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Probability Test - 30

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Probability Test - 30
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  • Question 1
    1 / -0
    Let A, B, C be three events such that A and B are independent and $$P(C) = 0$$, then events A, B, C are
    Solution
    Given three events $$A,B,C.$$
    A and B are independent $$\xi$$ $$P(C)=0.$$
    A term 'independent event' means probability of events. They have a common outcomes and each event is non-empty.
    As A and B are independent
    i.e, $$P(A)\neq 0$$ and $$P(B)\neq 0.$$
    For $$P(C)$$ to be independent,
    $$\Rightarrow P(A\cap B)=P(A).P(B)$$            $$P(A\cap B\cap C)=P(A).P(B).P(C)$$
    $$\Rightarrow P(A\cap C)=P(A).P(C)$$
    $$\Rightarrow P(B\cap C)=P(B).P(C)$$
    If all the given events are true, then A,B,C are independent.
    Hence, the answer is independent.
  • Question 2
    1 / -0
    You enter a chess tournament where your probability of winning a game is $$0.3$$ against half the players (call them type $$1$$), $$0.4$$ against a quarter of the players (call them type $$2$$), and $$0.5$$ against the remaining quarter of the players (call them type $$3$$). You play a game against a randomly chosen opponent. What is the probability of winning?
    Solution
    Let $$A_i$$ be the event of playing an opponent of type $$i$$.
    $$A_1$$ = playing the half of the opponent pool where the probability of winning $$P (winning|A_1) = 0.3$$
    $$A_2$$ = playing the half of the opponent pool type 2, where the probability of winning $$P (winning|A_2) = 0.4$$
    $$A_3$$ = playing the half of the opponent pool type 3, where the probability of winning $$P (winning|A_3) = 0.5$$
    By total probability theorem, we get, 
    $$P(winning) = P(A_1)P(winning|A_1) + P(A_2)P(winning|A_2) + P(A_3)P(winning|A_3)$$
    $$P (winning) = 50\% \times 0.3 + 25\% \times 0.4 + 25\% \times 0.5$$
    $$P (winning) = 0.375.$$
  • Question 3
    1 / -0
    Sample space for experiment in which two coins are tossed is
    Solution
    Two coins are tossed.
    Number of outcomes in sample space when two coins are tossed is given by $$2^2=4$$.
    Hence, sample space for experiment in which two coins are tossed is $$4$$.
  • Question 4
    1 / -0
    Choosing a birthdate is an example of .........
    Solution
    Let us see the sample space for choosing a birthdate.
    A person can be born in any date of a month and a month has maximum of $$30$$ or $$31$$ days.
    Therefore, $$S=\{1,2,3,4,5,6,7,8,9,10,11,12,13,...,30,31\}$$ which is a finite set.
    Thus choosing a birthdate is an example of finite sample space.
  • Question 5
    1 / -0
    A bag contains $$4$$ white, $$5$$ red and $$6$$ blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
    Solution
    Let S be the sample space.
    Then, $$n(S)$$ = number of ways of drawing $$3$$ balls out of $$15$$ $$= ^{15}C_3$$
    $$\implies =\dfrac {(15\times  14\times 13)}{3\times 2\times 1}=455$$
    Let E = event of getting all the $$3$$ red balls.
    $$ n(E) =^5C_3 = \dfrac {(5 \times 4\times 3)}{3\times 2\times 1}=10$$
    Hence, the probability that all of them are red, is:
    $$ P(E) =\dfrac {n(E)}{n(S)}=\dfrac {10}{455}=\dfrac 2{91}$$
  • Question 6
    1 / -0
    The term law of total probability is sometimes taken to mean the ____
    Solution
    The term law of total probability is sometimes taken to mean the law of alternatives, which is a special case of the law of total probability applying to discrete random variables.
    Hence, option B is correct.
  • Question 7
    1 / -0
    If $$P(A) = 0.25, P(B) = 0.50, P(A\cap B) = 0.14$$, then $$P$$ (neither $$A$$ nor $$B) =$$
    Solution
    $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

    $$\Rightarrow P(A\cup B)=0.25+0.50-0.14=0.61$$

    $$P(\text{neither A nor B})=P(\bar A\cap \bar B)=1-P(A \cup B)=1-0.61=0.39$$

    Option A is correct.
  • Question 8
    1 / -0
    The experiment is to randomly select a human and measure his or her length. Identify the type of the sample space.
    Solution
    The experiment is to randomly select a human and measure his or her length. 
    Depending of how far reaching our means of selection is it is possible to consider a sample space of about $$6.6$$ billion humans inhabiting the planet Earth. 
    In this case, the height of the selected person becomes a random variable. 
    However, it is also possible to consider the sample space consisting of all possible values of height measurements of the world population. 
    The tallest man ever measured lived in the United States and had a height of $$272$$ cm $$ (8'11'')$$. The height of the shortest person is more difficult to determine. Zero is clearly the low bound, but, for a living adult, it may be safely raised to, say, $$40 $$ cm. 
    This suggests a sample space which is a line segment $$[40,272]$$ in centimeters. 
    While at all times the human population is discrete, we may assume that in some height range near the normal average, all possible heights are realized making a continuous classification. 
    We know that, "A continuous sample space is one which takes values in one or more intervals."
    Thus the sample space is a continuous sample space.
  • Question 9
    1 / -0
    The experiment is to repeatedly toss a coin until first tail shows up. Identify the type of the sample space.
    Solution
    The experiment is to repeatedly toss a coin until first tail shows up.
    A coin is tossed. The possible outcomes are $$head-H, tail-T$$.
    Look into the following table:

     $$1^{st}$$ toss $$2^{nd}$$ toss $$3^{rd}$$ toss $$4^{th}$$ toss$$5^{th}$$ toss 
     $$T$$$$-$$$$-$$
    		$$-$$
    		 $$-$$
     $$H$$ $$T$$ $$-$$$$-$$ $$-$$
     $$H$$ $$H$$ $$T$$ $$-$$$$-$$
     $$H$$ $$H$$ $$H$$ $$T$$ $$-$$
     $$H$$ $$H$$ $$H$$ $$H$$ $$T$$
    If we get $$T$$ at first toss, then our experiment ends.
    Otherwise second toss. If we get $$T$$, out experiment ends. If not the process continues till we end up in tail.
    Hence the possible outcomes are sequences of $$H$$ that, if finite, end with a single $$T$$, and an infinite sequence of $$H$$.
    Therefore, $$ S=\{T,HT,HHT,HHHT,HHHHT,..., \{HHHH....\}\}$$
    Thus we get a infinite but countable(depends on the number of toss) sample space.
    As we shall see elsewhere, this is a remarkable space that contains a not impossible event whose probability is $$0$$. 
    We know that, "a discrete sample space is one with a finite or countably infinite number of possible values."
    Hence, it is a infinite discrete sample space.
  • Question 10
    1 / -0
    Given a circle of radius $$R$$, the experiment is to randomly select a chord in that circle. Identify the type of the sample space.
    Solution

    Given a circle of radius $$R$$, the experiment is to randomly select a chord in that circle.

    There are many ways to accomplish such a selection. 

    However the sample space is always the same:

    $$\{AB: A \:and\: B\: are\: points\: on\: a\: given\: circle\}$$.

    One natural random variable defined on this space is the length of the chord.

    Here the length of the chord takes more values depending on the point we choose in the circle.

    We know that, "A continuous sample space is one which takes values in one or more intervals."

    Therefore, this is a continuous sample space.

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