Probability Test

Result

Probability Test - 22

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

A certain player, say $$X$$, known to win with probability $$0.3$$ if the track is fast and $$0.4$$ if the track is slow. For Monday, there is a $$0.7$$ probability of a fast track and $$0.3$$ probability of a slow track. The probability that player $$X$$ will win a Monday, is

A
$$0.22$$

B
$$0.11$$

C
$$0.33$$

D
None of these

Solution

Let us define the events:
$$W:$$ Players wins on Monday
$$F:$$ Track is fast
$$S:$$ Track is slow
We have given: $$\displaystyle P\left( F \right) =0.7,P\left( S \right) =0.3,P\left( \frac { W }{ F } \right) =0.3,P\left( \frac { W }{ S } \right) =0.4$$
$$\therefore P($$ player $$X$$ will win on Monday $$)$$
$$\displaystyle =P\left( W \right) =P\left( W\cap F \right) +P\left( W\cap S \right) =P\left( F \right) \times P\left( \frac { W }{ F } \right) +P\left( S \right) \times P\left( \frac { W }{ S } \right) \\ =0.7\times 0.3+0.3\times 0.4=0.33$$
Question 2
1 / -0

A letter is known to have come eithe from London or Clifton; on the post only the consecutive letters ON are legible; what is the chance that it came from London?

A
$$\displaystyle \frac{12}{17}$$

B
$$\displaystyle \frac{5}{17}$$

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

D
$$\displaystyle \frac{7}{12}$$

Solution

If letter came from Clifton there are $$6$$ pairs of consecutive letters i.e., $$cl, li, if, ft, to$$, $$ON $$ in which $$ON$$ appears only once.
$$\displaystyle \therefore $$ the chance that this was the legible couple on the Clifton hypothesis $$\displaystyle = \frac{1}{6}$$
pairs of consecutive letters in the word London are $$lo, on, nd, do$$, $$ON $$ in which $$ON$$ occurs twice.
$$\displaystyle \therefore $$ the chance that this was the legible couple on the London hypothesis$$=2/5.$$
$$\displaystyle \therefore $$ The a posteriori chances that the letter was from Clifton or London are $$\displaystyle \frac{1/6}{\dfrac{1}{6}+\dfrac{2}{5}}$$ and $$\displaystyle \frac{2/5}{\dfrac{1}{6}+\dfrac{2}{5}}$$ respectively.
Thus the reqd.chance $$\displaystyle = \frac{12}{17}.$$
Question 3
1 / -0

A sample of size $$4$$ is drawn with replacement be the first part of the problem and without replacement be the second part of the problem, then from an urn containing $$12$$ balls, of which $$8$$ are white, what is the conditional probability that the ball drawn on the third draw was white, given that the sample contains $$3$$ white balls ?

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

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

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

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

Solution

Let A denote the event that the sample contains exactly three white balls and let B be the event that the ball taking the condition of with replacement.
Now $$\displaystyle p\left(A \right)=\left ( ^{4}C_{3} \right )\frac{8}{12}\times\frac{8}{12}\times\frac{8}{12}\times\frac{4}{12}$$

$$\displaystyle P\left(AB \right)=\left ( ^{3}C_{2} \right )\frac{8}{12}\times \frac{8}{12}\times \frac{8}{12}\times \frac{4}{12}$$

$$\displaystyle \therefore P\left [ B/A \right ]=\frac{P\left [ AB \right ]}{P\left ( A \right )}=\frac{^{3}C_{2}}{^{4}C_{3}}=\frac{3}{4}$$
In the case of sampling without replacement,
$$\displaystyle P\left(A \right)=\left ( ^{4}C_{3} \right )\frac{8}{12}\times\frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}$$

$$\displaystyle P\left (AB \right )=\left ( ^{3}C_{2} \right )\frac{8}{12}\times\frac{7}{11}\times \frac{6}{10}\times \frac{4}{9}$$

$$\displaystyle \therefore P\left [B/A \right ]=\frac{^{3}C_{2}}{^{4}C_{3}}=\frac{3}{4}$$
Question 4
1 / -0

A person is know to speak the truth 4 times out of 5. He throws a die and reports that it is a ace. The probability that it is actually a ace is

A
$$1/3$$

B
$$2/9$$

C
$$4/9$$

D
$$5/9$$

Solution

Let $$E_1$$ denote the event that an ace occurs and $$E_2$$ the event that it does not occur. Let $$A$$ denote the event that the person reports that it is an ace.
Then $$P(E_1)=1/6, P(E_2)=5/6, P(A|E_1)=4/5$$ an $$P(A|E_2)=1/5$$.
By Bayes' theorem,
$$P(E_1|A)=\dfrac {P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}$$
$$=\dfrac {4}{9}$$
Question 5
1 / -0

