Probability Test 15

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Probability Test 15

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Weekly Quiz Competition

Question 1
1 / -0

Suppose A and B are two events. Event B has occured and it is known that $$P(B) <1$$. What is $$P(A|B^c)$$ equal to?

A
$$\displaystyle\frac{P(A)-P(B)}{1-P(B)}$$

B
$$\displaystyle\frac{P(A)-P(A\cap B)}{1-P(B)}$$

C
$$\displaystyle\frac{P(A)+P(B^c)}{1-P(B)}$$

D
None of the above

Solution

Given that $$A$$ and $$B$$ are two events and event $$B$$ has occured,
From the division rule of probability we know that $$P\left(\dfrac { A }{ B } \right)=\dfrac { P(A\cap B) }{ P(B) } $$,
similarly,
$$P\left(\dfrac { A }{ { B }^{ c } }\right )=\dfrac { P(A\cap { B }^{ c }) }{ P({ B }^{ c }) } $$

$$\\ \Longrightarrow \dfrac { P(A)-P(A\cap B) }{ 1-P(B) } $$
Question 2
1 / -0

Probability of impossible event is

A
$$1$$

B
$$0$$

C
$$\dfrac {1}{2}$$

D
$$-1$$

Solution

(P) Of possible event is $$0$$
Question 3
1 / -0

A researcher conducted a survey to determine whether people in a certain town prefer watching sports on television to attending the sporting event. The researcher asked 117 people who visited a local restaurant on a Saturday, and 7 people refused to respond. Which of the following factors makes it least likely that a reliable conclusion can be drawn about the sports-watching preferences of all people in the town?

A
Sample size

B
Population size

C
The number of people who refused to respond

D
Where the survey was given.

Solution

Considering the population of a town, the number $$117$$ is very much low.
We can't take a survey of $$117$$ people and conclude the result for the whole town.
So, the $$sample\ size$$ of the survey is very low compared to the population. $$[A]$$
Question 4
1 / -0

For two independent events $$A$$ and $$B$$, what is $$P(A + B)$$, given $$P(A) = \dfrac{3}{5}$$ and $$P(B) = \dfrac{2}{3}$$?

A
$$\dfrac{11}{15}$$

B
$$\dfrac{13}{15}$$

C
$$\dfrac{7}{15}$$

D
$$0.65$$

Solution

Given: $$P(A)=\dfrac 35, P(B)=\dfrac 23$$, and A, B are independent.
To find: $$P(A+B)=?$$
Sol: As A, B are independent, $$\implies P(A\cap B)=P(A)P(B)$$
And also $$P(A+B)=P(A\cup B)$$
$$\therefore, P(A+B)=P(A)+P(B)-P(A\cap B)\\\implies P(A+B)=\dfrac 35+\dfrac 23-\left(\dfrac 35\times \dfrac 23\right)\\\implies P(A+B)=\dfrac {9+10-6}{15}=\dfrac {13}{15}$$
Question 5
1 / -0

If $$\dfrac{1+3p}{3},\dfrac{1-2p}{2} $$ are probabilities of two mutually exclusive event, then $$p$$ lies in the interval

A
$$\left [-\dfrac{1}{3},\dfrac{1}{2}\right]$$

B
$$\left (-\dfrac{1}{2},\dfrac{1}{2}\right)$$

C
$$\left [-\dfrac{1}{2},\dfrac{2}{3}\right]$$

D
$$\left (-\dfrac{1}{3},\dfrac{2}{3}\right)$$

Solution

$$Probabilities\quad are\quad \frac { 1+3p }{ 3 } \quad and\frac { 1-2p }{ 2 } \\ so,\quad 0\le \frac { 1+3p }{ 3 } \le 1\\ \Rightarrow 0\le 1+3p\le 3\\ \Rightarrow -1/3\le p\le 2/3\\ Alse,\quad 0\le \frac { 1-2p }{ 2 } \le 1\\ \Rightarrow 0\le 1-2p\le 2\\ \Rightarrow -1\le -2p\le 1\\ \Rightarrow -1\le 2p\le 1\\ \Rightarrow -1/2\le p\le 1/2\\ Intersection\quad range\quad of\quad p\quad for\quad both\quad probabilities\quad to\quad be\quad valid\quad =-1/3\le p\le 1/2$$
Question 6
1 / -0

