Probability Test

Result

Probability Test - 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$P(A\cup B)=P(A\cap B)$$ then the relation between $$P(A)$$ and $$P(B)$$ is..........

A
$$P(A) = P(B)$$

B
$$P(A) <P(B)$$

C
$$P(A) > P(B)$$

D
None of these

Solution

$$A\cup B=nA+nB-nA\cap B$$
$$=nA\cap B$$
Thus
$$nA+nB=2nA\cap B$$
This is true if $$A=B$$.
Hence
$$P(A\cup B)=P(A\cap B)$$
Implies $$P(A)=P(B)$$.
Question 2
1 / -0

$$A$$ and $$B$$ are two independent events. The probability that $$A$$ and $$B$$ occur is $$\displaystyle\frac{1}{6}$$ and the probability that at least one of them occurs is $$\displaystyle\frac{2}{3}$$. The probability of the occurrence of $$A = ....$$ if $$P(A) = 2P(B)$$

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

B
$$\displaystyle \frac{4}{9}$$

C
$$\displaystyle \frac{8}{9}$$

D
None of these

Solution

Here given $$P(A\cap B) = \cfrac{1}{6}, P(A\cup B) = \cfrac{2}{3}, P(A)=2P(B)$$

Thus using $$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$

$$\Rightarrow \cfrac{2}{3}=P(A)+\cfrac{1}{2}P(B)-\cfrac{1}{6}\Rightarrow P(A) = \cfrac{5}{9}$$
Question 3
1 / -0

The probability that at least one of the events $$A$$ and $$B$$ occur is $$0.6$$. If $$A$$ and $$B$$ occurs simultaneously with probability $$0.2$$, then $$P\left( \overline { A } \right) +P\left( \overline { B } \right) $$ is

A
$$0.4$$

B
$$0.8$$

C
$$1.2$$

D
$$1.4$$

E
None of these

Solution

Given, $$P\left( A\cup B \right) =0.6,P\left( A\cap B \right) =0.2$$
$$\therefore P\left( \overline { A } \right) +P\left( \overline { B } \right) =\left( 1-P\left( A \right) \right) +\left( 1-P\left( B \right) \right) \\ =2-\left( P\left( A \right) +P\left( B \right) \right) =2-\left( P\left( A\cup B \right) +P\left( A\cap B \right) \right) \\ =2-\left( 0.6+0.2 \right) =1.2$$
Question 4
1 / -0

For the three events $$A, B \& C,$$ probability of exactly one of the events $$ A$$ or $$B$$ occurs = probability of exactly one of the events $$C$$ or $$A$$ occurs $$= p$$ $$\&$$ $$P$$ (all the three events occur simultaneously) $$= p$$$$^2$$, where

A
$$\displaystyle \frac{3p+2p^2}{2}$$

B
$$\displaystyle \frac{p+3p^2}{4}$$

C
$$\displaystyle \frac{p+3p^2}{2}$$

D
$$\displaystyle \frac{3p+2p^2}{4}$$

Solution

$$P$$(exactly one of $$A$$ or $$B$$ occurs)
$$=P\left( A \right) +P\left( B \right) -2P\left( A\cap B \right) $$
Therefore, $$P\left( A \right) +P\left( B \right) -2P\left( A\cap B \right) =p $$ ...(1)
Similarly, $$P\left( B \right) +P\left( C \right) -2P\left( B\cap C \right) =p$$ ...(2)
and $$P\left( C \right) +P\left( A \right) -2P\left( C\cap A \right) =p $$ ...(3)
Adding $$(1),(2)$$ and $$(3)$$ and dividing by $$2$$
$$ \displaystyle \Rightarrow P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) -P\left( B\cap C \right) -P\left( C\cap A \right) =\frac { 3p }{ 2 }$$ ...(4)
Now,$$P$$(at least one of $$A,B$$ and $$C$$)
$$ \displaystyle =\frac { 3p }{ 2 } +{ p }^{ 2 }=\frac { 3p+{ 2p }^{ 2 } }{ 2 } $$
Put this equal to $$ \displaystyle \frac { 11 }{ 18 } $$ and solve for $$p$$.
Question 5
1 / -0

