Probability Test

Result

Probability Test - 20

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

Let $$A$$ and $$B$$ are events of an experiment and $$P(A)=\dfrac 14, P(A \cup B) = \dfrac 12$$, then value of $$ P(\dfrac B{A^c}) $$ is

A
$$\dfrac 23$$

B
$$\dfrac 13$$

C
$$\dfrac 56$$

D
$$\dfrac 12$$

Solution

Given $$P(A)=\cfrac { 1 }{ 4 } $$
$$P\left( A\bigcup { B } \right) =\cfrac { 1 }{ 2 } $$
$$P\left( \cfrac { B }{ { A }^{ c } } \right) =\cfrac { P\left( B\bigcap { { A }^{ c } } \right) }{ P\left( { A }^{ c } \right) } $$
$$=\cfrac { P(B)-P\left( B\bigcap { { A } } \right) }{ 1-P(A) } $$
$$\because P\left( A\bigcup { B } \right) =P(A)+P(B)-P\left( A\bigcap { { B } } \right) $$
$$\therefore P(B)-P\left( A\bigcap { { B } } \right) =P\left( A\bigcup { B } \right) -P(A)$$
$$=\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 4 } =\cfrac { 1 }{ 4 } $$
and $$1-P(A)=1-\cfrac { 1 }{ 4 } =\cfrac { 3 }{ 4 } $$
(Substituting the values)
$$\therefore P\left( \cfrac { B }{ { A }^{ c } } \right) =\cfrac { { 1 }/{ 4 } }{ { 3 }/{ 4 } } =\cfrac { 1 }{ 3 } $$
Question 2
1 / -0

There are 10 pairs of shoes in a cupboards, from which 4 shoes are picked at random. The probability that there is at least one pair, is..........

A
$$\dfrac{101}{323}$$

B
$$\dfrac{100}{323}$$

C
$$\dfrac{99}{323}$$

D
$$\dfrac{110}{323}$$

Solution

The total number of ways of choosing $$4$$ shoes (in order)
out of $$10$$ pairs (or $$20$$ shoes) is $$20\times 19\times 18\times 17$$
The number of ways in which pair is selected is $$20\times 18\times 16\times 14$$
Thus the probability of not getting a pair is $$\displaystyle\frac { 20\times 18\times 16\times 14 }{ 20\times 19\times 18\times 17 } =\frac { 224 }{ 323 } $$

Hence, the probability of getting atleast one pair is $$\displaystyle 1-\frac { 224 }{ 323 } =\frac { 99 }{ 323 }$$
Question 3
1 / -0

Three groups $$A,\ B,\ C$$ are contesting for positions on the Board of Directors of a company. The probabilities of their winning are $$0.5,\ 0.3, 0.2$$ respectively. If the group $$A$$ wins, the probability of introducing a new product is $$0.7$$ and the corresponding probabilities for groups $$B$$ and $$C$$ are $$0.6$$ and $$0.5$$ respectively. The probability that the new product will be introduced is given by

A
$$0.36$$

B
$$0.35$$

C
$$0.63$$

D
$$0.53$$

Solution

$$P\left( A \right) =0.5,P\left( B \right) =0.3,P\left( C \right) =0.2$$

If $$N$$ is for introducing new product, then

$$P\left( N|A \right) =0.7,P\left( N|B \right) =0.6,P\left( N|C \right) =0.5$$

Therefore,

$$P\left( N \right) =P\left( A \right) \times P\left( N|A \right) +P\left( B \right) \times P\left( N|B \right) +P\left( C \right) \times P\left( N|C \right) \\ =0.5\times 0.7+0.3\times 0.6+0.2\times 0.5=0.63$$
Question 4
1 / -0

A box contains $$100$$ tickets numbered $$1,2,3,...,100.$$ two tickets are chosen at random. If it is given that the maximum number on the two chosen tickets is not more than $$10$$, then the probability that the minimum number on them is not less than $$5$$ is

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

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

C
$$\displaystyle \frac { 152 }{ 165 } $$

D
None of these

Solution

Let $$E$$ be the event: ''the maximum number is not more than $$10$$'' and $$F:$$ ''minimum number is not less $$5$$''
Now $$\displaystyle P\left( E \right) =\dfrac { _{ }^{ 10 }{ { C }_{ 2 } } }{ _{ }^{ 100 }{ { C }_{ 2 } } } $$ and $$P\left( E\cap F \right) =P\quad $$ (number on the selected tickets is from $$5$$ to $$10$$)
$$=\displaystyle \dfrac { _{ }^{ 6 }{ { C }_{ 2 } } }{ _{ }^{ 100 }{ { C }_{ 2 } } } $$
$$(\because $$ there are six tickets with such numbers$$)$$
$$\therefore$$ Required probability$$\displaystyle =P\left( \dfrac { F }{ E } \right) $$
$$\displaystyle =\dfrac { P\left( E\cap F \right) }{ P\left( E \right) } =\dfrac { \dfrac { _{ }^{ 6 }{ { C }_{ 2 } } }{ _{ }^{ 100 }{ { C }_{ 2 } } } }{ \dfrac { _{ }^{ 10 }{ { C }_{ 2 } } }{ _{ }^{ 100 }{ { C }_{ 2 } } } } =\dfrac { 15 }{ 45 } =\dfrac { 1 }{ 3 } .$$
Question 5
1 / -0

