Probability Test

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Probability Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

For any two independent events $${ E }_{ 1 }$$ and $${ E }_{ 2 }$$, $$P\left\{ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \overline { { E }_{ 1 } } \right) \cap \left( \overline { { E }_{ 2 } } \right) \right\} $$ is

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

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

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

D
None of these

Solution

We have $$=P\left\{ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \overline { { E }_{ 1 } } \right) \cap \left( \overline { { E }_{ 2 } } \right) \right\} $$
$$=P\left\{ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \left( \overline { { E }_{ 1 } } \right) \cap \left( \overline { { E }_{ 2 } } \right) \right) \right\} =P\left\{ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \overline { { E }_{ 1 }\cup { E }_{ 2 } } \right) \right\} $$
$$\displaystyle =P\left( \phi \right) =0\le \frac { 1 }{ 4 } $$
Question 2
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Let $$A$$ and $$B$$ be two events such that $$P(\displaystyle \overline{A\cup B})=\frac{1}{6}, P(A\displaystyle \cap B)=\frac{1}{4}$$ and $$P(\displaystyle \overline{A})=\frac{1}{4}$$, then events $$A$$ and $$B$$ are

A
equally likely, but not independent

B
equally likely and mutually exclusive

C
mutually exclusive and independent

D
independent but not equally likely

Solution

Given, $$P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } $$
$$ P\left( A\cap B \right) =\dfrac { 1 }{ 4 } ,P\left( \overline { A } \right) =\dfrac { 1 }{ 4 } \\ \because P\left( \overline { A\cup B } \right) =\dfrac { 1 }{ 6 } $$

$$\Rightarrow 1-P\left( A\cup B \right) =\dfrac { 1 }{ 6 } $$

$$ \Rightarrow P\left( A\cup B \right) =1-\dfrac { 1 }{ 6 } =\dfrac { 5 }{ 6 } $$

$$ \Rightarrow P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) =\dfrac { 5 }{ 6 } $$

$$ \Rightarrow P\left( B \right) =\dfrac { 5 }{ 6 } -\dfrac { 3 }{ 4 } +\dfrac { 1 }{ 4 } \quad \quad \quad \quad \quad \quad \quad \because P\left( \overline { A } \right) =\dfrac { 1 }{ 4 } \Rightarrow P\left( A \right) =\dfrac { 3 }{ 4 } $$

$$ \Rightarrow P\left( B \right) =\dfrac { 1 }{ 3 }$$

Hence, $$ \\ \\ P\left( A\cap B \right) =P\left( A \right) P\left( B \right)$$ but $$P\left( A \right) \neq P\left( B \right) $$
Question 3
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If $$A$$ and $$B$$ are two events such that $$P\left ( A\cup B \right )= 0.65$$ and $$P\left ( A\cap B \right )= 0.15$$, then $$P\left ( \bar{A} \right )+P\left ( \bar{B} \right )=$$

A
$$0.6$$

B
$$0.8$$

C
$$1.2$$

D
$$1.4$$

Solution

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B)=1-P\left ( \bar{A} \right )- 1 - P\left ( \bar{B} \right )-P(A\cap B)$$
$$0.65=2-P\left ( \bar{A} \right )- P\left ( \bar{B} \right )-0.15$$ Given: $$\left [ {P(A\cup B)=0.65} ; {P(A\cap B)=0.15} \right ] $$
$$P\left ( \bar{A} \right )+ P\left ( \bar{B} \right )=2-0.65-0.15=1.2$$
Question 4
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An urn contains $$6$$ white and $$4$$ black balls. A fair die whose faces are numbered from $$1$$ to $$6$$ is rolled and number of balls equal to that of the number appearing on the die is drawn from the urn at random. The probability that all those are white is

A
$$\displaystyle \frac{1}{5}$$

B
$$\displaystyle \frac{2}{5}$$

C
$$\displaystyle \frac{3}{5}$$

D
$$\displaystyle \frac{4}{5}$$

Solution

Required probability $$=$$ probability of drawing 1 White or 2 or 3 or 4 or 5 or 6 White $$=(\dfrac { 1 }{ 6 } *\dfrac { _{ 1 }^{ 6 }{ C } }{ _{ 1 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 2 }^{ 6 }{ C } }{ _{ 2 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 3 }^{ 6 }{ C } }{ _{ 3 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 4 }^{ 6 }{ C } }{ _{ 4 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 5 }^{ 6 }{ C } }{ _{ 5 }^{ 10 }{ C } } )=\dfrac { 251 }{ 1260 } $$
This is closest to $$\dfrac { 1 }{ 5 } $$
Question 5
1 / -0

