Probability Test

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Probability Test - 21

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Weekly Quiz Competition

Question 1
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If two events $$A$$ and $$B$$ are such that $$P\left( \overline { A } \right) =0.3,P\left( B \right) =0.5$$ and $$P\left( A\cap B \right) =0.3$$, then $$\displaystyle P\left( \frac { B }{ A\cup \overline { B } } \right) $$ is equal to

A
$$\displaystyle \frac { 5 }{ 8 } $$

B
$$\displaystyle \frac { 7 }{ 8 } $$

C
$$\displaystyle \frac { 3 }{ 8 } $$

D
$$\displaystyle \frac { 1 }{ 2 } $$

Solution

We have, $$P\left( A\cup \overline { B } \right) =P\left( A \right) +P\left( \overline { B } \right) -P\left( A\cap \overline { B } \right) $$

$$\\ =\left( 1-P\left( \overline { A } \right) \right) +\left( 1-P\left( B \right) \right) -\left( P\left( A \right) -P\left( A\cap B \right) \right) \\ =\left( 1-0.3 \right) +\left( 1-0.5 \right) -\left( 0.7-0.3 \right) =0.8$$

$$\displaystyle \therefore P\left( \frac { B }{ A\cup \overline { B } } \right) =\frac { P\left( B\cap \left( A\cup \overline { B } \right) \right) }{ P\left( A\cup \overline { B } \right) } =\frac { P\left( \left( B\cap A \right) \cup \left( B\cap \overline { B } \right) \right) }{ P\left( A\cup \overline { B } \right) } $$

$$\displaystyle =\frac { P\left( A\cap B \right) }{ P\left( A\cup \overline { B } \right) } =\frac { 0.3 }{ 0.8 } =\frac { 3 }{ 8 } $$
Question 2
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The chances of defective screws in three boxes $$A,B,C$$ are $$\displaystyle \frac { 1 }{ 5 } ,\frac { 1 }{ 6 } ,\frac { 1 }{ 7 } $$ respectively. A box is selected at random and a screw drawn from it at random from it is found defective. The probability that it came from the box $$A$$ is

A
$$\displaystyle \frac { 16 }{ 29 } $$

B
$$\displaystyle \frac { 1 }{ 15 } $$

C
$$\displaystyle \frac { 27 }{ 59 } $$

D
$$\displaystyle \frac { 42 }{ 107 } $$

Solution

Let $$A$$ be the event that a screw drawn from the box and similar is for events $$B$$ and $$C$$.
Let $$D$$ be the event that a defective screw is drawn. Then
$$P\left( D \right) =P\left( A\cap D \right) +P\left( B\cap D \right) +P\left( C\cap D \right) $$
$$\displaystyle =\dfrac { 1 }{ 3 } .\dfrac { 1 }{ 5 } +\dfrac { 1 }{ 3 } .\dfrac { 1 }{ 6 } +\dfrac { 1 }{ 3 } .\dfrac { 1 }{ 7 } $$
$$\displaystyle P\left( \dfrac { A }{ D } \right) =\dfrac { P\left( A\cap D \right) }{ P\left( D \right) } =\dfrac { \dfrac { 1 }{ 3 } .\dfrac { 1 }{ 5 } }{ \dfrac { 1 }{ 15 } +\dfrac { 1 }{ 18 } +\dfrac { 1 }{ 21 } } =\dfrac { 42 }{ 107 } $$
Question 3
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There are two bags, one of which contains three black and four white balls while the other contains four black and three white balls. A die is cast: if the face 1 or 3 turns up, a ball is taken from the first bag; and if any other face turns up, a ball is chosen from the second bag. Find the probability of choosing a black ball.

