Probability Test

Result

Probability Test - 47

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 bag contains 6 balls. Two balls are drawn and found to be red. The probability that five balls in the bag are red

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

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

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

D
$$\displaystyle \frac{2}{7}$$

Solution

Let A be the event that we draw two red balls.
Let B be the event that five balls are red.
We need to find $$P(B|A)$$.
Required probability = $$P(\dfrac { B }{ A } )=\dfrac { P(\dfrac { A }{ B } )\times P(B) }{ P(A) } $$

Now, $$P(A)=P(\dfrac { A }{ 2R } )\times P(2R)+P(\dfrac { A }{ 3R } )\times P(3R)+P(\dfrac { A }{ 4R } )\times P(4R)+P(\dfrac { A }{ 5R } )\times P(5R)+P(\dfrac { A }{ 6R } )\times P(6R)$$
There can be $$5$$ cases: 6 Red or 5 Red or 4 Red or 3 Red or 2 Red. We assume that each case has a probability of $$\dfrac { 1 }{ 5 } $$.

Thus, $$P(A)=\dfrac { _{ 2 }^{ 2 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 3 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 4 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 5 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } +\dfrac { _{ 2 }^{ 6 }{ C } }{ _{ 2 }^{ 6 }{ C } } \times \dfrac { 1 }{ 5 } =\dfrac { 7 }{ 15 } $$

Again, $$P(\dfrac { A }{ B } )=\dfrac { _{ 2 }^{ 5 }{ C } }{ _{ 2 }^{ 6 }{ C } } =\dfrac { 2 }{ 3 } $$
Therefore, required probability $$=$$ $$\dfrac { \dfrac { 2 }{ 3 } \times \dfrac { 1 }{ 5 } }{ \dfrac { 7 }{ 15 } } =\dfrac { 2 }{ 7 } $$
Question 2
1 / -0

There are 3 boxes each having two drawers The first box contains a gold coin in each drawer. The second, a gold coin in one drawer and a silver coin in the other. The third box contains a silver coin in each drawer. A box is chosen at random and a drawer is opened. If a gold coin is found in that drawer, then the probability that the other drawer also contains a gold coin is

A
$$\dfrac{1}{3}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{2}{3}$$

D
$$\dfrac{3}{5}$$

Solution

if boxes are $${ B }_{ 1, }{ \quad B }_{ 2, }\quad { B }_{ 3, }\quad P({ B }_{ i, })\quad =\quad 1/3\quad for\quad i\quad =1,2,3$$
$$P(\dfrac { G }{ { B }_{ 1 } } )\quad =\quad 1,\quad P(\dfrac { G }{ { B }_{ 2 } } )\quad =\quad 1/2,\quad P(\dfrac { G }{ { B }_{ 3 } } )\quad =\quad 0$$
by Baye's theorem
$$P(\dfrac { { B }_{ 1 } }{ G } )\quad =\quad \dfrac { 1\times 1/3 }{ 1\times 1/3+1/2\times 1/3+0 } =2/3$$
Question 3
1 / -0

The contents of urn I and II are as follows,
Urn I: 4 white and 5 black balls
Urn II : 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn,
colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. The probability that the chosen urn was I is

A
$$\frac{125}{287}$$

B
$$\frac{64}{127}$$

C
$$\frac{25}{287}$$

D
$$\frac{79}{192}$$

E
$$\frac{250}{9^{4}}$$

Solution

Let I be the event that urn I is selected and II be the event that urn II is selected. Let E be the event that one ball drawn is white and three balls are black.
P(E) = P(I) P(E/I) + P(II) P(E/II)
$$=\frac{1}{2}.\frac{4}{9}\left ( \frac{5}{9} \right )^{3} \times 4 + \frac{1}{2}.\frac{3}{9}.\left ( \frac{6}{9} \right )^{3} \times 4=\frac{1148}{2 \times 9^{4}}\times 4$$
$$P(I/E)=\frac{\frac{500}{2\times 9 ^{4}}\times 4}{\frac{1148}{2 \times 9^{4}}\times 4}=\frac{125}{287}$$
Question 4
1 / -0

If $$P(A)=0.3$$ and $$P(A\cup B)=0.8, P(A-B)=0.1$$
then value of $$P(A/\overline{B})=$$