A box contains $$100$$ tickets numbered $$1,2,.....100$$. Two tickets are chosen at random. It is given that the minimum number on the two chosen tickets is not more than $$10$$. The maximum number on them is $$5$$ with probability.

A
$$\cfrac{1}{9}$$

B
$$\cfrac{1}{4}$$

C
$$\cfrac{1}{90}$$

D
$$\cfrac{1}{5}$$

Solution

Let $$A$$ be the event that the maximum number pn the two chosen tickets is not more than 10, and B be the event that the minimum number on them is 5.
$$\displaystyle \therefore P\left( A\cap B \right) =\frac { _{ }^{ 5 }{ { C }_{ 1 } } }{ ^{ 100 }{ { C }_{ 2 } } } $$ and $$\displaystyle P\left( A \right) =\frac { _{ }^{ 10 }{ { C }_{ 2 } } }{ ^{ 100 }{ { C }_{ 2 } } } $$

Then $$\displaystyle P\left( \frac { B }{ A } \right) =\frac { P\left( A\cap B \right) }{ P\left( A \right) } =\frac { _{ }^{ 5 }{ { C }_{ 1 } } }{ ^{ 10 }{ { C }_{ 2 } } } =\frac { 1 }{ 9 } $$
Question 6
1 / -0

An experiment has $$10$$ equally likely outcomes. Let $$A$$ and $$B$$ be two non-empty events of the experiment. If $$A$$ consists of $$4$$ outcomes, the number of outcomes that $$B$$ must have so that $$A$$ and $$B$$ are independent, is

A
$$2, 4$$ or $$8$$

B
$$3, 6$$ or $$9$$

C
$$4$$ or $$8$$

D
$$5$$ or $$10$$

Solution

Since the Set $$A$$ has $$4$$ out of $$10$$ outcomes, we have $$\displaystyle P\left( A \right) = \frac{4}{{10}}$$

Similarly, let set $$B$$ has $$k$$ outcomes, then we have $$\displaystyle P\left( B \right) = \frac{k}{{10}}$$
Now, let $$A$$ and $$B$$ have $$m$$ outcomes in common i.e., $$\left| {A \cap B} \right| =m$$, then $$m\le \left|A\right|=4$$.
Hence,
$$\displaystyle P\left( {A \cap B} \right)=\frac{m}{{10}}$$.
Now, $$\displaystyle P\left( {\left. A \right|B} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \dfrac{{\dfrac{m}{{10}}}}{{\dfrac{k}{{10}}}} = \dfrac{m}{k}$$
But, $$P\left( {A \cap B} \right)= P\left( A \right) \times P\left( B \right)$$ as $$A$$ and $$B$$ are independent.
$$\therefore \displaystyle P\left( {\left. A \right|B} \right) =P\left( A \right) = \frac{4}{{10}}$$
Hence, $$\displaystyle \frac{m}{k} = \frac{4}{{10}} = \frac{2}{5}$$
$$2k = 5m$$
The only integral answers for $$m$$ are $$0, 2,$$ and $$4$$.

If $$m=0 \Rightarrow k=0$$, contrary to the given fact that $$B$$ is a non-empty event.
Therefore, $$m$$ must be equal to either to either $$2$$ or $$4$$, which gives us values of $$k$$ as $$5$$ or $$10$$.
Question 7
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

$$\displaystyle P\left( \frac { E }{ F } \right) +P\left( \frac { \overline { E } }{ F } \right) $$
$$\displaystyle =\frac { P\left( E\cap F \right) }{ P\left( F \right) } +\frac { P\left( \overline { E } \cap F \right) }{ P\left( F \right) } =\frac { P\left( E\cap F \right) +P\left( \overline { E } \cap F \right) }{ P\left( F \right) } $$
$$\displaystyle =\frac { P\left\{ \left( E\cap F \right) \cup \left( \overline { E } \cap F \right) \right\} }{ P\left( F \right) } $$ $$\displaystyle [\because E\cap F$$ and $$\overline { E } \cap F$$ are disjoint$$]$$
$$\displaystyle =\frac { P\left\{ \left( E\cup \overline { E } \right) \cap F \right\} }{ P\left( F \right) } =\frac { P\left( S\cap F \right) }{ P\left( F \right) } =\frac { P\left( F \right) }{ P\left( F \right) } =1.$$
Question 8
1 / -0