There are two bags, one of which contains $$3$$ black and $$4$$ white balls,$$II$$ bag contain $$4$$ black and $$3$$ white balls.A die is cast, if the number $$3$$ or less than $$3$$ turns up, a ball is drawn from $$1$$ bag and if the face greater than $$3$$ turns up, a ball is drawn from $$II$$ bag. It is found that ball drawn is black. Find the probability that it is from bag I.

A
$$\dfrac {3}{14}$$

B
$$\dfrac {4}{7}$$

C
$$\dfrac {2}{7}$$

D
$$\dfrac {3}{7}$$

Solution

Let $$A_{1}$$ be the event $$\leq 3$$ (when a die is cast) $$A_{2}$$ be the event $$\geq 4$$ (when a die is cast) $$'E'$$ be the event drawn ball is Black.
$$\therefore P(A_{1}) = \dfrac {1}{2} = P(A_{2})$$
$$P\left (\dfrac {E}{A_{1}}\right ) = \dfrac {3}{7}, P\left (\dfrac {E}{A_{2}}\right ) = \dfrac {4}{7}$$
$$\therefore P\left (\dfrac {A_{1}}{E}\right ) = \dfrac {P(A_{1})P\left (\dfrac {E}{A_{1}}\right )}{P(A_{1}) P\left (\dfrac {E}{A_{1}}\right ) + P(A_{2}) P\left (\dfrac {E}{A_{2}}\right )}$$
$$= \dfrac {\dfrac {1}{2}\times \dfrac {3}{7}}{\dfrac {1}{2} \times \dfrac {3}{7} + \dfrac {1}{2}\times \dfrac {4}{7}} = \dfrac {3}{7}$$
Hence choice (d) is correct.
Question 7
1 / -0

The probability that an event does not happens in one trial is 0.8.The probability that the event happens atmost once in three trails is

A
$$0.896$$

B
$$0.791$$

C
$$0.642$$

D
$$0.592$$

Solution
Question 8
1 / -0

A box contains $$2$$ silver coins and 4 gold coins and the second box contains $$4$$ silver coins and $$3$$ gold coins. If a coin is selected from one of the box, what is the probability that it is a silver coin.

A
$$0.3$$

B
$$0.4$$

C
$$0.5$$

D
$$0.6$$

Solution

Box $$I:2$$ Silver coins $$+\;{4}$$ gold coins
Box $$II:4$$ Silver coins $$+\;{3}$$ gold coins
Let $$E_1$$ be the event of selecting a gold coin from box $$I$$
and $$E_2$$ be the event of selecting a gold coin from box $$II.$$
$$A:$$ Probability of gold coin
$$\Rightarrow P(E_1)={^6C_2}$$            $$P(E_2)={^7C_4}$$
$$\Rightarrow P\left( \dfrac { A }{ E_1} \right) =P\left( \dfrac { A }{ E_2} \right) =\dfrac{1}{2}$$
$$\Rightarrow P\left( \dfrac { E_1 }{ E_2} \right) =\dfrac{P(E_1)P\left( \dfrac { A }{ E_1} \right) }{P(E_1)P\left( \dfrac { A }{ E_1} \right) P(E_2)P\left( \dfrac { A }{ E_2} \right) }$$
                        $$=\dfrac{{^6C_2}\times\dfrac{1}{2}}{{^6C_2}\times\dfrac{1}{2}+{^7C_4}\times\dfrac{1}{2}}$$
                        $$=\dfrac{7.5}{7.5+17.5}$$
                        $$=0.3$$
Hence, the answer is $$0.3$$
Question 9
1 / -0

