be P(A)P(A).
    B\therefore B is subset of AA

  • Question 3
    1 / -0
    Lot AA consists of 3G3G and 2D2D articles. Lot BB consists of 4G4G 1D1D article. A new lot CC is formed by taking 33 articles from AA articles from AA and 22 form BB . The probability that an article chosen at random from CC is 
    Solution
    A] A : event it is chosen from lot A
    P(a)=35 P(a) = \frac{3}{5}
    P(B)=25 P(B) = \frac{2}{5}
    P (Defective) =P(A)×P(DA)+P(B)×P(DB)= P(A)\times P(\frac{D}{A})+P(B)\times P(\frac{D}{B})
    =35×25+25×15 = \frac{3}{5} \times \frac{2}{5}+\frac{2}{5}\times \frac{1}{5}
    =625+225 = \frac{6}{25}+\frac{2}{25}
    =825 = \frac{8}{25}
    P(D)=825 \therefore P(D) = \frac{8}{25}

  • Question 4
    1 / -0
    A and B are two independent  events such that P(AB)=215 { P\left( { A }^{ ' }\cap B \right) =\dfrac { 2 }{ 15 }  }then P(B) is equal to :
    Solution

  • Question 5
    1 / -0
    A bag contains 66 red, 44 white and 88 blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.
    Solution
    EE\rightarrow Event of getting one is red, one is white and one 66 red +4+4 white +8+8 blue balls =18=18 balls

    Total outcomes =( 18C3)=(\ ^{18}C_{3})

     R W B\underbrace { \bigodot  }_{ R } \underbrace { \bigodot  }_{ W } \underbrace { \bigodot  }_{ B } \rightarrow no. of fobourable element

    = 6C1× 4C1× 8C1=\ ^{6}C_{1}\times \ ^{4}C_{1}\times \ ^{8}C_{1}

    =6×4×8=6\times 4\times 8

    P(E)=6×4×8 18C3=6×4×8×3×2×118×17×16\therefore P(E)=\dfrac{6\times 4\times 8}{\ ^{18}C_{3}}=\dfrac{6\times 4\times 8\times 3\times 2\times 1}{18\times 17\times 16}

    =417=\dfrac{4}{17}
  • Question 6
    1 / -0
    If A and B are such events that P(A)>0P(A)>0 and P(B)1 P(B)\neq 1 then P(AˉBˉ)P\left(\dfrac{\bar{A}}{\bar{B}}\right) is equal to-
    Solution

  • Question 7
    1 / -0
    If 1+4p4,1+p4,12p2\dfrac {1+4p}{4},\dfrac {1+p}{4},\dfrac {1-2p}{2} re probabilities of three mutually exclusive events, then 
    Solution

  • Question 8
    1 / -0

    Directions For Questions

    88 players compete in a tournament, every one plays everyone else just once. The winner of a game gets 11, the loser 00 or each gets 12\dfrac{1}{2} if the game is drawn. The final result is that every one gets a different score and the player playing placing second gets the same as the total of four bottom players.

    ...view full instructions

    The score  of the player placing II was 
    Solution

    The total no.of matches held is given by 8×72=28\dfrac{8\times 7}{2}=28

    So the winner is the one who has the score in between [28,14][28,14]

    The second is the one who has less score than the first one.

    Since it has many possibilities it cannot be determined.


  • Question 9
    1 / -0
    If 1+4pp,1p4,12p2\dfrac{1+4p}{p},\dfrac{1-p}{4}, \dfrac{1-2p}{2} are propabilities of three mutually exclusive events, then-
    Solution

  • Question 10
    1 / -0
    A,B and C are three mutually exclusive and exhaustive events and P(B)=32P(A),P(C)=13P(B)P(B)=\dfrac{3}{2}P(A), P(C)=\dfrac{1}{3}P(B) then the value of P(A)P(A) is
    Solution
    Given, A,BA, B and CC are three mutually exclusive and exhaustive events and P(B)=32P(A),P(C)=13P(B)P(B)=\dfrac{3}{2}P(A), P(C)=\dfrac{1}{3}P(B).

    Since A,B,CA, B, C are mutually exclusive and exhaustive events then we've, 

    P(A)+P(B)+P(C)=1P(A)+P(B)+P(C)=1

    or, P(A)+32P(A)+12P(A)=1P(A)+\dfrac{3}{2}P(A)+\dfrac{1}{2}P(A)=1

    or, 3P(A)=13P(A)=1

    or, P(A)=13P(A)=\dfrac{1}{3}.
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