Probability Test

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Probability Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The probability distribution of a random variable is given below :
$$X = x$$ 0 1 2 3 4 5 6 7
$$P(X = x)$$ $$0$$ $$K$$ $$2K$$ $$2k $$ $$3K $$ $$K^2$$ $$2K^2 $$ $$7K^2 + k$$
Then $$P(0 < X < 5) = $$

A
$$\dfrac{1}{10}$$

B
$$\dfrac{3}{10}$$

C
$$\dfrac{8}{10}$$

D
$$\dfrac{7}{10}$$

Solution

$$\sum p(X=x_i)=1$$

$$9K+10K^{2}=1$$

$$10K^{2}+9K-1=0$$

$$10K^{2}+10k-k-1=0$$

$$10K(K+1)-1(K+1)=0$$

$$K=\dfrac{1}{10}$$

$$P(0< X< 5)=p(x=1)+p(x=2)+p(x=3)+p(x=4)=K+2K+2K+3K=8K=\dfrac{8}{10}$$

$$\therefore P(0< X< 5)=\dfrac{8}{10}$$
Question 2
1 / -0

A man is known to speak the truths $$3$$ out of $$4$$ times. He throw a die and report that it is six. The probability that it is actually a six, is

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

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

C
$$\dfrac {3}{4}$$

D
None of these

Solution

Let $$E$$ denote the event that a six occurs and $$A$$ be the event that the man reports that it is a $$6$$.
We have,
$$P(E) = \dfrac {1}{6}, P(E') = \dfrac {5}{6}$$
$$P\left (\dfrac {A}{E}\right ) =\dfrac {3}{4}$$ and $$P\left (\dfrac {A}{E'}\right ) = \dfrac {1}{4}$$
Using Baye's theorem, we get
$$P\left (\dfrac {E}{A}\right ) = \dfrac {P(E)\cdot P\left (\dfrac {A}{E}\right )}{P(E) \cdot P\left (\dfrac {A}{E}\right ) + P(E') \cdot P\left (\dfrac {A}{E}\right )}$$
$$= \dfrac {\dfrac {1}{6}\times \dfrac {3}{4}}{\dfrac {1}{6}\times \dfrac {3}{4} + \dfrac {5}{6}\times \dfrac {1}{4}} = \dfrac {3}{8}$$.
Question 3
1 / -0

Let $$u_1$$ and $$u_2$$ be two urns such that $$u_1$$ contains 3 white, 2 red balls and $$u_2$$ contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from urn $$u_1$$ and put into $$u_2$$. However, if tail appears, then 2 balls are drawn at random from $$u_1$$ and put into $$u_2$$. Now, 1 ball is drawn at random from $$u_2$$. Then, probability of the drawn ball from $$u_2$$ being white is

A
$$\dfrac{13}{30}$$

B
$$\dfrac{23}{30}$$

C
$$\dfrac{19}{30}$$

D
$$\dfrac{11}{30}$$

Solution

$${\textbf{Step - 1: Finding probabilities of different possible events}}$$

$${\text{Case - 1: Head appears on the coin, }}$$

$${\text{P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ = Probability of drawing white ball from urn 2, when head appears on the coin}}$$
  $$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ = }}\left( {\dfrac{{^{\text{3}}{{\text{C}}_{\text{1}}}}}{{^{\text{5}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{(when white is tranferred) }}$$
  $${\text{ + }}\left( {\dfrac{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{{^{\text{5}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{1}}{{\text{C}}_{\text{1}}}}}{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when red is transferred)}}$$

  $$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ = }}\left( {\dfrac{{\text{3}}}{{\text{5}}}{\text{ }} \times {\text{ 1 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + }}\left( {\dfrac{{\text{2}}}{{\text{5}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ = }}\dfrac{{\text{3}}}{{{\text{10}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{10}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{5}}}$$

  $${\text{Case - 2: Tail appears on the coin,}}$$

  $${\text{P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = Probability of drawing white ball from urn 2, when tail appears on the coin}}$$