Two events $$A$$ and $$B$$ are such that
$$P(A)=\cfrac { 1 }{ 4 } ,P(A|B)=\cfrac { 1 }{ 4 } $$ and $$P(B|A)=\cfrac { 1 }{ 2 } $$
Consider the following statements:
(I) $$P(\overline { A } |\overline { B } )=\cfrac { 3 }{ 4 } $$
(II) $$A$$ and $$B$$ are mutually exclusive
(III) $$P(A|B)+P(A|\overline { B } )=1$$
Then

A
Only (I) is correct

B
Only (I) and (II) are correct

C
Only (I) and (III) are correct

D
Only (II) and (III) are correct

Solution

$$P(\overline{A}|\overline{B})=1-\dfrac{1}{4}=\dfrac{3}{4}$$

for A to B mutually Exclusive , $$P(A|B)$$x$$P(B|A)=1$$ but here $$P(A|B)$$x$$P(B|A)=\dfrac18$$

$$(III)$$ is also incorrect because $$(I)$$ is the correct property that is

$$P(\overline{A}|\overline{B})+P(A|B)=1$$
Question 10
1 / -0

$$A, B$$ and $$C$$ are three mutually exclusive and exhaustive events such that $$P(A) = 2P(B) = 3P(C)$$.
What is $$P(B)$$?

A
$$6/11$$

B
$$6/22$$

C
$$1/6$$

D
$$1/4$$

Solution

Given: A, B, C are mutually exclusive and exhaustive events, so this means
$$P(A\cup B\cup C)=P(S)=1$$
And ,$$P(A\cap B\cap C)=P(A\cap B)=P(B\cap C)=P(A\cap C)=0$$
We know,
$$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\\\implies 1=P(A)+P(B)+P(C)-0-0-0+0\\\implies P(A)+P(B)+P(C)=1$$
Give, $$P(A)=2P(B)=2P(C)$$
Hence, $$P(A)+P(B)+P(C)=1 $$ becomes
$$2P(B)+P(B)+P(B)=1\\\implies 4P(B)=1\\\implies P(B)=\dfrac 14$$

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Re-Attempt Weekly Quiz Competition

Probability Test 15

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Probability Test 15

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Question 1

$$\displaystyle\frac{P(A)-P(B)}{1-P(B)}$$

$$\displaystyle\frac{P(A)-P(A\cap B)}{1-P(B)}$$

$$\displaystyle\frac{P(A)+P(B^c)}{1-P(B)}$$

None of the above

Solution

Question 2

$$1$$

$$0$$

$$\dfrac {1}{2}$$

$$-1$$

Solution

Question 3

Sample size

Population size

The number of people who refused to respond

Where the survey was given.

Solution

Question 4

$$\dfrac{11}{15}$$

$$\dfrac{13}{15}$$

$$\dfrac{7}{15}$$

$$0.65$$

Solution

Question 5

$$\left [-\dfrac{1}{3},\dfrac{1}{2}\right]$$

$$\left (-\dfrac{1}{2},\dfrac{1}{2}\right)$$

$$\left [-\dfrac{1}{2},\dfrac{2}{3}\right]$$

$$\left (-\dfrac{1}{3},\dfrac{2}{3}\right)$$

Solution

Question 6

$$\dfrac {3}{14}$$

$$\dfrac {4}{7}$$

$$\dfrac {2}{7}$$

$$\dfrac {3}{7}$$

Solution

Question 7

$$0.896$$

$$0.791$$

$$0.642$$

$$0.592$$

Solution

Question 8

$$0.3$$

$$0.4$$

$$0.5$$

$$0.6$$

Solution

Question 9

Only (I) is correct

Only (I) and (II) are correct

Only (I) and (III) are correct

Only (II) and (III) are correct

Solution

Question 10

$$6/11$$

$$6/22$$

$$1/6$$

$$1/4$$

Solution

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