Probability Test

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Probability Test - 56

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Weekly Quiz Competition

Question 1
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A man has $$3$$ coins $$A, B$$ & $$C$$. $$A$$ is fair coin. $$B$$ is biased such that the probability of occurring head on it is $$2/3$$. $$C$$ is also biased with the probability of occurring head as $$1/3$$. If one coin is selected and tossed three times, giving two heads and one tail, find the probability that the chosen coin was $$A$$

A
$$9/25$$

B
$$3/5$$

C
$$27/125$$

D
$$1/3$$

Solution

Outcome HHT
Coin A
$$P(HHT)_A = (\cfrac{1}{2})^3 = 1/8 $$
Coin B
$$P(HHT)_B = (\cfrac{2}{3})^2.\cfrac{1}{3} = 4/27 $$
Coin C
$$P(HHT)_C = (\cfrac{1}{3})^2.\cfrac{2}{3} = 2/27 $$.
$$P(HHT)_{total} = 25/72$$
P(chosen coin was A) = $$ \cfrac{1/8}{25/72} = 9/25 $$
Question 2
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Let $$A$$ and $$B$$ be two events such that $$P(\overline { A\cup B } )=\cfrac { 1 }{ 6 } ,P(A\cap B)=\cfrac { 1 }{ 4 } $$ and $$P(\overline { A } )=\cfrac { 1 }{ 4 } $$, where $$\overline { A } $$ stands for complement of event $$A$$. Then, the events $$A$$ and $$B$$ are

A
Mutually exclusive and independent

B
Independent, but not equally likely

C
Equally likely but not independent

D
Equally likely and mutually exclusive

Solution

Given, $$P(\overline { A\cup B } )=\cfrac { 1 }{ 6 } $$
We know $$1-P(A\cup B)=\cfrac { 1 }{ 6 } $$
$$ \Rightarrow 1-P(A)-P(B)+P(A\cap B)=\cfrac { 1 }{ 6 } $$
$$\Rightarrow \cfrac { 1 }{ 4 } -P(B)+\cfrac { 1 }{ 4 } =\cfrac { 1 }{ 6 } $$
$$\Rightarrow P(B)=\cfrac { 1 }{ 3 } $$
Therefore, $$P(A)=1-P(\overline { A } )=1-\cfrac { 1 }{ 4 } =\cfrac { 3 }{ 4 } $$
Question 3
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If two events $$A$$ and $$B$$ are such that $$P({ A }^{ c })=0.3\quad $$ and $$P(B)=0.4$$ and $$P\left( { A\cap B } \right)^c =0.5$$, then $$P\left[ B|\left( A\cup { B }\right)^c \right] $$ is equal to

A
$$\cfrac { 1 }{ 2 } $$

B
$$\cfrac { 1 }{ 3 } $$

C
$$\cfrac { 1 }{ 4 } $$

D
$$\cfrac { 1 }{ 5 } $$

Solution
Question 4
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Directions For Questions

Choose the correct answer:

...view full instructions

The probabilities of three mutually exclusive events A, B and C are given by $$\frac{1}{3}, \frac{1}{4}$$, and $$\frac{5}{12}$$. Then $$(P \cup B \cup C)$$ is.

A
$$\frac{9}{12}$$

B
$$\frac{11}{12}$$

C
$$\frac{7}{12}$$

D
$$1$$
Question 5
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If $$\dfrac {1 + 3p}{3}, \dfrac {1 - p}{4}$$ and $$\dfrac {1 - 2p}{2}$$ are mutually exclusive events. Then, range of $$p$$ is

A
$$\dfrac {1}{3} \leq p \leq \dfrac {1}{2}$$

B
$$\dfrac {1}{2} \leq p \leq \dfrac {1}{2}$$

C
$$\dfrac {1}{3} \leq p \leq \dfrac {2}{3}$$

D
$$\dfrac {1}{3} \leq p \leq \dfrac {2}{5}$$

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 6
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Probability of any event $$x$$ lies

A
$$0 < x < 1$$

B
$$0\leq x < 1$$

C
$$0\leq x \leq 1$$

D
$$1 < x < 2$$

Solution
Question 7
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Directions For Questions

Choose the correct answer:

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Let A and B be any two events and S be the corresponding sample space. Then $$P(\bar{A}\cap B)$$

A
$$P(B)-P(A\cap B)$$

B
$$P(A\cap B)-P(B)$$

C
$$P(S)$$

D
$$P[A(A\cup B)']$$

Solution
Question 8
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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..

A
$$[\frac{-1}{5}, \frac{1}{2}]$$

B
$$[\frac{-1}{5}, \frac{1}{4}]$$

C
$$[\frac{-1}{5}, \frac{1}{3}]$$

D
$$[\frac{-1}{3}, \frac{1}{3}]$$

Solution
Question 9
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There are eight delegates $$4$$ of them are Americans, $$1$$ British, $$1$$ Chinese, $$1$$ Dutch and $$1$$ Egyptian. These delegates are paired randomly.The probability that no two delegates of the same country are paired is

A
$$\dfrac{19}{28}$$

B
$$\dfrac{5}{14}$$

C
$$\dfrac{11}{14}$$

D
$$\dfrac{9}{14}$$

Solution
Question 10
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Consider the following events.
$$E_1$$: Six fair dice are rolled and at least one die shows six.
$$E_2$$: Twelve fair dice are rolled and at least two dice show six.
Let $$p_1$$ be the probability of $$E_1$$ and $$p_2$$ be the probability of $$E_2$$. Which of the following is true?

A
$$p_1 > p_2$$

B
$$p_1=p_2=0.6651$$

C
$$p_1 < p_2$$

D
$$p_1 = p_2 = 0.3349$$

Solution

$$p_1 = 1-(\text{no die shows six})$$
$$p_1 = 1-(\dfrac{5}{6})^6 = 0.665102$$

$$p_2 = 1-(\text{no die show six, and one die shows six})$$

$$p_2 = 1-(\dfrac{5}{6})^{12} -\dfrac{12!}{1! \times 11!} (\dfrac{5}{6})^{11} \times \dfrac{1}{6} =0.618667$$

$$p_1 >p_2$$