  $$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = }}\left( {\dfrac{{^{\text{3}}{{\text{C}}_{\text{2}}}}}{{^{\text{5}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{3}}{{\text{C}}_{\text{1}}}}}{{^{\text{3}}{{\text{C}}_{\text{2}}}}} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when 2 white are transferred) }}$$
     $${\text{+ }}\left( {\dfrac{{^{\text{2}}{{\text{C}}_{\text{2}}}}}{{^{\text{5}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{1}}{{\text{C}}_{\text{1}}}}}{{^{\text{3}}{{\text{C}}_{\text{1}}}}} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when 2 red is transferred)}}$$

     $${\text{ + }}\left( {\dfrac{{^{\text{3}}{{\text{C}}_{\text{1}}}{\text{ }} \times {{\text{ }}^{\text{2}}}{{\text{C}}_{\text{1}}}}}{{^{\text{5}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{^{\text{2}}{{\text{C}}_{\text{1}}}}}{{^{\text{3}}{{\text{C}}_{\text{2}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)({\text{when 1 red and 1 white is transferred)}}$$

$$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = }}\left( {\dfrac{{{\text{3 }} \times {\text{ 2}}}}{{{\text{5 }} \times {\text{ 4}}}}{\text{ }} \times {\text{ 1 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{ + }}\left( {{\text{ }}\dfrac{{{\text{1 }} \times {\text{ 2}}}}{{{\text{5 }} \times {\text{ 4}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{3}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$$
$$+\left( {\dfrac{{{\text{3 }} \times {\text{ 2 }} \times {\text{ 2}}}}{{{\text{5 }} \times {\text{ 4}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$$

$$ \Rightarrow {\text{ P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right){\text{ = }}\dfrac{{\text{3}}}{{{\text{20}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{60}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{5}}}{\text{ = }}\dfrac{{{\text{11}}}}{{{\text{30}}}}$$

$${\textbf{Step - 2: Finding total probability}}$$

  $${\text{P = P}}\left( {\dfrac{{\text{W}}}{{\text{H}}}} \right){\text{ + P}}\left( {\dfrac{{\text{W}}}{{\text{T}}}} \right)$$

  $$ \Rightarrow {\text{ P = }}\dfrac{{\text{2}}}{{\text{5}}}{\text{ + }}\dfrac{{{\text{11}}}}{{{\text{30}}}}{\text{ = }}\dfrac{{{\text{23}}}}{{{\text{30}}}}$$

$$\mathbf{{\text{Hence, the probability that a white ball is drawn from Urn 2 is }}\dfrac{{{\text{23}}}}{{{\text{30}}}}}$$
Question 4
1 / -0

If r.v$$X$$: waiting time in minutes for bus and p.d.f of $$X$$ is given by
$$f(x)=\begin{cases} \cfrac { 1 }{ 5 } ,0\le x\le 5 \\ 0,\quad otherwise \end{cases}$$
then probability of waiting time not more than $$4$$ minutes is $$=$$...........

A
$$0.3$$

B
$$0.8$$

C
$$0.2$$

D
$$0.5$$

Solution

Given :
$$f(x)=\begin{cases} \cfrac { 1 }{ 5 } ,0\le x\le 5 \\ 0,\quad otherwise \end{cases}$$
Probability of waiting time not more than $$4$$ is same as probability of waiting time less than $$4$$ minutes and it is given by
$$P(X<4)=P(X=3)+P(X=2)+P(X=1)+P(X=0)$$
$$=\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}$$ .... [From the definition of $$f$$]
$$=4\times \dfrac { 1 }{ 5 } =0.8$$
Question 5
1 / -0

The probability distribution of $$X$$ is
$$X$$ 0 1 2 3
$$P(x)$$ 0.3 k 2k 3k
The value of $$k$$ is