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

B
$$\displaystyle \frac{11}{21}$$

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

D
$$\displaystyle \frac{10}{21}$$

Solution

Let $$\displaystyle E_{1}$$ be the event that a ball is drawn from first bag.

$$\displaystyle E_{2}$$ be the event that a ball is drawn form the second bag.

$$E$$ be the event a black ball is chosen.

$$P(E_1)=\dfrac{2}{6}=\dfrac{1}{3}$$

$$P(E_2)=\dfrac{4}{6}=\dfrac{2}{3}$$

Now, $$\displaystyle P\left ( E \right )=P\left ( E_{1} \right )P\left (
E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{2} \right )$$

$$=\displaystyle \frac{1}{3}.\frac{3}{7}+\frac{2}{5}.\frac{4}{7}$$

$$=\dfrac{11}{21}.$$
Question 4
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Two coins are tossed. What is the conditional probability that two heads result, given that there is at least one head ?

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

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

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

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

Solution

Let A be the event that two heads result and B the event that there is at least one head.
If S denote the sample space, then $$\displaystyle S=\left \{\left ( H,H \right ),\left ( H,T \right )\left ( T,H \right ) \left ( T,T \right )\right \}$$
$$\displaystyle A=\left \{ \left ( H,H \right )\right \}$$
$$\displaystyle B=\left \{ \left ( H,H \right ),\left ( H,T \right )\left ( T,H \right ) \right \}$$
So, $$\displaystyle A\cap B=\left \{ H,H \right \}$$
$$\displaystyle P\left ( B \right )=\frac{3}{4},$$
$$\displaystyle P\left ( A\cap B \right )=\frac{1}{4}$$
Hence $$\displaystyle P\left ( A|B \right )=\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$$
$$=\dfrac{1/4}{3/4}=\dfrac{1}{3}$$
Question 5
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Consider a family with two children. Assume that each child is as likely to be a boy as it is to be a girl. Find the conditional probability that both children are boys, given that one child is a boy

A
$$1/2$$

B
$$1/4$$

C
$$3/4$$

D
$$2/3$$

Solution

$$P(boy) = 1/2 = P(girl) $$
$$ P(2\ boys) = (\cfrac{1}{2})^2 = \cfrac{1}{4}$$
$$P(2\ boys | 1\ boy) = P(2\ boy)/P(1\ boy) = 1/2 $$
Question 6
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In a bolt factory, machines A, B and C manufacture 25%, 35%, 40% respectively. Of the total of their output 5, 4 and 2% are defective. A bolt is drawn and is found to be defective. What are the probabilities that it was manufactured by the machines A, B and C?

A
$$\displaystyle \frac{25}{69}, \frac{28}{69}, \frac{16}{69}.$$

B
$$\displaystyle \frac{25}{69}, \frac{27}{69}, \frac{17}{69}.$$

C
$$\displaystyle \frac{28}{69}, \frac{25}{69}, \frac{16}{69}.$$

D
$$\displaystyle \frac{27}{69}, \frac{26}{69}, \frac{16}{69}.$$

Solution

Here $$\displaystyle P\left( A \right) =\frac { 25 }{ 100 } ,P\left( B \right) =\frac { 35 }{ 100 } ,P\left( C \right) =\frac { 40 }{ 100 } $$

$$\displaystyle P\left( \frac { D }{ A } \right) =\frac { 5 }{ 100 } ,P\left( \frac { D }{ B } \right) =\frac { 4 }{ 100 } ,P\left( \frac { D }{ C } \right) =\frac { 2 }{ 100 } $$

where $$D$$ denotes defective bolts
Now $$\displaystyle P\left( D \right) =P\left( A \right) .P\left( \frac { D }{ A } \right) +P\left( B \right) .P\left( \frac { D }{ B } \right) +P\left( C \right) .P\left( \frac { D }{ C } \right) $$

$$\displaystyle =\frac { 25 }{ 100 } .\frac { 5 }{ 100 } +\frac { 35 }{ 100 } .\frac { 4 }{ 100 } +\frac { 40 }{ 100 } .\frac { 2 }{ 100 } =0.0345$$

$$\displaystyle P\left( \frac { A }{ D } \right) =\frac { P\left( A \right) .P\left( \frac { D }{ A } \right) }{ P\left( A \right) .P\left( \frac { D }{ A } \right) +P\left( B \right) .P\left( \frac { D }{ B } \right) +P\left( C \right) .P\left( \frac { D }{ C } \right) } $$