Mr. Dupont is a professional wine taster. When given a French wine, he will identify it with probability $$0.9$$ correctly as French and will mistake it for a Californian wine with probability $$0.1$$. When given a Californian wine, he will identify it with probability $$0.8$$ correctly as Californian and will mistake it for a French wine with probability $$0.2$$. Suppose that Mr. Dupont is given ten unlabelled glasses of wine, three with French and seven with Californian wines. He randomly picks a glass, tries the wine and solemnly says : "French". The probability that the wine he tasted was Californian, is nearly equal to

A
$$0.14$$

B
$$0.24$$

C
$$0.34$$

D
$$0.44$$

Solution

Given,There are 7 californian wine glasses and 3 french wine glasses.
The probability of selecting French wine glass, $$P(FG)=\frac{3}{10}$$
The probability of selecting California wine glass, $$P(CG)=\frac{7}{10}$$ When given french wine,
The probability of Dupont to say correctly as french wine, $$P(F)=0.9$$ When given french wine,
The probability of Dupont to say wrongly as Californian wine, $$P(\overline F)=0.1$$ When given Californian wine,
The probability of Dupont to say correctly as Californian wine, $$P(F)=0.8$$ When given Californian wine,
The probability of Dupont to say wrongly as french wine, $$P(\overline C)=0.2$$$$\therefore$$
The probability that Dupont says selected glass as French wine, $$P(A)=$$The probability of selecting french wine glass and will say $$correctly$$ as $$french$$ wine $$+$$ Probability of selecting $$californian$$ wine glass and saying $$wrongly$$ it as $$French$$ wine. $$=P(FG)*P(F)+P(CG)*P(\overline C)$$
$$=\displaystyle\frac{3}{10}*0.9+\displaystyle\frac{7}{10}*0.2=0.041$$
$$\therefore$$ The probability that Dupont says selected glass as French wine Given it as Californian=$$\displaystyle\frac{P(CG)*P(\overline C)}{P(A)}=\displaystyle\frac{(\frac{7}{10}*0.2)}{0.41}=0.341$$
Question 6
1 / -0

Twelve players $${ S }_{ 1 },{ S }_{ 2 },...,{ S }_{ 12 }$$ play in a chess tournament. They are divided into six pairs at random. From each pair a winner is decided. It is assumed that all players are of equal strength. The probability that both $${S}_{1}$$ and $${S}_{2}$$ are among the six winners is :

A
$$\displaystyle \frac { 12 }{ 33 } $$

B
$$\displaystyle \frac { 13 }{ 33 } $$

C
$$\displaystyle \frac { 5 }{ 22 } $$

D
None of these

Solution

Consider the probability for the given statement where S$$_1$$ and S$$_2$$ both are winners

Events where both S$$_1$$ and S$$_2$$ are there in 6 winners = $$^{10}C_4$$

Number of ways to select the 6 winners = $$^{12}C_6$$

Now, lets evaluate the probability

P(Both are winners)=$$\dfrac {^{10}C_4}{^{12}C_6}=\dfrac {210}{924}=\dfrac{5}{22}$$
Question 7
1 / -0

An artillery target may be either at point $$I$$ with the probability $$\displaystyle \frac { 8 }{ 9 } $$ or at the point $$II$$ with probability $$\displaystyle \frac { 1 }{ 9 } $$. We have $$21$$ shells each of which can be fired either at point $$I$$ or $$II$$. Each shell may hit the target independently of the other shell with probability $$\displaystyle \frac { 1 }{ 2 } $$. The number of shells which be fired at point $$I$$ to hit the target with maximum probability is

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Let $$A$$ denote the event that the target is hit when $$x$$ shells are fired at point $$I$$.
Let $${ E }_{ 1 },{ E }_{ 2 }$$ denote the events hotting $$I$$ and $$II$$ respectively.
$$\displaystyle \therefore P\left( { E }_{ 1 } \right) =\frac { 8 }{ 9 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 9 } $$
Now $$\displaystyle P\left( \frac { A }{ { E }_{ 1 } } \right) =1-{ \left( \frac { 1 }{ 2 } \right) }^{ x }$$ and $$\displaystyle P\left( \frac { A }{ { E }_{ 2 } } \right) =1-{ \left( \frac { 1 }{ 2 } \right) }^{ 21-x }$$
Now $$\displaystyle P\left( A \right) =\frac { 8 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 } \right) }^{ x } \right] +\frac { 1 }{ 9 } \left[ 1-{ \left( \frac { 1 }{ 2 } \right) }^{ 21-x } \right] $$
$$\displaystyle \therefore \frac { dP\left( A \right) }{ dx } =\frac { 8 }{ 9 } \left[ { \left( \frac { 1 }{ 2 } \right) }^{ x }\log { 2 } \right] +\frac { 1 }{ 9 } \left[ -{ \left( \frac { 1 }{ 2 } \right) }^{ 21-x }\log { 2 } \right] $$
For maximum probability $$\displaystyle \frac { dP\left( A \right) }{ dx } =0$$
$$\therefore x=12\quad \quad \quad \left[ \because { 2 }^{ 3-x }={ 2 }^{ x-21 }\Rightarrow 3-x=x-21\Rightarrow x=12 \right] $$
Since $$\displaystyle \frac { { d }^{ 2 }P\left( A \right) }{ d{ x }^{ 2 } } <0$$ for $$x=12$$
$$\therefore P(A) $$ is maximum when $$x=12$$.
Question 8
1 / -0

