Probability Test

Result

Probability Test - 35

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

$$X$$ has three children in his family. The probability of one girl and two boys is......

A
$$\dfrac{1}{8}$$

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

C
$$\dfrac{1}{4}$$

D
$$\dfrac{3}{8}$$

Solution

$$X$$ has $$3$$ children in his family.
All possible outcomes are if G represents girl child and $$B$$ represent boy child,$$\{(BBB), (BBG), (BGB), (BGG), (GBB), (GBG), (GGB), (GGG)\}$$, hence $$n(S)=8$$
Out of which favorable outcomes for one girl and two boys are, $$n(E)=3$$
Hence, the probability that there are one girl and two boys is $$\dfrac {n(E)}{n(S)}=\dfrac 38$$
Question 2
1 / -0

A student appears for tests I, II and III. The student is successful if he passes in tests I, II or I, III. The probabilities of the student passing tests I, II and III are respectively $$p,q$$ and $$\cfrac{1}{2}$$. If the probability of the student to be successful is $$\cfrac{1}{2}$$. Then

A
$$p(1+q)=1$$

B
$$q(1+p)=1$$

C
$$pq=1$$

D
$$\cfrac{1}{p}+\cfrac{1}{q}=1$$

Solution

$$\quad \cfrac { 1 }{ 2 } =P(I)P(II)P(III')+P(I)P(II')P(III)+P(I)P(II)P(III)$$
$$=p.q\left( 1-\cfrac { 1 }{ 2 } \right) +P.(1-q)\cfrac { 1 }{ 2 } +p.q\cfrac { 1 }{ 2 } $$
$$1=pq+p-pq+pq$$
$$\Rightarrow 1=pq+p=p(q+1)$$
Question 3
1 / -0

A car is parked by an owner in a parking lot of $$25$$ cars in a row, including his car not at either end. On his return he finds that exactly $$15$$ placed are still occupied. The probability that both the neighboring places are empty is

A
$$\dfrac {91}{276}$$

B
$$\dfrac {15}{184}$$

C
$$\dfrac {15}{92}$$

D
None of these

Solution

Fixing $$15$$ cars in $$25$$ places, there are $$14$$ cars other than his own in $$24$$ cars.

Therefore, total number of ways = $$^{24}C_{14}=\dfrac {24!}{14!(25-15)!}$$
Now the $$14$$ cars must be parked in $$(25-3=22)$$ places, since neighboring places are empty.

Therefore, number of favorable ways = $$^{22}C_{14}=\dfrac {22!}{14!(22-14)!}$$

Hence the required probability = $$\dfrac {\text{Favourable ways}}{\text{Total ways}}=\dfrac {22!}{14!8!}\times \dfrac {14!10!}{24!}=\dfrac {10\times 9}{23\times 24}=\dfrac {15}{92}$$
Question 4
1 / -0

Urn $$A$$ contains $$6$$ red and $$4$$ black balls and urn $$B$$ contains $$4$$ red and $$6$$ black balls. One ball is drawn at random from urn $$A$$ and placed in urn $$B$$. Then one ball is drawn at random from urn $$B$$ and placed in urn $$A$$. If one ball is now drawn at random from urn $$A$$, the probability that it is red is

A
$$\dfrac {19}{55}$$

B
$$\dfrac {32}{55}$$

C
$$\dfrac {41}{55}$$

D
$$\dfrac {9}{55}$$

Solution

There are four scenarios that can occur where final draw gives us red ball

Scenario 1
1st red drawn from urn a put into urn b
then black drawn from urn b put into urn a
then red drawn from urn a

Probability = $$\dfrac6{10}\times\dfrac{6}{11}\times\dfrac{5}{10} =\dfrac{180}{1100}$$

Scenario 2
1st black drawn from urn a put into urn b
then black drawn from urn b put into urn a
then red drawn from urn a

Probability = $$\dfrac4{10}\times\dfrac{7}{11}\times\dfrac{6}{10}=\dfrac{168}{1100}$$

Scenario 3
1st red drawn from urn a put into urn b
then red drawn from urn b put into urn a
then red drawn from urn a

Probability = $$\dfrac6{10}\times\dfrac{5}{11}\times\dfrac{6}{10}=\dfrac{180}{1100}$$

Scenario 4
1st black drawn from urn a put into urn b
then red drawn from urn b put into urn a
then red drawn from urn a

Probability = $$\dfrac4{10}\times\dfrac{4}{11}\times\dfrac{7}{10}=\dfrac{112}{1100}$$