$$\displaystyle =\frac { \frac { 25 }{ 100 } .\frac { 5 }{ 100 } }{ 0.0345 } =\frac { 25 }{ 69 } $$

$$\displaystyle P\left( \frac { B }{ D } \right) =\frac { P\left( B \right) .P\left( \frac { D }{ B } \right) }{ P\left( A \right) .P\left( \frac { D }{ A } \right) +P\left( B \right) .P\left( \frac { D }{ B } \right) +P\left( C \right) .P\left( \frac { D }{ C } \right) } $$

$$\displaystyle =\frac { \frac { 35 }{ 100 } .\frac { 4 }{ 100 } }{ 0.0345 } =\frac { 28 }{ 69 } $$
$$\displaystyle P\left( \frac { C }{ D } \right) =\frac { P\left( C \right) .P\left( \frac { D }{ C } \right) }{ P\left( A \right) .P\left( \frac { D }{ A } \right) +P\left( B \right) .P\left( \frac { D }{ B } \right) +P\left( C \right) .P\left( \frac { D }{ C } \right) } $$

$$\displaystyle =\frac { \frac { 40 }{ 100 } .\frac { 2 }{ 100 } }{ 0.0345 } =\frac { 16 }{ 69 } $$
Question 7
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If in Q. 104, we are told that a white ball has been drawn, find the probability that it was drawn from the first urn.

A
$$\displaystyle \frac{5}{9}.$$

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

C
$$\displaystyle \frac{2}{9}.$$

D
$$\displaystyle \frac{7}{9}.$$

Solution

Here we have to find $$\displaystyle P\left ( A_{1}/B \right ).$$
By Baye's theorem
$$\displaystyle P\left ( A_{1}/B \right )= \frac{P\left ( A_{1} \right )P\left ( B/A_{1} \right )}{P\left ( A_{1} \right )P\left ( B/A_{1} \right )+P\left ( A_{2} \right )P\left ( B/A_{2} \right )+P\left ( A_{3} \right )P\left ( B/A_{3} \right )}$$
$$\displaystyle = \dfrac{\dfrac{1}{3}.\dfrac{2}{5}}{\dfrac{3}{5}},$$
$$\displaystyle = \frac{2}{9}.$$
Question 8
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A is known to tell the truth in $$5$$ cases out of $$6$$ and he states that a white ball was drawn from a bag containing $$8$$ black and $$1$$ white ball. The probability that the white ball was drawn, is

A
$$\displaystyle \frac { 7 }{ 13 } $$

B
$$\displaystyle \frac { 5 }{ 13 } $$

C
$$\displaystyle \frac { 9 }{ 13 } $$

D
None of these

Solution

Let $$W$$ denote the event that $$A$$ draws a white ball and $$T$$ the event that $$A$$ speak truth.
In the usual notations, we are given that
$$\displaystyle P\left( W \right) =\frac { 1 }{ 9 } ,P\left( \frac { T }{ w } \right) =\frac { 5 }{ 6 } $$
so that $$\displaystyle P\left( \overline { W } \right) =1-\frac { 1 }{ 9 } =\frac { 8 }{ 9 } ,P\left( \frac { T }{ \overline { W } } \right) =1-\frac { 5 }{ 6 } =\frac { 1 }{ 6 } $$.
Using Baye's theorem required probability is given by
$$\displaystyle P\left( \frac { W }{ T } \right) =\frac { P\left( W\cap T \right) }{ P\left( T \right) } =\frac { P\left( W \right) P\left( \frac { T }{ w } \right) }{ P\left( W \right) P\left( \frac { T }{ w } \right) +P\left( \overline { W } \right) P\left( \frac { T }{ \overline { W } } \right) } $$
$$\displaystyle =\frac { \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 } }{ \dfrac { 1 }{ 9 } \times \dfrac { 5 }{ 6 } +\dfrac { 8 }{ 9 } \times \dfrac { 1 }{ 6 } } =\frac { 5 }{ 13 } $$
Question 9
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If two events A and B such that $$P(A')=0.3, P(B)=0.5$$ and $$P(A\cap B)=0.3$$, then $$P(B|A \cup B')$$ is

A
$$3/8$$

B
$$2/3$$

C
$$5/6$$

D
$$1/4$$

Solution

We have $$P(A\cup B')$$
$$=P(A)+P(B')-P(A\cap B')$$
$$=[1-P(A')]+[1-P(B)]-[P(A)-P(A\cap B)]$$
$$=(1-0.3)+(1-0.5)-(0.7-0.3)=0.8$$.

