Probability Test

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Probability Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

Suppose $$A$$ and $$B$$ are two events with $$P(A)=0.5$$ and $$P(A \cup B) =0.8$$. Let $$P(B)=p$$ if $$A$$ and $$B$$ are mutually exclusive and $$P(B)=q$$ if $$A$$ and $$B$$ are independent events, then the value of $$q/p$$ is

A
$$2$$

B
$$1$$

C
$$3$$

D
$$0.1$$

Solution

$$P(A)=0.5$$ and $$P(AB)=0.8$$
When A and B are independent events,
$$ P(A\cap B) = P(A) \times P(B)$$
$$\Rightarrow P(A) + P(B) - P(A \cup B) = 0.5 \times q $$
$$\Rightarrow 0.5 + q -0.8 =0.5q $$
$$\Rightarrow 0.5 q =0.3 $$
$$\Rightarrow q =0.6 $$

When A and B are mutually exclusive events
$$ P(A \cap B) =0 $$
$$ P(A) + P(B) = P(A \cup B) $$
$$ 0.5 + p =0.8 $$
$$ \Rightarrow p =0.3 $$
Hence the ratio of $$q$$ and $$p$$ is 2
Question 2
1 / -0

Let A and B be two events. Suppose $$P(A)=0.4, P(B)=p$$, and $$P(A \cup B) = 0.7 $$. The value of $$p$$ for which A and B are independent is

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

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

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

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

Solution

Given, $$P(A)=0.4,P(B)=p$$, and $$P(A\;U\;B)=0.7$$
A and B are independent events.
$$ \Rightarrow P(A\;U\;B)=P(A) +P(B) -P(A) \times P(B) $$
$$0.7 = 0.4 + p - 0.4 p$$
$$p = \displaystyle \frac{1}{2} $$
Question 3
1 / -0

If $$P (A \cup B) = \dfrac{2}{3} , P (A \cap B) = \dfrac{1}{6} $$ and $$ P(A) = \dfrac{1}{3}$$ then -

A
$$A$$ and $$B$$ are independent events

B
$$A$$ and $$B$$ are disjoint events

C
$$A$$ and $$B$$ are dependent events

D
None of these

Solution

Given: $$P\left( A\bigcup { B } \right) =\cfrac { 2 }{ 3 } $$
$$P\left( A\bigcap { B } \right) =\cfrac { 1 }{ 6 } $$
$$P\left( A \right) =\cfrac { 1 }{ 3 } $$
Using the formula:
$$P\left( A\bigcup { B } \right) =P(A)+P(B)-P\left( A\bigcap { B } \right) $$
$$\therefore P(B)=P\left( A\bigcup { B } \right) -P(A)+P\left( A\bigcap { B } \right) $$
$$=\cfrac { 2 }{ 3 } -\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 } $$
$$=\cfrac { 4-2+1 }{ 6 } =\cfrac { 3 }{ 6 } =\cfrac { 1 }{ 2 } $$
$$\therefore P(B)=\cfrac { 1 }{ 2 } \quad P(A)\neq P(B)$$
and $$P(A)=\cfrac { 1 }{ 3 } $$
Thus A and B are independent events.
Question 4
1 / -0

If odds against solving a question by three students are 2:1, 5:2 and 5:3 , respectively, then probability that the question is solved only by one student is

A
31/56

B
24/56

C
25/56

D
none of these

Solution

Given, $$P(A)=\dfrac{1}{3}, P(B)=\dfrac{2}{7}, P(C)=\dfrac{3}{8}$$

Probability that the question is solved only by one student is
$$=P(A)\times P(B)^c\times P(C)^c+P(A)^c\times P(B)\times P(C)^c + P(A)^c\times P(B)^c\times P(C)$$
$$=\left(\dfrac{1}{3}\times\dfrac{5}{7}\times\dfrac{5}{8}\right) +\left(\dfrac{2}{3}\times\dfrac{2}{7}\times\dfrac{5}{8}\right) +\left(\dfrac{2}{3}\times\dfrac{5}{7}\times\dfrac{3}{8}\right) = \dfrac{25}{56}$$
Question 5
1 / -0

If $$P( A \cup B )=\displaystyle\frac{3}{4}$$ and $$P( \bar A)= \displaystyle\frac{2}{3}$$, then find the value of $$P( \bar A \cap B )$$.

