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Probability Test 25

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Probability Test 25
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  • Question 1
    1 / -0
    There are three boxes, each containing a different number of light bulbs. The first box has 10 bulbs, of which four are dead, the second box has six bulbs, of which one is dead, and the third box has eight bulbs of which three are dead. What is the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes?
    Solution
    Let $$𝐴_1, 𝐴_2, 𝐴_3$$ denotes the events of selecting bulbs from bags $$1,2\: and\: 3$$ respectively. 

    Let $$𝐵$$ denotes the event the bulbs selected are dead.

    $$ 𝑃(𝐴_1) = 𝑃(𝐴_2)  = 𝑃(𝐴_3)  = \dfrac{1}{3} $$

    Also $$P(B|A_1)=\dfrac{4}{10}, P(B|A_2)=\dfrac{1}{6}, P(B|A_3)=\dfrac{3}{8}$$

    By law of total probability,

    $$P(B)=P(A_1)P(B|A_1)+P(A_2)P(B|A_2)+P(A_3)P(B|A_3)$$

    Substituting the values we get,

    $$P(B)=\dfrac{1}{3} \times \dfrac{4}{10}+ \dfrac{1}{3} \times \dfrac{1}{6}+ \dfrac{1}{3} \times \dfrac{3}{8}$$

    $$\Rightarrow P(B)=\dfrac{113}{360}$$

    Thus the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes is $$\dfrac{113}{360}$$.


  • Question 2
    1 / -0
    Suppose that two factories supply light bulbs to the market. Factory X's bulbs work for over $$5000$$ hours in $$99\%$$ of cases, whereas factory Y's bulbs work for over $$5000$$ hours in $$95\%$$ of cases. It is known that factory X supplies $$60\%$$ of the total bulbs available. What is the chance that a purchased bulb will work for longer than $$5000$$ hours?
    Solution
    Let $$X$$ be the event "comes from factory $$X$$" and $$Y$$ be the event "comes fom factory $$Y$$ " and Let $$H$$ be the event "works over $$5000$$ hours." 

    Therefore $$P(X)=60 \%=0.60 \Rightarrow P(Y)=1-0.60=0.40$$

    Given that, Factory $$X's$$ bulbs work for over $$5000$$ hours in $$99 \%$$ of cases.

    $$\therefore P(H|X)=99 \%=0.99$$

    Also given, factory Y's bulbs work for over $$$000$$5000$$ hours in %$$95\%$$  of cases.

    $$\therefore P(H|Y)=95 \%=0.95$$

    Then by the Law of Total Probability we have

     $$P(H) = P (H | X) P(X) + P (H | Y ) P(Y ) $$

                $$= (.99) (.6) + (.95) (.4)$$

                $$ = .974$$

    Thus $$P(H)=0.974=\dfrac{974}{1000}$$.


  • Question 3
    1 / -0
    Probability of any event $$x$$ lies
    Solution

  • Question 4
    1 / -0
    For the events $$A$$ and $$B, P(A) = \dfrac {3}{4}, P(B) = \dfrac {1}{5}, P(A\cap B) = \dfrac {1}{20}$$ then $$P(A/B) =$$ ___________.
    Solution
    Given $$P(A)=\dfrac{3}{4}$$, $$P(B)=\dfrac{1}{5}$$, $$P(A\cap B)=\dfrac{1}{20}$$.

    We have to find $$P(A/B)$$.

    We know that $$P(A/B)=\dfrac{P(A\cap B)}{P(B)}$$.

    Substituting the values we get

    $$P(A/B)=\dfrac{\dfrac{1}{20}}{\dfrac{1}{5}}$$

                    $$=\dfrac{5}{20}$$

    $$\therefore P(A/B)=\dfrac{1}{4}$$
  • Question 5
    1 / -0
    In throwing a pair of dice, the events 'coming up of 6 on Ist dice' and 'a total of 7 on both the dice' are 
    Solution

  • Question 6
    1 / -0
    If $$\dfrac {1 + 3p}{3}, \dfrac {1 - p}{4}$$ and $$\dfrac {1 - 2p}{2}$$ are mutually exclusive events. Then, range of $$p$$ is
    Solution
    Since, the probability lies between $$0$$ and $$1$$.
    $$0\leq \dfrac {1 + 3p}{3}\leq 1, 0\leq \dfrac {1 - p}{4}\leq 1, 0\leq \dfrac {1 - 2p}{2}\leq 1$$
    $$\Rightarrow 0\leq 1 + 3p\leq 3, 0\leq 1 - p\leq 4, 0\leq 1 - 2p \leq 2$$
    $$\Rightarrow -\dfrac {1}{3} \leq p\leq \dfrac {2}{3}, -3 \leq p\leq 1, -\dfrac {1}{2} \leq p\leq \dfrac {1}{2} ..... (i)$$
    Again, the events are mutually exclusive
    $$0\leq \dfrac {1 + 3p}{3} + \dfrac {1 - p}{4} + \dfrac {1 - 2p}{2}\leq 1$$
    $$\Rightarrow 0\leq 13 - 3p \leq 12$$
    $$\Rightarrow \dfrac {1}{3}\leq p\leq \dfrac {13}{3} .... (ii)$$
    From Eqs. (i) and (ii),
    $$max\left \{-\dfrac {1}{3}, -3, \dfrac {-1}{2}, \dfrac {1}{3}\right \} \leq p\leq min \left \{\dfrac {2}{3}, 1, \dfrac {1}{2}, \dfrac {13}{3}\right \}$$
    $$\Rightarrow \dfrac {1}{3} \leq p\leq \dfrac {1}{2}$$.
  • Question 7
    1 / -0
    If $$\frac{1+4p}{p};\frac{1-p}{4};\frac{1-2p}{2}$$ are probabilities of three mutually exclusive events, then the possible values of $$'p'$$ belong to the set is..
    Solution

  • Question 8
    1 / -0
    A bag contains a white and black balls. Two players $$A$$ and $$B$$ alternately draw a ball from the bag.replacing the ball each form after the draw. The person who first draw a white ball will wins the bag.game. If $$A$$ 'begins the game and $$\frac { P ( B ) } { P ( A ) } = \frac { 1 } { 2 }$$ then $$a : b$$
  • Question 9
    1 / -0
    If $$E_{1}$$ and $$E_{2}$$ are mutually exclusice events, then 
    Solution

  • Question 10
    1 / -0
    The total numbers of outcomes when three coins tossed once is .....
    Solution
    When  three coins tossed ones then total number of outcomes is : $$8$$
    $$\{ TTT,HHH,TTH,THT,HTT,HHT,HTH,THH\} $$
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