A and B are two independent events such that $$P(A'\cap B')=\dfrac{1}{6}$$ and $$P(A')=\dfrac{5}{24}$$. Then P(B) is equal to

A
$$\dfrac{4}{5}$$

B
$$\dfrac{1}{6}$$

C
$$\dfrac{1}{5}$$

D
$$\dfrac{5}{6}$$

Solution

Given: A and B are two independent events and
$$P(\bar { A } \bigcap { \bar { B } } )=\cfrac { 1 }{ 6 } ,P\left( \bar { A } \right) =\cfrac { 5 }{ 24 } $$
$$\therefore $$ Using
$$P(\bar { A } \bigcap { \bar { B } } )=P(\bar { A } )P(\bar { B } )$$
$$\cfrac { 1 }{ 6 } =\cfrac { 5 }{ 24 } P(\bar { B } )$$
i.e., $$P(\bar { B } )=\cfrac { 4 }{ 5 } $$ then
$$P(B)=1-P(\bar { B } )=1-\cfrac { 4 }{ 5 } =\cfrac { 1 }{ 5 } $$
Question 6
1 / -0

The probability that at least one of the events $$A$$ and $$B$$ occurs is $$\displaystyle \frac {3}{5}$$. If $$A$$ and $$B$$ occur simultaneously with probability $$\displaystyle \frac {1}{5}$$ then $$\displaystyle P(A') + P(B')$$ is

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

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

C
$$\displaystyle \frac {6}{5}$$

D
$$\displaystyle \frac {7}{5}$$

Solution

$$P\left( A\cup B \right) =\dfrac { 3 }{ 5 } \quad and\quad P\left( A\cap B \right) =\dfrac { 1 }{ 5 } \\ P\left( A\cup B \right)
=P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \\ \Rightarrow P\left( A \right) +P\left( B \right)
=P\left( A\cup B \right) +P\left( A\cap B \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad =\dfrac { 3 }{ 5 } +\dfrac { 1 }{ 5 }
=\dfrac { 4 }{ 5 } \\ \Rightarrow 1-P\left( \bar A \right) +1-P\left(\bar B \right) =\dfrac { 4 }{ 5 }
\Rightarrow P\left(\bar A \right) +P\left(\bar B \right)=2-\dfrac { 4 }{ 5 }
=\dfrac{6}{5}\\ \Rightarrow P\left( { A }^{ c } \right) +P\left( { B }^{ c } \right) =\dfrac { 6 }{ 5 } $$
Question 7
1 / -0

Probability is $$0.45$$ that a dealer will sell at least $$20$$ television sets during a day, and the probability is $$0.74$$ that he will sell less that $$24$$ televisions. The probability that he will sell $$20,21,22$$ or $$23$$ televisions during the day, is

A
$$0.19$$

B
$$0.32$$

C
$$0.21$$

D
None of these

Solution

Let $$A$$ be the event that the sale is at least $$20$$ televisions, i.e. $$20,21,22,...$$
and $$B$$ be the event that sale is less than $$24$$ i.e. $$0.1.2.3...23$$.
Then $$A\cap B$$ will denote the sale of $$20,21,22$$ and $$23$$ televisions,
We are given $$P(A)=0.45$$ and $$P(B)=0.74$$.
It is required to find $$P(A\cap B)$$.
Also $$P\left( A\cup B \right) =P($$ sale of $$0,1,2,3,...,20,21,22,23$$ televisions$$)=P(S)=1$$
From addition rule, required probability is
$$P\left( A\cap B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cup B \right) =0.45+0.74-1=0.19$$
Question 8
1 / -0

A man alternatively tosses a coin and throws a die. The probability of getting a head on the coin before he gets 4 on the die is