Total probability$$ = \dfrac{640}{1100} = \dfrac{32}{55}$$
Question 5
1 / -0

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is

A
$$0.06$$

B
$$\dfrac { 7 }{ 22 } $$

C
$$0.2$$

D
$$0.7$$

Solution

$${\textbf{Step -1: We have probabilities of I and II scoring a hit correctly are 0}}{\textbf{.3 and 0}}{\textbf{.2}}{\text{.}}$$

$${\text{Let us assume that the probability of aeroplane I hitting correctly be P}}\left( A \right)$$

$$ \Rightarrow P\left( A \right) = 0.3$$ $$\left( {{\textbf{Given}}} \right)$$

$${\text{Therefore, The probability of aeroplane I missing is,}}$$

$$P\left( {\overline A } \right) = 1 - P\left( A \right)$$

$$ \Rightarrow P\left( {\overline A } \right) = 1 - 0.3 = 0.7$$

$${\text{Let us assume that the probability of aeroplane II hitting correctly be P}}\left( B \right)$$

$$ \Rightarrow P\left( B \right) = 0.2$$ $$\left( {{\textbf{Given}}} \right)$$

$${\text{Therefore, The probability of aeroplane II missing is,}}$$

$$P\left( {\overline B } \right) = 1 - P\left( B \right)$$

$$ \Rightarrow P\left( {\overline A } \right) = 1 - 0.2 = 0.8$$

$${\textbf{Step -2: Find the probability of aeroplane II hitting the target.}}$$

$${\text{The Probability of aeroplane II hit the target}}$$

$$ = P\left( {\overline A } \right) \times P\left( B \right) + P\left( {\overline A } \right) \times P\left( {\overline B } \right) \times P\left( {\overline A } \right) \times P\left( B \right) + \ldots $$

$${\text{Substituting the known values in the above equation we have,}}$$

$${\text{The Probability of aeroplane II hit the target}}$$

$$ = 0.7 \times 0.2 + 0.7 \times 0.8 \times 0.7 \times 0.2 + \ldots $$

$$ = 0.7 \times 0.2\left[ {1 + 0.56 + 2\left( {0.56} \right) + \ldots } \right]$$

$$ = 0.14\left[ {1 + 0.56 + 2\left( {0.56} \right) + \ldots } \right]$$

$${\text{The above equation is in G}}{\text{.P}}{\text{.}}$$ $${\text{Where}}$$ $$a = 1,$$ $$r = 0.56$$

$$\therefore $$ $${\text{Probability of aeroplane II hit the target}}$$ $$ = 0.14\left[ {\dfrac{1}{{1 - 0.56}}} \right]$$ $$\left( {\because \mathbf{{S_\mathbf {\infty }}} = \mathbf{\dfrac{a}{{1 - r}}} }\right)$$

$$\therefore $$ $${\text{Probability of aeroplane II hit the target}}$$ $$ = 0.14\left[ {\dfrac{1}{{0.44}}} \right]$$

$$\therefore $$ $${\text{Probability of aeroplane II hit the target}}$$ $$ = \dfrac{7}{{22}}$$

$${\textbf{Hence, the probability that the target is hit by the second plane is}}$$ $$\mathbf{\dfrac{7}{{22}}.}$$ $${\textbf{Therefore, option B}}{\textbf{. is the correct answer.}}$$
Question 6
1 / -0

Directions For Questions

* A coin of Diameter $$D$$ is thrown randomly on a floor.
* The floor is tiled with squares of side $$'L'$$.
* Player $$'A'$$ bets that the coin will land Exactly on one Tile
* Player $$'B'$$ bets that the coin will land on More than One Tile.

...view full instructions

Player $$A$$ is more likely to win if

A
$$L = 10$$ cm, $$D = 3$$ cm

B
$$L = 20$$ cm, $$D = 58$$ mm

C
$$D < \left (L - \dfrac {L}{\sqrt {2}}\right )$$

D
$$D < \dfrac {L}{\sqrt {2}}$$

Solution

$$\rightarrow$$ For $$A$$ to win, coin should fully Lie inside outer square (see figure).
$$\rightarrow$$ Thus, coin center must lie inside the inner square of side length $$\left (L - \left (\dfrac {D}{2}\right )\right )$$, for the coin to fully lie inside outer square.
$$\rightarrow$$ When the coin in thrown randomly, it's centre falls inside inner square with probability,
$$P = \dfrac {\text {Area of Inner Square}}{\text {Area of Outer Square Tile}} = \dfrac {(L - 2\left (\dfrac {D}{2}\right ))^{2}}{L^{2}} = \dfrac {(L - D)^{2}}{L^{2}}$$
$$\rightarrow$$ Now, for $$A$$ to win $$P > \dfrac {1}{2}$$
ie $$\dfrac {(L - D)^{2}}{L^{2}} > \dfrac {1}{2}$$
$$\Rightarrow \dfrac {D}{L} < \left (1 - \dfrac {\sqrt {2}}{2}\right )$$ ie $$\dfrac {D}{L} < 1 - 0.707$$
ie $$\dfrac {D}{L} < 0.293$$ for $$A$$ to win
Question 7
1 / -0

The probabilities that a student will solve Question A and Question B are $$0.4$$ and $$0.5$$ respectively. What is the probability that he solves at least one of the two questions?