Let $$A,B,C$$ be three events such that $$P\left( A \right) =0.3,P\left( B \right) =0.4,P\left( C \right) =0.8,P\left( A\cap B \right) =0.08,$$
$$P\left( A\cap C \right) =0.28,P\left( A\cap B\cap C \right) =0.09$$. If $$P\left( A\cup B\cup C \right) \ge 0.75$$, then

A
$$0.23\le P\left( B\cap C \right) \le 0.48$$

B
$$0.23\le P\left( B\cap C \right) \le 0.75$$

C
$$0.48\le P\left( B\cap C \right) \le 0.75$$

D
None of these

Solution

Since $$P\left( A\cup B\cup C \right) \ge 0.75$$, therefore
$$0.75\le P\left( A\cup B\cup C \right) \le 1\\ \Rightarrow 0.75\le P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) -P\left( B\cap C \right) \\ -P\left( A\cap C \right) +P\left( A\cap B\cap C \right) \le 1\\ \Rightarrow 0.75\le 0.3+0.4+0.8-0.08-P\left( B\cap C \right) -0.28+0.09\le 1\\ \Rightarrow 0.75\le 1.23-P\left( B\cap C \right) \le 1\\ \Rightarrow -0.48\le -P\left( B\cap C \right) \le -0.23\\ \Rightarrow 0.23\le P\left( B\cap C \right) \le 0.48$$
Question 6
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A coin is tossed until a head appears or it has been tossed 3 times. Given that head does not appear on the first toss, the probability that the coin is tossed 3 times is

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

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

C
$$\displaystyle \frac{1}{8}$$

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

Solution

Possible cases: (i) TH, (ii) TTH, (iii) TTT.
(i): $$\dfrac { 1 }{ 2 } *\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 4 } $$

(ii): $$\dfrac { 1 }{ 2 } *\dfrac { 1 }{ 2 } *\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 8 } $$

(iii): $$\dfrac { 1 }{ 2 } *\dfrac { 1 }{ 2 } *\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 8 } $$

Hence, probability $$= \dfrac { \dfrac { 1 }{ 8 } +\dfrac { 1 }{ 8 } }{ \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 8 } +\dfrac { 1 }{ 8 } } =\dfrac { \dfrac { 1 }{ 4 } }{ \dfrac { 1 }{ 2 } } =\dfrac { 1 }{ 2 } $$
Question 7
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A and B be two independent events of a sample space such that $$P(A) = 0.2, P(B) = 0.5$$
LIST-I LIST-II
A) $$P(B/A)$$ 1$$. 0.2$$
B) $$P(A/B)$$ 2 $$. 0.1$$
C) $$P(A\cap B)$$ 3$$ . 0.3$$
D) $$P(A\cup B)$$ 4 $$. 0.6$$
5$$. 0.5$$
The correct match for List-I from List-II
A B C D

A
$$4 ,5, 3, 1$$

B
$$5, 1, 2, 4$$

C
$$3, 1 ,2, 4$$

D
$$2, 1 ,4, 5$$

Solution

$$A$$ and $$B$$ both are independent event.

A)$$P(B/A)=P(B)=0.5$$

B)$$P(A/B)=P(A)=0.2$$

C)$$P(A\cup B)=P(A)\times P(B)=0.5\times 0.2=0.1$$

D)$$P(A\cap B)=P(A)+P(B)-P(A)\times P(B)=0.5+0.2-(0.5\times 0.2)=0.6$$

$$\therefore\ A B C D \rightarrow 5 1 2 4$$
Question 8
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A letter is taken out at random from the word ASSISTANT and an other from STATISTICS. The probability that they are the same letters is

A
$$\displaystyle \frac{13}{90}$$

B
$$\displaystyle \frac{17}{90}$$

C
$$\displaystyle \frac{19}{90}$$

D
$$\displaystyle \frac{15}{90}$$

Solution

The common letters are $$A,S,T,I$$
$$P(S) = \dfrac{3}{9}\times\dfrac{3}{10}$$

$$P(A) = \dfrac{2}{9}\times\dfrac{1}{10}$$

$$P(T) = \dfrac{2}{9}\times\dfrac{3}{10}$$

$$P(I) = \dfrac{1}{9}\times\dfrac{2}{10}$$
$$\Rightarrow P = P(A)+P(S)+P(T)+P(I)=\dfrac{9}{90}$$
Question 9
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If $${ A }_{ 1 },{ A }_{ 2 },{ A }_{ 3 },...,{ A }_{ n }$$ are any $$n$$ events, then

