Probability Test

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Probability Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

If A and B are two events in a sample space S such that $$P\left ( A \right )\neq 0$$ and $$P\left ( B \right )\neq 0$$, then $$P\left ( A\cap B \right )$$

A
$$P\left ( A \right ).P\left ( B \right )$$

B
$$P\left ( A \right ).P\left ( \dfrac{B}{A} \right )$$

C
$$P\left ( B \right )$$

D
$$P\left ( A \right )$$

Solution

we know that
$$P(\dfrac{B}{A}) = \dfrac{P(A \cap B)}{P(A)}$$
From this, we get $$P(A \cap B) = P(A).P(\dfrac{B}{A})$$
So, B is the correct answer.
Question 2
1 / -0

If A and B are two events such that $$P\left ( A \right )> 0$$ and $$P\left ( B \right )\neq 1$$, then $$P\left ( \bar{A}/\bar{B} \right )= $$

A
$$1-\dfrac{P\left ( A\cup B \right )}{P\left ( \bar{B} \right )}$$

B
$$1-\dfrac{P\left ( A\cup B \right )}{P\left ( B \right )}$$

C
$$\dfrac{1-P\left ( A\cup B \right )}{1-P\left ( B \right )}$$

D
$$\dfrac{P\left ( A\cup B \right )}{P\left ( B \right )}$$

Solution

$$P( {A}'/{B}' )= \dfrac {P({A}'\cap {B}')}{P({B}')}$$

According to De Morgan's Law: $${P({A}'\cap {B}')}={P({A}\cup {B})}'$$

So $$=P( {A}'/{B}' )= \dfrac {P({A}\cup {B})'}{P({B}')}$$
$$=\dfrac {1-P({A}\cup {B})}{1-P{(B)}}$$
Question 3
1 / -0

If A and B are two events in a sample space S such that $$P\left ( A \right )\neq 0$$, then $$P\left ( \frac{B}{A} \right )= $$

A
$$P\left ( A \right ).P\left ( B \right )$$

B
$$\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$$

C
$$\frac{P\left ( A\cap B \right )}{P\left ( A \right )}$$

D
$$P\left ( B \right )$$

Solution

$$P(\dfrac{B}{A})$$ means probability of B given that A has occurred.
So, $$P(\dfrac{B}{A}) = \dfrac{P(A \cap B)}{P(A)}$$
So, C is the correct answer.
Question 4
1 / -0

If $$P\left ( {B \ |A} \right )< P\left ( B \right )$$ the relationship between $$P ( {A \ |B} )$$ and $$P\left ( A \right )$$ is

A
$$P ( {A \ |B} )< \frac{1}{2}P\left ( A \right )$$

B
$$P(A \ | B )> P\left ( A \right )$$

C
$$P(A \ |B )< P\left ( A \right )$$

D
$$P(A \ |B )< 2P\left ( A \right )$$

Solution

First we note that
$$P(\dfrac{B}{A}) = \dfrac{P(A \cap B)}{P(A)} < P(B)$$ (given)
This means that
$$ \dfrac{P(A \cap B)}{P(A)} < P(B) $$
So, $$P(A \cap B) < P(A).P(B) $$
divide throughout by P(B), we get
$$\dfrac{P(A \cap B)}{P(B)} < P(A)$$
Now ,we know that
$$P(\dfrac{A}{B}) = \dfrac{P(A \cap B)}{P(B)}$$
So, we conclude that
$$P(\dfrac{A}{B}) < P(A)$$
So, C is the correct answer.
Question 5
1 / -0

The probability that a bulb produced by a factory will fuse after $$100$$ days of use is $$0.05$$. The probability that out of $$5$$ such bulbs not more than one will fuse after $$100$$ days is

A
$$^{ 5 }{ C }_1{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$

B
$${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }+^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$

C
$$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$

D
$$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }-^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$

Solution

Probability that bulb will fuse after 100 days $$=0.5=\dfrac{1}{20}$$
Probability that bulb will not fuse after 100 days $$=1-\dfrac{1}{20}=\frac{19}{20}$$
Probability that out of 5 such bulbs at least one will fuse $$= 1- $$Probability that out of 5 such bulbs none of them fuse
$$1- ^{5}C_{0}\times (\dfrac{1}{20})^0 \times (\dfrac{19}{20})^5=1-(\dfrac{19}{20})^5$$
Question 6
1 / -0

If A and B are mutually exclusive events with $$P\left ( B \right )\neq 1$$, then $$P\left ( A\mid \bar{B} \right )= $$