A
$$0.6$$

B
$$0.7$$

C
$$0.8$$

D
$$0.9$$

Solution

Given probabilities that a student will solve Question A and Question B are $$0.4$$ and $$0.5$$ respectively

Probability the student will not solve any are $$1-0.4=-0.6$$ and $$1-0.5=0.5$$ respectively
and

probability that he solves at least one of the two questions
$$\Rightarrow (total \ probability) -(\text{not solved at least one})= \ 1-(0.6\times0.5)=1-0.3=0.7$$

Therefore, Answer is $$B$$
Question 8
1 / -0

It is given that a family of $$3$$ children has two girls, what is the probability that the other child is also a girl?

A
$$0.50$$

B
$$0.75$$

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

D
$$\dfrac {1 }{ 4}$$

Solution

Probability that child is girl=1/2
Probability that the other child is also a girl given that two are already girls=
$$\frac { Probabilty\quad that\quad child\quad is\quad a\quad girl\quad given\quad two\quad are\quad already\quad girl }{ Probabilty\quad that\quad child\quad is\quad girl } $$
=$$\frac { ^{ 3 }{ C_{ 1 } }*1/2*1/2*1/2 }{ ^{ 3 }{ C_{ 1 } }*1/2 } =1/4$$
Question 9
1 / -0

If $$A$$ and $$B$$ are two events such that $$2P(A)=3P(B)$$, where $$0< P(A)< P(B)< 1$$, then which one of the following is correct?

A
$$P(A|B)<P(B|A)<P(A\cap B)$$

B
$$P(A\cap B)<P(B|A)<P(A|B)$$

C
$$P(B|A)<P(A|B)<P(A\cap B)$$

D
$$P(A\cap B) < P(A|B)<P(B|A)$$

Solution

Given$$:-$$
$$2P(A)=3P(B)$$

$$\Rightarrow$$ $$P(A)=\dfrac{3P(B)}{2}$$ $$\rightarrow \left \langle 1 \right \rangle$$

$$\Rightarrow \therefore P(A)>P(B)$$

Now we know$$:-$$
$$P(A|B)=\dfrac{P(A\cap B)}{P(B)} $$

Similarly,
$$P(B|A)=\dfrac{P(A\cap B)}{P(A)}$$

$$\Rightarrow P(B|A)=\dfrac{2P(A\cap B)}{3P(B)} $$ $$\left [ Using \left \langle 1 \right \rangle \right ]$$

$$\therefore$$ From above we conclude $$P(A|B)>P(B|A)$$

Now also we know, $$0<P(A)<1$$ and $$0<P(B)<1$$

$$\therefore P(B|A)=\dfrac{P(A\cap B)}{P(A)}>P(A\cap B) $$

Similarly,
$$P(A|B)=\dfrac{P(A\cap B)}{P(B)}>P(A\cap B) $$

$$\therefore P(A|B)>P(B|A)>P(A\cap B)$$
Question 10
1 / -0

A number is selected at random from the first $$1,000$$ natural numbers. What is the probability that the number so selected would be a multiple of $$7$$ or $$11$$?

A
$$0.25$$

B
$$0.32$$

C
$$0.23$$

D
$$0.33$$

Solution

Number of multiples of 7 from 1 to 100 $$= 142$$
Number of multiples of 11 from 1 to 100 $$= 99$$
Number of multiples of 77 = 12
So, number of multiples of 7 or 11
$$=142+99-12$$
$$=229$$
Probability $$=\quad \cfrac { 229 }{ 1000 } $$ = 0.229

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 35

Result

Probability Test - 35

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac{1}{8}$$

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

$$\dfrac{1}{4}$$

$$\dfrac{3}{8}$$

Solution

Question 2

$$p(1+q)=1$$

$$q(1+p)=1$$

$$pq=1$$

$$\cfrac{1}{p}+\cfrac{1}{q}=1$$

Solution

Question 3

$$\dfrac {91}{276}$$

$$\dfrac {15}{184}$$

$$\dfrac {15}{92}$$

None of these

Solution

Question 4

$$\dfrac {19}{55}$$

$$\dfrac {32}{55}$$

$$\dfrac {41}{55}$$

$$\dfrac {9}{55}$$

Solution

Question 5

$$0.06$$

$$\dfrac { 7 }{ 22 } $$

$$0.2$$

$$0.7$$

Solution

Question 6

$$L = 10$$ cm, $$D = 3$$ cm

$$L = 20$$ cm, $$D = 58$$ mm

$$D < \left (L - \dfrac {L}{\sqrt {2}}\right )$$

$$D < \dfrac {L}{\sqrt {2}}$$

Solution

Question 7

$$0.6$$

$$0.7$$

$$0.8$$

$$0.9$$

Solution

Question 8

$$0.50$$

$$0.75$$

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

$$\dfrac {1 }{ 4}$$

Solution

Question 9

$$P(A|B)<P(B|A)<P(A\cap B)$$

$$P(A\cap B)<P(B|A)<P(A|B)$$

$$P(B|A)<P(A|B)<P(A\cap B)$$

$$P(A\cap B) < P(A|B)<P(B|A)$$

Solution

Question 10

$$0.25$$

$$0.32$$

$$0.23$$

$$0.33$$

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.