A
$$0.116$$

B
$$0.7$$

C
$$1$$

D
$$0.3$$

Solution

$$\displaystyle \sum _{ x=0 }^{ \infty }{ P(x) } =1$$

$$\displaystyle \sum _{ x=0 }^{ 3 }{ P(x) } =1$$

$$P\left( 0 \right) +P\left( 1 \right) +P\left( 2 \right) +P\left( 3 \right) =1$$

$$\Rightarrow0.3+k+2k+3k=1$$

$$\Rightarrow 6k=1-0.3$$

$$\Rightarrow 6k=0.7$$

$$k=\cfrac { 0.7 }{ 6 }; \ $$$$k=0.116$$
Question 6
1 / -0

If two events $$A$$ and $$B$$ are such that $$P({ A }^{ C })=0.3,P(B)=0.4$$ and $$P({ AB }^{ C })=0.5$$, then $$P\left[ B/\left( A\cup { B }^{ C } \right) \right] $$ is equal to

A
$$\cfrac { 1 }{ 2 } $$

B
$$\cfrac { 1 }{ 3 } $$

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

D
None of these

Solution

Given, $$P({ A }^{ C })=0.3,P(B)=0.4$$ and $$P({ AB }^{ C })=0.5$$
Therefore, $$P\left[ { B }^{ C }|\left( A\cup { B }^{ C } \right) \right] =\cfrac { P\left\{ B\cap \left( A\cup { B }^{ C } \right) \right\} }{ P\left( A\cup { B }^{ C } \right) } $$
$$=\cfrac { P(A\cap { B }^{ C }) }{ P(A)+P(B)-P(A\cap { B }^{ C }) } $$
$$=\cfrac { P(A)-P(A\cap { B }^{ C }) }{ P(A)+P({ B }^{ C })-P(A\cap { B }^{ C }) } $$
$$=\cfrac { 0.7-0.5 }{ 0.8 } =\cfrac { 1 }{ 4 } $$

Question 7

1 / -0

For the following distribution function $$F(x)$$ of a r.v $$X$$ is given

$$x$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$	$$6$$
$$F(x)$$	$$0.2$$	$$0.37$$	$$0.48$$	$$0.62$$	$$0.85$$	$$1$$

Then $$P(3 < x\leq 5) =$$

$$0.48$$

$$0.37$$

$$0.27$$

$$1.47$$

Solution

$$x$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$	$$6$$
$$f(x)$$	$$0.2$$	$$0.37$$	$$0.48$$	$$0.62$$	$$0.85$$	$$1$$
$$p(x)$$	$$0.2$$	$$0.17$$	$$0.11$$	$$0.14$$	$$0.23$$	$$0.15$$

$$P(3 < x\leq 5) = p(x = 4) + p(x = 5)$$
$$= 0.14 + 0.23$$
$$= 0.37$$

Question 8
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A biased coin with probability P, (0 < p < 1 ) of heads is tossed until a head appear for the first time. If the probability that the number of tosses required is even is $$\frac{2}{5}$$ then P =

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

B
$$\frac{3}{5}$$

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

D
$$\frac{1}{3}$$

Solution

Chance of doing 2 toss=(1-p)*p
Chance of doing 4 toss=(1-p)*(1-p)*(1-p)*p
Chance of doing 6 toss=(1-p)*(1-p)*(1-p)*(1-p)*(1-p)*p
Probability that number of coins is even=chance of 2+chance of 4+chance of 6+......
$$=(1-p)*p(1+(1-p)*(1-p)+(1-p)*(1-p)*(1-p)*(1-p)+....)\\ \Rightarrow (1-p)*p(\frac { 1 }{ 1-{ (1-p })^{ 2 } } )=2/5\\ \Rightarrow (1-p)*p(\frac { 1 }{ p{ (2-p }) } )=2/5\\ \Rightarrow (1-p)*(\frac { 1 }{ { (2-p }) } )=2/5\\ On\quad solving\quad we\quad get\quad p=1/3\\ \\ $$
Question 9
1 / -0

From an urn containing six balls, $$3$$ white and $$3$$ black ones, a person selects at random an even number of balls (all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number).
Then the probability that there will be the same number of black and white balls among them is?