A
$$P\left( { A }_{ 1 }\cup { A }_{ 2 }\cup ...\cup { A }_{ n } \right) =P\left( { A }_{ 1 } \right) +P\left( { A }_{ 2 } \right) +...+P\left( { A }_{ n } \right) $$

B
$$P\left( { A }_{ 1 }\cup { A }_{ 2 }\cup ...\cup { A }_{ n } \right) >P\left( { A }_{ 1 } \right) +P\left( { A }_{ 2 } \right) +...+P\left( { A }_{ n } \right) $$

C
$$P\left( { A }_{ 1 }\cup { A }_{ 2 }\cup ...\cup { A }_{ n } \right) \le P\left( { A }_{ 1 } \right) +P\left( { A }_{ 2 } \right) +...+P\left( { A }_{ n } \right) $$

D
None of these

Solution

For any two events $$A$$ and $$B$$, we have
$$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \Rightarrow P\left( A\cup B \right) \le P\left( A \right) +P\left( B \right) $$
Using principle of mathematical induction it can be easily established that
$$\displaystyle P\left( \bigcup _{ i=1 }^{ n }{ { A }_{ i } } \right) \le \sum _{ i=1 }^{ n }{ P\left( { A }_{ i } \right) } $$
Question 10
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$$A,B,C$$ are any three events. If $$P\left( S \right) $$ denotes the probability of $$S$$ happening, then $$P\left( A\cap \left( B\cup C \right) \right) =$$

A
$$P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) -P\left( A\cap C \right) $$

B
$$P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( B \right) P\left( C \right) $$

C
$$P\left( A\cap B \right) +P\left( A\cap C \right) -P\left( A\cap B\cap C \right) $$

D
None of these

Solution

$$P\left( A\cap \left( B\cup C \right) \right) =P\left[ \left( A\cap B \right) \cup \left( A\cap C \right) \right] \\ =P\left( A\cap B \right) +P\left( A\cap C \right) -P\left[ \left( A\cap B \right) \cap \left( A\cap C \right) \right] \\ =P\left( A\cap B \right) +P\left( A\cap C \right) -P\left( A\cap B\cap C \right) $$

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Re-Attempt Weekly Quiz Competition

Probability Test - 23

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Probability Test - 23

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SHARING IS CARING

Question 1

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

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

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

None of these

Solution

Question 2

equally likely, but not independent

equally likely and mutually exclusive

mutually exclusive and independent

independent but not equally likely

Solution

Question 3

$$0.6$$

$$0.8$$

$$1.2$$

$$1.4$$

Solution

Question 4

$$\displaystyle \frac{1}{5}$$

$$\displaystyle \frac{2}{5}$$

$$\displaystyle \frac{3}{5}$$

$$\displaystyle \frac{4}{5}$$

Solution

Question 5

$$0.23\le P\left( B\cap C \right) \le 0.48$$

$$0.23\le P\left( B\cap C \right) \le 0.75$$

$$0.48\le P\left( B\cap C \right) \le 0.75$$

None of these

Solution

Question 6

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

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

$$\displaystyle \frac{1}{8}$$

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

Solution

Question 7

$$4 ,5, 3, 1$$

$$5, 1, 2, 4$$

$$3, 1 ,2, 4$$

$$2, 1 ,4, 5$$

Solution

Question 8

$$\displaystyle \frac{13}{90}$$

$$\displaystyle \frac{17}{90}$$

$$\displaystyle \frac{19}{90}$$

$$\displaystyle \frac{15}{90}$$

Solution

Question 9

$$P\left( { A }_{ 1 }\cup { A }_{ 2 }\cup ...\cup { A }_{ n } \right) =P\left( { A }_{ 1 } \right) +P\left( { A }_{ 2 } \right) +...+P\left( { A }_{ n } \right) $$

$$P\left( { A }_{ 1 }\cup { A }_{ 2 }\cup ...\cup { A }_{ n } \right) >P\left( { A }_{ 1 } \right) +P\left( { A }_{ 2 } \right) +...+P\left( { A }_{ n } \right) $$

$$P\left( { A }_{ 1 }\cup { A }_{ 2 }\cup ...\cup { A }_{ n } \right) \le P\left( { A }_{ 1 } \right) +P\left( { A }_{ 2 } \right) +...+P\left( { A }_{ n } \right) $$

None of these

Solution

Question 10

$$P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) -P\left( A\cap C \right) $$

$$P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( B \right) P\left( C \right) $$

$$P\left( A\cap B \right) +P\left( A\cap C \right) -P\left( A\cap B\cap C \right) $$

None of these

Solution

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