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

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

C
$$\displaystyle\frac{5}{12}$$

D
None of these

Solution

Since $$ \bar A \cap B $$ and A are mutually exclusive events such that
$$ A \cup B = (\bar A \cap B) \cup A$$
or $$ P( A \cup B )=P( \bar A \cap B )+P(A)$$
or $$\displaystyle\frac{3}{4}=P(\bar A \cap B ) + 1 - \displaystyle\frac{2}{3}$$
or $$P(\bar A \cap B ) = \displaystyle\frac{5}{12}$$
Question 6
1 / -0

A bag contains $$3$$ white, $$3$$ black, and $$2$$ red balls. One by one three balls are drawn without replacing them, then find the probability that the third ball is red.

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

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

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

D
None of these

Solution

$${\textbf{Step -1: Find the probabilities of one by one 3 balls are drawn replacing them.}}$$
$${\text{Let R stand for drawing red ball, B for black ball and W for drawing white ball.}}$$
$${\text{Then, the probabilities that one by one 3 balls are drawn replacing them is,}}$$
$$\Rightarrow P(WWR)= \dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{2}{6}= \dfrac{3 \times 2 \times 2}{8 \times 7 \times 6}=\dfrac{2}{56}$$

$$\Rightarrow P(BBR)= \dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{2}{6}= \dfrac{3 \times 2 \times 2}{8 \times 7 \times 6}=\dfrac{2}{56}$$

$$\Rightarrow P(WBR)= \dfrac{3}{8} \times \dfrac{3}{7} \times \dfrac{2}{6}= \dfrac{3 \times 3 \times 2}{8 \times 7 \times 6}=\dfrac{3}{56}$$

$$\Rightarrow P(BWR)= \dfrac{3}{8} \times \dfrac{3}{7} \times \dfrac{2}{6}= \dfrac{3 \times 3 \times 2}{8 \times 7 \times 6}=\dfrac{3}{56}$$

$$\Rightarrow P(WRR)= \dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{1}{6}= \dfrac{3 \times 2 \times 1}{8 \times 7 \times 6}=\dfrac{1}{56}$$

$$\Rightarrow P(BRR)= \dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{1}{6}= \dfrac{3 \times 2 \times 1}{8 \times 7 \times 6}=\dfrac{1}{56}$$

$$\Rightarrow P(RWR)= \dfrac{2}{8} \times \dfrac{3}{7} \times \dfrac{1}{6}= \dfrac{2 \times 3 \times 1}{8 \times 7 \times 6}=\dfrac{1}{56}$$

$$\Rightarrow P(RBR)= \dfrac{2}{8} \times \dfrac{3}{7} \times \dfrac{1}{6}= \dfrac{2 \times 3 \times 1}{8 \times 7 \times 6}=\dfrac{1}{56}$$
$${\textbf{Step -2: Find probability that the third ball is red.}}$$
$${\text{Probability that the third ball is red}}$$ $$=\dfrac{2}{56}+ \dfrac{2}{56}+ \dfrac{3}{56} + \dfrac{3}{56} + \dfrac{1}{56} + \dfrac{1}{56} + \dfrac{1}{56} + \dfrac{1}{56}$$

$${\text{Probability that the third ball is red}}$$ $$=\dfrac{1}{4}$$
$${\textbf{Thus, option A is correct.}}$$
Question 7
1 / -0

Let $$A$$ and $$B$$ be two events such that $$\displaystyle P(\overline{A\cup B})=\frac{1}{6},P(A\cap B)=\frac{1}{4}$$ and $$P(\bar{A})=\dfrac{1}{4}$$ where $$\bar{A}=$$ complementary of event $$A$$. Then $$A$$ and $$B$$ are