A
$$\dfrac{1}{P\left ( B \right )}$$

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

C
$$\dfrac{P\left ( A \right )}{P\left ( B \right )}$$

D
$$\dfrac{P\left ( A \right )}{1-P\left ( B \right )}$$

Solution

If A and B are mutually exclusive events then: $$P(A\cap B)=0$$
$$P(\dfrac{A}{\bar B})=\dfrac{P(A\cap \bar B)}{P(\bar B)}$$
Because A and B are mutually exclusive events so $$P(A\cap \bar B)=P(A)$$
$$P(\dfrac{A}{\bar B})=\dfrac{P(A)}{1-P(B)}$$
Question 7
1 / -0

If $$P\left ( \dfrac{A}{C} \right )> P\left ( \dfrac{B}{C} \right )$$ and $$P\left ( \dfrac{A}{\bar{C}} \right )> P\left ( \dfrac{B}{\bar{C}} \right )$$ then the relationship between $$P(A)$$ and $$P(B)$$ is

A
$$P\left ( A \right )=P\left ( B \right )$$

B
$$P\left ( A \right )\leq P\left ( B \right )$$

C
$$P\left ( A \right )> P\left ( B \right )$$

D
$$P\left ( A \right )\geq P\left ( B \right )$$

Solution

$$P(\dfrac{A}{C})=\dfrac{P(A\cap C)}{P(C)}$$
Now $$0<P(C)<1$$
Also,
$$P(\dfrac{B}{C})=\dfrac{P(B\cap C)}{P(C)}$$
Now
$$P(\dfrac{A}{C})>P(\dfrac{B}{C})$$
$$\dfrac{P(A\cap C)}{P(C)}>\dfrac{P(B\cap C)}{P(C)}$$
Hence
$$P(A\cap C)>P(B\cap C)$$ ...(i)
$$P(A)-P(A\cap C)>P(B)-P(B\cap C)$$
Now $$P(A)>P(B)+P(A\cap C)-P(B\cap C)$$
$$P(A)>P(B)$$
Question 8
1 / -0

A couple has $$2$$ children. The probability that both are boys, if it is known that at least one of the children is a boy is

A
$$\displaystyle \frac{2}{3}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{1}{4}$$

D
$$\displaystyle \frac{3}{4}$$

Solution

Let $$B$$ stand for Boys and $$G$$ for girls.
Then possibility of two children will of the following types
$$BB,BG,GB,GG$$.
Now it is given that atleast $$1$$ is a boy.
Hence total number of possibilities
$$=[BB,BG,GB,GG]-GG$$
$$=BB,BG,GB$$
$$=3$$.
Now in the sample set there is only $$1$$ element where both the children are Boys.
Hence the required probability is
$$=\dfrac{1}{3}$$
Question 9
1 / -0

If $$\overline { E } $$ and $$\overline { F } $$ are the complementary events of events $$E$$ and $$F$$ respectively and if $$0<P(F)<1$$, then

A
$$\displaystyle P\left( \frac { E }{ F } \right) +P\left( \frac { \overline { E } }{ F } \right) =1$$

B
$$\displaystyle P\left( \frac { E }{ F } \right) +P\left( \frac { E }{ \overline { F } } \right) =1$$

C
$$\displaystyle P\left( \frac { \overline { E } }{ F } \right) +P\left( \frac { E }{ \overline { F } } \right) =1$$

D
$$\displaystyle P\left( \frac { E }{ \overline { F } } \right) +P\left( \frac { \overline { E } }{ \overline { F } } \right) =1$$

Solution

$$\displaystyle P\left( \frac { E }{ F } \right) +P\left( \frac { \overline { E } }{ F } \right) $$
$$\displaystyle =\frac { P\left( E\cap F \right) }{ P\left( F \right) } +\frac { P\left( \overline { E } \cap F \right) }{ P\left( F \right) } =\frac { P\left( E\cap F \right) +P\left( \overline { E } \cap F \right) }{ P\left( F \right) } =\frac { P\left\{ \left( E\cap F \right) \cup \left( \overline { E } \cap F \right) \right\} }{ P\left( F \right) } $$ $$[\therefore E\cap F$$ and $$\overline { E } \cap F$$ aredisjoint$$]$$
$$\displaystyle =\frac { P\left\{ \left( E\cup \overline { E } \right) \cap F \right\} }{ P\left( F \right) } =\frac { P(S\cap F\} }{ P\left( F \right) } =\frac { P\left( F \right) }{ P\left( F \right) } =1.$$
Question 10
1 / -0