A
$$4/5$$

B
$$11/15$$

C
$$11/30$$

D
$$2/5$$

Solution

Probability of selecting 2,4, or 6 balls $$=\dfrac{1}{3}$$ each
Total ways of selecting 2 balls from 6 balls$$=\binom{6}{2}$$

Ways of selecting 1 white and 1 black ball$$=\binom{3}{1}\times\binom{3}{1}$$
Probability that 2 balls are picked having same no. of black and white balls$$=P_1=\dfrac{1}{3}\times\dfrac{\binom{3}{1}\times\binom{3}{1}}{\binom{6}{2}}$$
$$\Rightarrow \dfrac{3\times3\times2}{3\times6\times5}$$

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

Total ways of selecting 4 balls from 6 balls$$=\binom{6}{4}$$

Ways of selecting 2 white and 2 black balls$$=\binom{3}{2}\times\binom{3}{2}$$
Probability that 4 balls are picked having same no. of black and white balls$$=P_2=\dfrac{1}{3}\times\dfrac{\binom{3}{2}\times\binom{3}{2}}{\binom{6}{4}}$$
$$\Rightarrow \dfrac{3\times3\times2}{3\times6\times5}$$

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

Total ways of selecting 6 balls from 6 balls$$=\binom{6}{6}$$

Ways of selecting 3 white and 3 black balls$$=\binom{3}{3}\times\binom{3}{3}$$
Probability that 6 balls are picked having same no. of black and white balls$$=P_3=\dfrac{1}{3}\times\dfrac{\binom{3}{3}\times\binom{3}{3}}{\binom{6}{6}}$$
$$\Rightarrow \dfrac{1\times1}{3\times1}$$

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

Total probability$$=P_1+P_2+P_3$$
$$\Rightarrow \dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{3}$$

$$\Rightarrow \dfrac{11}{15}$$

Hence, answer is option $$B$$
Question 10
1 / -0

The letters of the work "Questions" are arranged in a row at random.The probability that there are exactly two letters between Q and S is.

A
$$\dfrac{1}{14}$$

B
$$\dfrac{5}{7}$$

C
$$\dfrac{1}{7}$$

D
$$\dfrac{5}{28}$$

Solution

$$n(E)=^6 P_2\times 2! \times 5!$$
$$n(S)=8!$$
$$\therefore$$ Required Probablility
$$\dfrac {n(E)}{n(S)} $$ $$=$$ $$ \dfrac{6^P_2 \times2!\times5!}{8!}$$
$$= \dfrac5{28}$$

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$$X = x$$	0	1	2	3	4	5	6	7
$$P(X = x)$$	$$0$$	$$K$$	$$2K$$	$$2k $$	$$3K $$	$$K^2$$	$$2K^2 $$	$$7K^2 + k$$

$$X$$	0	1	2	3
$$P(x)$$	0.3	k	2k	3k

Probability Test - 33

Result

Probability Test - 33

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SHARING IS CARING

Question 1

$$\dfrac{1}{10}$$

$$\dfrac{3}{10}$$

$$\dfrac{8}{10}$$

$$\dfrac{7}{10}$$

Solution

Question 2

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

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

$$\dfrac {3}{4}$$

None of these

Solution

Question 3

$$\dfrac{13}{30}$$

$$\dfrac{23}{30}$$

$$\dfrac{19}{30}$$

$$\dfrac{11}{30}$$

Solution

Question 4

$$0.3$$

$$0.8$$

$$0.2$$

$$0.5$$

Solution

Question 5

$$0.116$$

$$0.7$$

$$1$$

$$0.3$$

Solution

Question 6

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { 1 }{ 3 } $$

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

None of these

Solution

Question 7

$$0.48$$

$$0.37$$

$$0.27$$

$$1.47$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$4/5$$

$$11/15$$

$$11/30$$

$$2/5$$

Solution

Question 10

$$\dfrac{1}{14}$$

$$\dfrac{5}{7}$$

$$\dfrac{1}{7}$$

$$\dfrac{5}{28}$$

Solution

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