A
$$\dfrac{1}{2}$$

B
$$\dfrac{2}{3}$$

C
$$1$$

D
$$\dfrac{1}{3}$$

Solution

Given $$P(A)=0.3$$ and $$P(A\cup B)=0.8, P(A-B)=0.1$$
Now, $$P(A/\overline { B } )=\frac { P(A\cap \overline { B } ) }{ P(\overline { B } ) } $$ ....(1)
Since, $$P(A-B)=P(A\cap \overline B)$$
$$\Rightarrow P(A\cap \overline B)=0.1$$ .....(2)
Also, we have $$P(A\cap B)=P(A)-P(A\cap \overline B)$$
$$\Rightarrow P(A\cap B)=0.3-0.1=0.2$$

Also, we have $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$\Rightarrow P(B)=0.7$$
$$\Rightarrow P(\overline B)=0.3$$ ....(3)
Put the values from (2) and (3) in (1)
$$\displaystyle P(A/\overline { B } )=\frac { 0.1 }{ 0.3 } =\frac { 1 }{ 3 } $$
Question 5
1 / -0

A bag contains 4 white and 5 black balls. A second bag contains 3 (identical) white and 6 (identical) black balls. One bag is chosen at random and a ball is drawn. Its colour is noted and the ball replaced. This is repeated four times. It was found that of these four, one was white and 3 were black. The $$n$$ the probability that the first bag was chosen is

A
$$\dfrac{162}{287}$$

B
$$\dfrac{125}{287}$$

C
$$\dfrac{250}{280}$$

D
$$\dfrac{125}{574}$$

Solution

If $$A$$ is the event of choosing the first bag and $$B$$ the event of choosing the second bag. $$P(A) = P(B) = 12 $$
Let $$C$$ be the event of one white and $$3$$ black being drawn .
Using Baye's theorem
$$P(A|C)=\dfrac { P(C|A)*P(A) }{ P(C|A)*P(A)+P(C|B)*P(B) } =\dfrac { 1/2\times (4/9)^{ 2 }\times (5/9)^{ 3 } }{ 1/2\times (4/9)^{ 2 }\times (5/9)^{ 3 }+1/2\times (1/3)^{ 2 }\times (2/3)^{ 3 } } =125/287$$
Question 6
1 / -0

Die $$A$$ has $$4$$ red and $$2$$ white faces where as die $$B$$ has two red and $$4$$ white faces. $$A$$ fair coin is tossed. If head turns up, the game continues by throwing die $$A$$, if tail turns up then die $$B$$ is to be used. If the first two throws resulted in red, what is the probability of getting red face at the third throw ?

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

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

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

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

Solution

Let $$E_{1}$$ be the event that die $$A$$ is used and $$E_{2}$$ be the event that die $$B$$ is used

Let $$C$$ be the event that a red face appears in any throw.
$$ P(E_{1}) = \displaystyle\frac{1}{2} = P(E_{2})$$

$$ P(C/E_{1}) = \displaystyle\frac{^{4}C_{1}}{^{6}C_{1}} = \displaystyle\frac{2}{3}$$,

$$P(C/E_{2}) = \displaystyle\frac{^{2}C_{1}}{^{6}C_{1}} = \displaystyle\frac{1}{3}$$

$$\displaystyle P(C) = P(E_{1}) P (C/E_{1}) + P(E_{2}). P(C/E_{2}) = 1/2 \times 2/3 + 1/2 \times 1/3 = 1/2$$

Let $$ D$$ be the event that red face appears in third throw
Let $$E$$ be the even that red face appears in first two throws
$$ P(E_{1}) = P(E_{2}) = \displaystyle\frac{1}{2}$$

$$ P(E/E1) = \displaystyle\frac{2}{3}.\displaystyle\frac{2}{3} =\left(\displaystyle\frac{2}{3}\right)^{2}$$,

$$P(D/EE_{1}) = \displaystyle\frac{2}{3}$$

$$ P(E/E2) = \displaystyle\frac{1}{3} \times \displaystyle\frac{1}{3} =\left(\displaystyle\frac{1}{3}\right)^{2}$$