A
equally likely but not indepenent

B
equally likely and mutually exclusive

C
mutually exclusive and independent

D
independent but not equally likely

Solution

$$\displaystyle P\left( \overline { A\cup B } \right) =\frac { 1 }{ 6 } ;P\left( A\cap B \right) =\frac { 1 }{ 4 } $$
$$\displaystyle P\left( \overline { A } \right) =\frac { 1 }{ 4 } \Rightarrow P\left( A \right) =\frac { 3 }{ 4 } $$
Now $$P\left( \overline { A\cup B } \right) =1-P\left( A\cup B \right) =1-P\left( A \right) -P\left( B \right) +P\left( A\cap B \right) $$
$$\displaystyle \Rightarrow \frac { 1 }{ 6 } =\frac { 1 }{ 4 } -P\left( B \right) +\frac { 1 }{ 4 } \Rightarrow P\left( B \right) =\frac { 1 }{ 3 } $$
Since $$P\left( A\cap B \right) =P\left( A \right) \times P\left( B \right) $$ and $$P\left( A \right) \neq P\left( B \right) $$, $$A$$ and $$B$$ are independent but not equally likely.
Question 8
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
$$\displaystyle <\frac { 1 }{ 4 } $$

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

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

D
None of these

Solution

Since $$\overline { { E }_{ 1 } } \cap \overline { { E }_{ 2 } } =\overline { { E }_{ 1 }\cup { E }_{ 2 } } $$ and $$\left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \overline { { E }_{ 1 }\cup { E }_{ 2 } } \right) =\phi $$
$$\displaystyle \therefore P\left[ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \overline { { E }_{ 1 } } \cap \overline { { E }_{ 2 } } \right) \right] =P\left( \phi \right) =0<\frac { 1 }{ 4 } $$
Question 9
1 / -0

$$A$$ and $$B$$ are two events where $$P(A) = 0.25$$ and $$P(B) = 0.5$$. The probability of both happening together is $$0.14$$. The probability of both $$A$$ and $$B$$ not happening is

A
$$\displaystyle 0.39$$

B
$$\displaystyle 0.25$$

C
$$\displaystyle 0.11$$

D
none of these

Solution

$$P\left( A\cap B \right) =0.14\quad \\ \because \left( { A }^{ c }\cap { B }^{ c } \right) =1-\left( A\cup B \right) \\ Then\\ P\left( { A }^{ c }\cap { B }^{ c } \right) =1-P\left( A\cup B \right) \\ \quad \quad \quad \quad \quad \quad \quad =1-\left\{ P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \right\} $$
Putting the given values
$$P\left( { A }^{ c }\cap { B }^{ c } \right) =1-\left\{ 0.25+0.5-0.14 \right\} \\ \quad \quad \quad \quad \quad \quad \quad =0.39$$
Question 10
1 / -0

If $$\displaystyle P\left ( B \right )= \frac{3}{4}, P\left ( A\cap B\cap \bar{C} \right )=\frac{1}{3}$$ and $$\displaystyle P\left ( \bar{A}\cap B\cap \bar{C} \right )=\frac{1}{3}$$ then $$\displaystyle P\left ( B\cap C \right )$$ is

A
$$\displaystyle \frac{1}{12}$$

B
$$\displaystyle \frac{1}{6}$$

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

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

Solution

$$\displaystyle P\left( B\cap C \right) =P\left( B \right) -\left\{ P\left( A\cap B\cap { C }^{ c } \right) +P\left( { A }^{ c }\cap B\cap { C }^{ c } \right) \right\} $$
$$ \Rightarrow P\left( B\cap C \right) =\displaystyle \frac { 3 }{ 4 } -\left\{ \frac { 1 }{ 3 } +\frac { 1 }{ 3 } \right\} $$
$$ \Rightarrow P\left( B\cap C \right) =\displaystyle \frac { 1 }{ 12 } $$

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Re-Attempt Weekly Quiz Competition

Probability Test - 26

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Probability Test - 26

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SHARING IS CARING

Question 1

$$2$$

$$1$$

$$3$$

$$0.1$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$$A$$ and $$B$$ are independent events

$$A$$ and $$B$$ are disjoint events

$$A$$ and $$B$$ are dependent events

None of these

Solution

Question 4

31/56

24/56

25/56

none of these

Solution

Question 5

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

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

$$\displaystyle\frac{5}{12}$$

None of these

Solution

Question 6

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

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

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

None of these

Solution

Question 7

equally likely but not indepenent

equally likely and mutually exclusive

mutually exclusive and independent

independent but not equally likely

Solution

Question 8

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

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

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

None of these

Solution

Question 9

$$\displaystyle 0.39$$

$$\displaystyle 0.25$$

$$\displaystyle 0.11$$

none of these

Solution

Question 10

$$\displaystyle \frac{1}{12}$$

$$\displaystyle \frac{1}{6}$$

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

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

Solution

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