Now, $$P(B|A \cup B')=\dfrac {P[B\cap A(\cup B')]}{P(A\cup B')}$$

$$=\dfrac {P[(B\cap A)\cup (B\cap B')]}{P(A\cup B')}$$

$$=\dfrac {P(A\cap B)}{P(A\cup B')}=\dfrac {0.3}{0.8}=\dfrac {3}{8}$$
Question 10
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If $$A$$ and $$B$$ are mutually exclusive events, then $$\displaystyle P\left ( A \right )> 0$$ and $$\displaystyle P\left ( B \right )\neq 1,$$ $$\displaystyle P\left ( \overline{A/B} \right )$$ is equal to

A
$$\displaystyle 1-P\left ( A/B \right )$$

B
$$\displaystyle 1-p\left ( \bar{A}/b \right )$$

C
$$\displaystyle \frac{1-P\left ( A \cup B \right )}{P\left ( \bar{B} \right )}$$

D
$$\displaystyle \frac{P\left ( \bar{A} \right )}{P\left ( \bar{B} \right )}.$$

Solution

$$\displaystyle P\left ( \overline{A/B} \right )=P\left ( \bar{A}/\bar{B} \right )=\frac{P\left (

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

\overline{A\cup B} \right )}{P\left ( \overline{B} \right

)}=\frac{1-P\left ( A\cup B \right )}{P\left ( \overline{B} \right )}$$

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Probability Test - 21

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Probability Test - 21

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SHARING IS CARING

Question 1

$$\displaystyle \frac { 5 }{ 8 } $$

$$\displaystyle \frac { 7 }{ 8 } $$

$$\displaystyle \frac { 3 }{ 8 } $$

$$\displaystyle \frac { 1 }{ 2 } $$

Solution

Question 2

$$\displaystyle \frac { 16 }{ 29 } $$

$$\displaystyle \frac { 1 }{ 15 } $$

$$\displaystyle \frac { 27 }{ 59 } $$

$$\displaystyle \frac { 42 }{ 107 } $$

Solution

Question 3

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

$$\displaystyle \frac{11}{21}$$

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

$$\displaystyle \frac{10}{21}$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$1/2$$

$$1/4$$

$$3/4$$

$$2/3$$

Solution

Question 6

$$\displaystyle \frac{25}{69}, \frac{28}{69}, \frac{16}{69}.$$

$$\displaystyle \frac{25}{69}, \frac{27}{69}, \frac{17}{69}.$$

$$\displaystyle \frac{28}{69}, \frac{25}{69}, \frac{16}{69}.$$

$$\displaystyle \frac{27}{69}, \frac{26}{69}, \frac{16}{69}.$$

Solution

Question 7

$$\displaystyle \frac{5}{9}.$$

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

$$\displaystyle \frac{2}{9}.$$

$$\displaystyle \frac{7}{9}.$$

Solution

Question 8

$$\displaystyle \frac { 7 }{ 13 } $$

$$\displaystyle \frac { 5 }{ 13 } $$

$$\displaystyle \frac { 9 }{ 13 } $$

None of these

Solution

Question 9

$$3/8$$

$$2/3$$

$$5/6$$

$$1/4$$

Solution

Question 10

$$\displaystyle 1-P\left ( A/B \right )$$

$$\displaystyle 1-p\left ( \bar{A}/b \right )$$

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

$$\displaystyle \frac{P\left ( \bar{A} \right )}{P\left ( \bar{B} \right )}.$$

Solution

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