$$ P(D/EE_{2}) = \displaystyle\frac{1}{3}$$

$$ P(D/E) = \displaystyle\frac{P(E_{1} ) P(E/E_{1} )P(D/EE_{1} ) + P(E_{2} ).P(E/E_{2} )P(D/EE_{2} )}{P(E_{1} ).P(E/E_{1} ) + P(E_{2} ).P(E/E_{2} )}$$

$$= \displaystyle\frac{1/2(2/3)^{2} \times 2/3 + 1/2(1/3)^{2} \times 1/3}{1/2 \times (2/3)^{2} + 1/2 \times (1/3)^{2}} = \displaystyle\frac{3}{5}$$
Question 7
1 / -0

Let $$E^{c}$$ denote the complement of an event Let$$E,F,G$$ be pairwise independent events with $$P(G)>0$$ and $$P(E\cap F\cap G)=0$$ Then $$P(E^{c}\cap F^{c}|G)$$ equals

A
$$P(E^{c})+P( F^{c})$$

B
$$P(E^{c})-P( F^{c})$$

C
$$P(E^{c})-P( F)$$

D
$$P(E)-P( F^{c})$$

Solution

$$P\left( \overline { E } \cap \overline { F } /G \right) =\dfrac { P\left( \overline { E } \cap \overline { F } \cap G \right) }{ P\left( G \right) } =\dfrac { P\left( \overline { E } \right) P\left( \overline { F } \right) P\left( G \right) }{ P\left( G \right) } =P\left( \overline { E } \right) P\left( \overline { F } \right) \\ =P\left( \overline { E } \right) \left\{ 1-P\left( F \right) \right\} =P\left( \overline { E } \right) -P\left( \overline { E } \right) P\left( F \right) \\ =P\left( \overline { E } \right) -\left\{ 1-P\left( E \right) \right\} P\left( F \right) =P\left( \overline { E } \right) -P\left( F \right) +P\left( E \right) P\left( F \right) $$
now given that $$E,F,G\quad are\quad independent\quad and\quad P\left( E\cap F\cap G \right) =0\\ \Rightarrow P\left( E \right) P\left( F \right) P\left( G \right) =0\quad but\quad P\left( G \right) \neq 0\\ \quad \Rightarrow P\left( E \right) P\left( F \right) =0\\ hence\quad P\left( \overline { E } \cap \overline { F } /G \right) =P\left( \overline { E } \right) -P\left( F \right) $$
Question 8
1 / -0

A bag contains some white and some black balls, all combinations of ball being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is

A
$$14/55$$

B
$$12/55$$

C
$$2/11$$

D
$$8/55$$

Solution
Question 9
1 / -0

In a class of $$10$$ students, probability of exactly $$i$$ students passing examination is directly proportional to $$i^2$$. If a student selected at random is found to have passed the examination, then the probability that he was the only student who has passed the examination is

A
$$\dfrac1{3025}$$

B
$$\dfrac1{605}$$

C
$$\dfrac1{275}$$

D
$$\dfrac1{121}$$

Solution

Let constant of proportionality be $$k$$.
Let the event that $$n$$ students pass the exam be $${ E }_{ n }$$
Thus, $${ E }_{ n }=k{ n }^{ 2 }$$
Thus, $${ E }_{ 0 }+{ E }_{ 1 }+{ E }_{ 2 }...{ E }_{ 10 }=1$$
Let $$A$$ be the event that a student passes the exam.
We need to know $$P\left(\dfrac { { E }_{ 1 } }{ A } \right)$$
Now,
$$P\left(A\right)=P\left({ E }_{ 1 }\right)P\left(\dfrac { A }{ { E }_{ 1 } } \right)+P\left({ E }_{ 2 }\right)P\left(\dfrac { A }{ { E }_{ 2 } } \right)+...+P\left({ E }_{ 10 }\right)P\left(\dfrac { A }{ { E }_{ 10 } } \right)$$
Thus,
$$P\left(A\right)=\left(k{ 1 }^{ 2 }\dfrac { 1 }{ 10 } \right)+\left(k2^{ 2 }\dfrac { 2 }{ 10 } \right)+...+\left(k9^{ 2 }\dfrac { 9 }{ 10 } \right)+\left(k{ 10 }^{ 2 }\dfrac { 10 }{ 10 } \right)=\dfrac { k }{ 10 } \left({ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ 10 }^{ 3 }\right)$$
Thus,
$$P\left(A\right)=\dfrac { k }{ 10 } { \left(\dfrac { 10\left(10+1\right) }{ 2 } \right) }^{ 2 }=\dfrac { 3025k }{ 10 }$$
Thus,
$$P\left(\dfrac { { E }_{ 1 } }{ A } \right)=\dfrac { P\left({ E }_{ 1 }\right)P\left(\dfrac { A }{ { E }_{ 1 } } \right) }{ P\left(A\right) }=\dfrac { \left(k{ 1 }^{ 2 }\right)\left(\dfrac { 1 }{ 10 } \right) }{ \dfrac { 3025k }{ 10 } } =\dfrac { 1 }{ 3025 }$$
Question 10
1 / -0