If two events $$A$$ and $$B$$ are such that $$P(A')=0.3,P(B)=0.4$$ and $$P\left( A\cap B' \right) =0.5$$, then $$\displaystyle P\left( \frac { B }{ A\cup B' } \right) $$ equala

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

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

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

D
$$\displaystyle \frac { 3 }{ 7 } $$

Solution

$$\displaystyle P\left( \frac { B }{ A\cup B' } \right) =\frac { P\left( B\cap \left( A\cup B' \right) \right) }{ P\left( A\cup B' \right) } =\frac { P\left( \left( B\cap A \right) \cup \left( B\cap B' \right) \right) }{ P\left( A \right) +P\left( B' \right) -P\left( A\cap B' \right) } $$
But $$\\ P\left( \left( B\cap A \right) \cup \left( B\cap B' \right) \right) =P\left( B\cap A \right) =P\left( A \right) -P\left( A\cap B' \right) =0.2$$
and $$P\left( A \right) +P\left( B' \right) -P\left( A\cap B' \right) =0.7+0.6-0.5=0.8$$
$$\displaystyle \therefore P\left( \frac { B }{ A\cup B' } \right) =\frac { 0.2 }{ 0.8 } =\frac { 1 }{ 4 } $$
Question 9
1 / -0

Directions For Questions

Let S and T are two events defined on a sample space with probabilities
$$P(S) = 0.5, P(T) = 0.69, P(S/T) = 0.5$$

...view full instructions

The value of $$P(S \cap T)$$

A
0.3450

B
0.2500

C
0.6900

D
0.350

Solution

Given, S and T are two events in a sample place
$$P(S)=0.5,P(T)=0.69,P(S | T)=0.5$$
$$P(S | T)=P(S \;\cap\; T)/P(T)$$
$$\Rightarrow 0.5=\frac{P(S \cap T)}{0.69}$$
$$P(S \cap T)=0.69*0.5=0.345$$
Question 10
1 / -0

A man is know to speak truth $$3$$ out of $$4$$ times. He throws a dice and reports that it is six. The probability that it is actually six is

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

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

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

D
None of these

Solution

Let $$A$$ denotes the event that a six occurs and $$B$$ the events that the man reports that it is a six.
Then the probability that it is actually a six is given by
$$\displaystyle P\left( \frac { A }{ B } \right) =\frac { P\left( A\cap B \right) }{ P\left( B \right) } $$
Now $$\displaystyle P\left( A\cap B \right) =\frac { 1 }{ 6 } \times \frac { 3 }{ 4 } =\frac { 3 }{ 24 } $$
$$\displaystyle P\left( B \right) =P\left( A\cap B \right) +P\left( \overline { A } \cap B \right) =\frac { 1 }{ 6 } \times \frac { 3 }{ 4 } +\frac { 5 }{ 6 } \times \frac { 1 }{ 4 } =\frac { 8 }{ 24 } $$
Hence $$\displaystyle P\left( \frac { A }{ B } \right) =\frac { \frac { 3 }{ 24 } }{ \frac { 8 }{ 24 } } =\frac { 3 }{ 8 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 22

Result

Probability Test - 22

Score

-

Rank

-

SHARING IS CARING

Question 1

$$0.22$$

$$0.11$$

$$0.33$$

None of these

Solution

Question 2

$$\displaystyle \frac{12}{17}$$

$$\displaystyle \frac{5}{17}$$

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

$$\displaystyle \frac{7}{12}$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$1/3$$

$$2/9$$

$$4/9$$

$$5/9$$

Solution

Question 5

$$0.14$$

$$0.24$$

$$0.34$$

$$0.44$$

Solution

Question 6

$$\displaystyle \frac { 12 }{ 33 } $$

$$\displaystyle \frac { 13 }{ 33 } $$

$$\displaystyle \frac { 5 }{ 22 } $$

None of these

Solution

Question 7

$$9$$

$$10$$

$$11$$

$$12$$

Solution

Question 8

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

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

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

$$\displaystyle \frac { 3 }{ 7 } $$

Solution

Question 9

0.3450

0.2500

0.6900

0.350

Solution

Question 10

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

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

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

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.