A
$$ \displaystyle \frac{6}{7} $$

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

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

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

Solution

Let the desired event be called E
then $$P\left( E \right) =P\left( H \right) +P\left( T \right) P\left( \overline { 4 } \right) P\left( H \right) +P\left( T \right) P\left( \overline { 4 } \right) P\left( T \right) P\left( \overline { 4 } \right) P\left( H \right) +.............\infty $$
where $$P\left( \overline { 4 } \right) =\dfrac { 5 }{ 6 } ,\quad P\left( H \right) =P\left( T \right) =0.5$$
thus we can see that an G.P sum is formed with first term =P(H) and common ratio =$$P\left( T \right) P\left( \overline { 4 } \right) $$
hence $$P\left( E \right) =\dfrac { P\left( H \right) }{ 1-P\left( T \right) P\left( \overline { 4 } \right) } =\dfrac { 0.5\times 12 }{ 7 } =\dfrac { 6 }{ 7 } $$
Question 9
1 / -0

A, B, C take one shot each at a target. Their probabilities of hitting the target are respectively 0.4, 0.5 and 0.8. The probability that at least two of them hit the target, is

A
$$ \displaystyle \frac{13}{25} $$

B
$$ \displaystyle \frac{9}{25} $$

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

D
None of these

Solution

Let the desired event be called E
Also let A,B and C hitting the target be called A,B and respectively and missing the target be called $$\overline { A } ,\overline { B } \quad and\quad \overline { C } $$
$$P\left( E \right) =P\left( A \right) P\left( B \right) P\left( C \right) +P\left( \overline { A } \right) P\left( B \right) P\left( C \right) +P\left( A \right) P\left( \overline { B } \right) P\left( C \right) +P\left( A \right) P\left( B \right) P\left( \overline { C } \right) \\ \Rightarrow P\left( E \right) =0.4\times 0.5\times 0.8+0.6\times 0.5\times 0.8+0.4\times 0.5\times 0.8+0.4\times 0.5\times 0.2\\ \quad \quad \quad \quad \quad \quad =0.16+0.24+0.16+0.04=0.6=\dfrac { 3 }{ 5 } $$
Question 10
1 / -0

If $$E$$ and $$F$$ are events with $$P(E)\le P(F)$$ and $$P(E\cap F)>0$$, then

A
occurrence of $$E\Rightarrow $$ occurrenece of $$F$$

B
occurrence of $$F\Rightarrow $$ occurrenece of $$E$$

C
non occurrence of $$E\Rightarrow $$ non occurrenece of $$F$$

D
none of the above imlications holds

Solution

Considering the set $$S$$.
$$E$$ is subset of $$S$$ and $$F$$ is subset of $$S, F$$ being the larger set.
As $$P(E\cap F)$$ is not zero then this means that $$E \& F$$ have something in common. But still they can be independent or independent. $$E $$ might be completely inside of $$F $$ or some very little part of $$E$$ is in $$F$$. Hence all the first three options are incorrect.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 28

Result

Probability Test - 28

Score

-

Rank

-

SHARING IS CARING

Question 1

$$P(A) = P(B)$$

$$P(A) <P(B)$$

$$P(A) > P(B)$$

None of these

Solution

Question 2

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

$$\displaystyle \frac{4}{9}$$

$$\displaystyle \frac{8}{9}$$

None of these

Solution

Question 3

$$0.4$$

$$0.8$$

$$1.2$$

$$1.4$$

None of these

Solution

Question 4

$$\displaystyle \frac{3p+2p^2}{2}$$

$$\displaystyle \frac{p+3p^2}{4}$$

$$\displaystyle \frac{p+3p^2}{2}$$

$$\displaystyle \frac{3p+2p^2}{4}$$

Solution

Question 5

$$\dfrac{4}{5}$$

$$\dfrac{1}{6}$$

$$\dfrac{1}{5}$$

$$\dfrac{5}{6}$$

Solution

Question 6

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

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

$$\displaystyle \frac {6}{5}$$

$$\displaystyle \frac {7}{5}$$

Solution

Question 7

$$0.19$$

$$0.32$$

$$0.21$$

None of these

Solution

Question 8

$$ \displaystyle \frac{6}{7} $$

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

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

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

Solution

Question 9

$$ \displaystyle \frac{13}{25} $$

$$ \displaystyle \frac{9}{25} $$

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

None of these

Solution

Question 10

occurrence of $$E\Rightarrow $$ occurrenece of $$F$$

occurrence of $$F\Rightarrow $$ occurrenece of $$E$$

non occurrence of $$E\Rightarrow $$ non occurrenece of $$F$$

none of the above imlications holds

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.