Each of the urns contains $$4$$ white and $$6$$ blackballs. The (n + 1)th urn contains $$5$$ white and $$5$$black balls. One of the (n+1) urns is chosen atrandom and two balls are drawn from it withoutreplacement and both the balls turn out to beblack. Then the probability that the (n+1)th urn was chosen to draw the balls is $$1/16$$, the value of n is

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

$$P(\dfrac { { (n+1) }^{ th }\quad urn }{ both\quad black } )=\quad \dfrac { P({ (n+1) }^{ th }urn\cap both\quad black) }{ P(both\quad black) } \\ =\dfrac { \dfrac { 1 }{ n+1 } \dfrac { _{ 2 }^{ 5 }{ C } }{ _{ 2 }^{ 10 }{ C } } }{ \dfrac { 1.n }{ n+1 } \dfrac { _{ 2 }^{ 6 }{ C } }{ _{ 2 }^{ 10 }{ C } } +\dfrac { 1 }{ n+1 } \dfrac { _{ 2 }^{ 5 }{ C } }{ _{ 2 }^{ 10 }{ C } } } =\dfrac { _{ 2 }^{ 5 }{ C } }{ n_{ 2 }^{ 6 }{ C }+_{ 2 }^{ 5 }{ C } } =\dfrac { 10 }{ 15n+10 } =\dfrac { 1 }{ 16 }$$

$$ \\ \Rightarrow 160=15n+10\\ \Rightarrow n=10$$

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Probability Test - 18

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Probability Test - 18

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Question 1

$$P\left ( A \right ).P\left ( B \right )$$

$$P\left ( A \right ).P\left ( \dfrac{B}{A} \right )$$

$$P\left ( B \right )$$

$$P\left ( A \right )$$

Solution

Question 2

$$1-\dfrac{P\left ( A\cup B \right )}{P\left ( \bar{B} \right )}$$

$$1-\dfrac{P\left ( A\cup B \right )}{P\left ( B \right )}$$

$$\dfrac{1-P\left ( A\cup B \right )}{1-P\left ( B \right )}$$

$$\dfrac{P\left ( A\cup B \right )}{P\left ( B \right )}$$

Solution

Question 3

$$P\left ( A \right ).P\left ( B \right )$$

$$\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$$

$$\frac{P\left ( A\cap B \right )}{P\left ( A \right )}$$

$$P\left ( B \right )$$

Solution

Question 4

$$P ( {A \ |B} )< \frac{1}{2}P\left ( A \right )$$

$$P(A \ | B )> P\left ( A \right )$$

$$P(A \ |B )< P\left ( A \right )$$

$$P(A \ |B )< 2P\left ( A \right )$$

Solution

Question 5

$$^{ 5 }{ C }_1{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$

$${ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }+^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$

$$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }$$

$$1-{ \left( \cfrac { 19 }{ 20 } \right) }^{ 5 }-^{ 5 }{ C }_{1}{ \left( \cfrac { 19 }{ 20 } \right) }^{ 4 }{ \left( \cfrac { 1 }{ 20 } \right) }^{ 1 }$$

Solution

Question 6

$$\dfrac{1}{P\left ( B \right )}$$

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

$$\dfrac{P\left ( A \right )}{P\left ( B \right )}$$

$$\dfrac{P\left ( A \right )}{1-P\left ( B \right )}$$

Solution

Question 7

$$P\left ( A \right )=P\left ( B \right )$$

$$P\left ( A \right )\leq P\left ( B \right )$$

$$P\left ( A \right )> P\left ( B \right )$$

$$P\left ( A \right )\geq P\left ( B \right )$$

Solution

Question 8

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{3}{4}$$

Solution

Question 9

$$\displaystyle P\left( \frac { E }{ F } \right) +P\left( \frac { \overline { E } }{ F } \right) =1$$

$$\displaystyle P\left( \frac { E }{ F } \right) +P\left( \frac { E }{ \overline { F } } \right) =1$$

$$\displaystyle P\left( \frac { \overline { E } }{ F } \right) +P\left( \frac { E }{ \overline { F } } \right) =1$$

$$\displaystyle P\left( \frac { E }{ \overline { F } } \right) +P\left( \frac { \overline { E } }{ \overline { F } } \right) =1$$

Solution

Question 10

$$10$$

$$11$$

$$12$$

$$13$$

Solution

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