A man is known to speak the truth $$3$$ out of $$4$$ times. He throws a die and reports that it is a six. The probability that it is really a six is

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

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

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

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

Solution

Let $${ E }_{ 1 },{ E }_{ 2 }$$ and $$A$$ be the events defined as follows:
$${ E }_{ 1 }=$$ die shows six, i.e., six has occured.
$${ E }_{ 2 }=$$ dei does not show six i.e., six has not occured.
$$A=$$ the man reports that six has occured.
We have to calculate that six has actually occured
given that the man reports that six occurs i.e., $$\displaystyle P\left( \dfrac { { E }_{ 1 } }{ A } \right) $$
Now, $$\displaystyle P\left( { E }_{ 1 } \right) =\dfrac { 1 }{ 6 } ,P\left( { E }_{ 2 } \right) =\dfrac { 5 }{ 6 } $$
$$\displaystyle P\left( \dfrac { A }{ { E }_{ 1 } } \right)= $$ probability that the man reports that six occured given that six has occured.
$$=$$ probability that the man is reporting the truth.$$\displaystyle=\dfrac { 3 }{ 4 } $$
And $$\displaystyle P\left( \dfrac { A }{ { E }_{ 2 } } \right) =$$probability that the man reports that six occured given that six has not occured.
$$=$$ probability that the man does not speak truth $$\displaystyle=\dfrac { 1 }{ 4 } $$
By Bayes' Theorem,
$$\displaystyle P\left( \dfrac { { E }_{ 1 } }{ A } \right) =\dfrac { P\left( { E }_{ 1 } \right) P\left( \dfrac { A }{ { E }_{ 1 } } \right) }{ P\left( { E }_{ 1 } \right) P\left( \dfrac { A }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) P\left( \dfrac { A }{ { E }_{ 2 } } \right) } $$
$$\displaystyle=\dfrac { \dfrac { 1 }{ 6 } .\dfrac { 3 }{ 4 } }{ \dfrac { 1 }{ 6 } .\dfrac { 3 }{ 4 } +\dfrac { 5 }{ 6 } .\dfrac { 1 }{ 4 } } =\dfrac { 3 }{ 8 } $$
Question 9
1 / -0

Suppose $$n(\ge 3)$$ persons are sitting in row. Two of them are selected at random. The probability that they are not together is

A
$$\displaystyle 1-\frac { 2 }{ n } $$

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

C
$$\displaystyle 1-\frac { 1 }{ n } $$

D
None of these

Solution

The total number of ways of choosing $$2$$ persons out of $$n$$ is $$_{ }^{ n }{ { C }_{ 2 } }$$
After selecting two persons when the remaining $$n-2$$ persons sit in a row $$(n-1)$$ places are created between them in which $$2$$ persons can be arranged in $$_{ }^{ n-1 }{ { C }_{ 2 } }.2!$$ ways.
So, the required probability $$\displaystyle =\frac { ^{ n-1 }{ { C }_{ 2 } }.2! }{ ^{ n }{ { C }_{ 2 } } } =\frac { n-2 }{ n } =1-\frac { 2 }{ n } $$
Question 10
1 / -0

Let $$A = {2, 3, 4, ... , 20, 21}$$. A number is chosen at random from the set $$A$$ and it is found to be a prime number. The probability that it is more than $$10$$ is

A
$$\displaystyle \frac {9}{10}$$

B
$$\displaystyle \frac {1}{10}$$

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

D
none of these

Solution

P ( A|B ) = P(A & B)/ P(B)
B = getting prime number
A = getting number greater than 10
P ( A|B ) = $$\dfrac { 4/20 }{ 8/20 } $$ = 1/2

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 20

Result

Probability Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac 23$$

$$\dfrac 13$$

$$\dfrac 56$$

$$\dfrac 12$$

Solution

Question 2

$$\dfrac{101}{323}$$

$$\dfrac{100}{323}$$

$$\dfrac{99}{323}$$

$$\dfrac{110}{323}$$

Solution

Question 3

$$0.36$$

$$0.35$$

$$0.63$$

$$0.53$$

Solution

Question 4

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

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

$$\displaystyle \frac { 152 }{ 165 } $$

None of these

Solution

Question 5

$$\cfrac{1}{9}$$

$$\cfrac{1}{4}$$

$$\cfrac{1}{90}$$

$$\cfrac{1}{5}$$

Solution

Question 6

$$2, 4$$ or $$8$$

$$3, 6$$ or $$9$$

$$4$$ or $$8$$

$$5$$ or $$10$$

Solution

Question 7

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

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

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

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

Solution

Question 8

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

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

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

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

Solution

Question 9

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

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

$$\displaystyle 1-\frac { 1 }{ n } $$

None of these

Solution

Question 10

$$\displaystyle \frac {9}{10}$$

$$\displaystyle \frac {1}{10}$$

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

none of these

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.