Let A and B be two events such that $$ P(A \cap B^{c})=0.20, P( A^{c} \cap B)=0.15, P(A^{c} \cap B^{c})=0.1$$, then $$P(A/B)$$ is equal to

A
$$11/14$$

B
$$2/11$$

C
$$2/7$$

D
$$1/7$$

Solution

$$\because P\left( A\cup B \right) =1-P\left( { A }^{ c }\cap { B }^{ c } \right) \\ \Rightarrow P\left( A\cup B \right) =1-0.1=0.9$$
We know that
$$P\left( A\cap { B }^{ c } \right) =P\left( A \right) -P\left( A\cap B \right) \\ and\quad P\left( { A }^{ c }\cap B \right) =P\left( B \right) -P\left( A\cap B \right)$$

Adding the above two equations
$$P\left( A\cap { B }^{ c } \right) +P\left( A^{c}\cap { B } \right) =P\left( A \right) -P\left( A\cap B \right) +P\left( B \right) -P\left( A\cap B \right) \\ \Rightarrow 0.2+0.15=P\left( A \right) +P\left( B \right) -2P\left( A\cap B \right) \\ \Rightarrow P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) -P\left( A\cap B \right) =0.35\\ \Rightarrow P\left( A\cup B \right) -P\left( A\cap B \right) =0.35$$

putting the value of $$P\left( A\cup B \right) $$obtained earlier in the above result
$$\Rightarrow \quad P\left( A\cap B \right) =0.9-0.35=0.55\\ Also\quad P\left( B \right) =P\left( { A }^{ c }\cap B \right) +P\left( { A }\cap { B } \right) =0.15+0.55=0.70\\ \Rightarrow P\left( B \right) =0.70\\ \because P\left( A/B \right) =\dfrac { P\left( A\cap B \right) }{ P\left( B \right) } $$
hence putting the values
$$P\left( A/B \right) =\dfrac { 0.55 }{ 0.70 } =\dfrac { 11 }{ 14 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 47

Result

Probability Test - 47

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

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

$$\displaystyle \frac{2}{7}$$

Solution

Question 2

$$\dfrac{1}{3}$$

$$\dfrac{2}{5}$$

$$\dfrac{2}{3}$$

$$\dfrac{3}{5}$$

Solution

Question 3

$$\frac{125}{287}$$

$$\frac{64}{127}$$

$$\frac{25}{287}$$

$$\frac{79}{192}$$

$$\frac{250}{9^{4}}$$

Solution

Question 4

$$\dfrac{1}{2}$$

$$\dfrac{2}{3}$$

$$1$$

$$\dfrac{1}{3}$$

Solution

Question 5

$$\dfrac{162}{287}$$

$$\dfrac{125}{287}$$

$$\dfrac{250}{280}$$

$$\dfrac{125}{574}$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$P(E^{c})+P( F^{c})$$

$$P(E^{c})-P( F^{c})$$

$$P(E^{c})-P( F)$$

$$P(E)-P( F^{c})$$

Solution

Question 8

$$14/55$$

$$12/55$$

$$2/11$$

$$8/55$$

Solution

Question 9

$$\dfrac1{3025}$$

$$\dfrac1{605}$$

$$\dfrac1{275}$$

$$\dfrac1{121}$$

Solution

Question 10

$$11/14$$

$$2/11$$

$$2/7$$

$$1/7$$

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.