Probability Test

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Probability Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).

A
$$\frac{14}{3}$$

B
$$\frac{13}{3}$$

C
$$\frac{12}{3}$$

D
$$\frac{11}{3}$$

Solution

E(X) is nothing but mean of X
$$\rightarrow $$ first sic natural numbers are
$$1,2,3,4,5,6$$
$$\rightarrow $$We can select two positive no.s in
$$6\times 5=30$$ ways
$$\rightarrow 2$$ numbers are selected at random among which X denotes the larger of the two no.s
$$\rightarrow $$But as X is largest among the two ,
it can be $$2,3,4,5$$ or $$6$$
$$\therefore P(X=2)=P$$ (larger no. is $$2$$) $${(1,2),(2,1)}$$
$$=\cfrac { 2 }{ 30 } $$
$$\rightarrow P\left( X=3 \right) =P$$ (Larger no. is $$3$$)
$$=\left\{ \left( 1,3 \right) ,\left( 3,1 \right) ,\left( 2,3 \right) ,\left( 3,2 \right) \right\} $$
$$=\cfrac { 4 }{ 30 } $$
$$\rightarrow P\left( X=4 \right) =P$$ (larger no. is $$4$$)
$$=\left\{ \left( 1,4 \right) ,\left( 4,1 \right) ,\left( 2,4 \right) ,\left( 4,2 \right) ,\left( 3,4 \right) ,\left( 4,3 \right) \right\} $$
$$=\cfrac { 6 }{ 30 } $$
$$\rightarrow P\left( X=5 \right) =P$$ (Larger no. is $$5$$)
$$=\left\{ \left( 1,5 \right) ,\left( 5,1 \right) ,\left( 2,5 \right) ,\left( 5,2 \right) ,\left( 3,5 \right) ,\left( 5,3 \right) ,\left( 4,5 \right) ,\left( 5,4 \right) \right\} $$
$$=\cfrac { 8 }{ 30 } $$
$$\rightarrow P\left( X=6 \right) =P$$ (Larger no. is $$6$$)
$$=\left\{ \left( 1,6 \right) ,\left( 6,1 \right) ,\left( 2,6 \right) ,\left( 6,2 \right) ,\left( 3,6 \right) ,\left( 6,3 \right) ,\left( 4,6 \right) ,\left( 6,4 \right) ,\left( 5,6 \right) ,\left( 6,50 \right) \right\} $$
$$=\cfrac { 10 }{ 30 } $$
$$\therefore E\left( X \right) =$$ mean $$=\sum { P\left( { X }_{ i } \right) } .{ X }_{ i }$$
$$=2\left( \cfrac { 2 }{ 30 } \right) +3\left( \cfrac { 4 }{ 30 } \right) +4\left( \cfrac { 6 }{ 30 } \right) +5\left( \cfrac { 8 }{ 30 } \right) +6\left( \cfrac { 10 }{ 30 } \right) $$
$$=\cfrac { 140 }{ 30 } =\cfrac { 14 }{ 3 } $$
Question 2
1 / -0

A bag contains $$5$$ red balls, $$3$$ black balls and $$4$$ white balls. Three balls are drawn at random. The probability that they are not of same colour is

A
$$\cfrac { 37 }{ 44 } $$

B
$$\cfrac { 31 }{ 44 } $$

C
$$\cfrac { 21 }{ 44 } $$

D
$$\cfrac { 41 }{ 44 } $$

Solution

To find the probability that they are not of same colour
Given ,
Bag Contains 5 red balls , 3 black balls and 4 white balls
First let's Find the Probability that they are of same colour .

Probability of drawing 5 red balls is

$$ P_{1} =\dfrac{\binom{5}{3} }{\binom{12}{3}}$$

Now ,  Probability of drawing 3 black balls is

$$ P_{2} =\dfrac{\binom{3}{3} }{\binom{12}{3}}$$

Also , Probability of drawing 4 white balls is

$$ P_{3} =\dfrac{\binom{4}{3} }{\binom{12}{3}}$$

Hence, Total Probability that they are of same colour . =

$$ P =\dfrac{\binom{5}{3} }{\binom{12}{3}} + \dfrac{\binom{3}{3} }{\binom{12}{3}}+ \dfrac{\binom{4}{3} }{\binom{12}{3}}$$

$$ P =\dfrac{\binom{5}{3} + \binom{3}{3} +\binom{4}{3} }{\binom{12}{3}}$$

Now,  Probability  that they are not of same colour can be calculated as

$$ P = 1 - \left [ \dfrac{\binom{5}{3} }{\binom{12}{3}} + \dfrac{\binom{3}{3} }{\binom{12}{3}}+ \dfrac{\binom{4}{3} }{\binom{12}{3}} \right ]$$

$$ P = 1 - \left [ \dfrac{\binom{5}{3} + \binom{3}{3} + \binom{4}{3}}{\binom{12}{3}}\right ]$$

$$ P = 1 - \left [ \dfrac{5\times 2 + 1 + 4}{220}\right ]$$

$$ P = 1 - \left [ \dfrac{15}{220}\right ]$$

$$ P = \dfrac{41}{44}$$
Question 3
1 / -0

Suppose four balls labelled $$1, 2, 3, 4$$ are randomly placed in boxes $$B_{1}, B_{2}, B_{3}, B_{4}$$. The probability that exactly one box is empty is

A
$$\dfrac {8}{256}$$

B
$$\dfrac {9}{16}$$

C
$$\dfrac {27}{256}$$

D
$$\dfrac {9}{64}$$

Solution

Total number of ways in which $$4$$ different balls can be placed in $$4$$ distinct boxes$$=4^4$$

number of ways in which exactly $$1$$ box is empty$$=^4C_1\times \dfrac{4!}{2!}\times 3$$

$$\therefore$$ The probability that exactly one box is empty is$$=\dfrac{4\times 4\times 3\times 3}{4^4}=\dfrac{9}{16}$$
Question 4
1 / -0

There are two bags, one of which contains $$3$$ black and $$4$$ white balls,$$II$$ bag contain $$4$$ black and $$3$$ white balls.A die is cast, if the number $$3$$ or less than $$3$$ turns up, a ball is drawn from $$1$$ bag and if the face greater than $$3$$ turns up, a ball is drawn from $$II$$ bag. It is found that ball drawn is black. Find the probability that it is from bag I.

A
$$\dfrac {3}{14}$$

B
$$\dfrac {4}{7}$$

C
$$\dfrac {2}{7}$$

D
$$\dfrac {3}{7}$$

Solution

Let $$A_{1}$$ be the event $$\leq 3$$ (when a die is cast) $$A_{2}$$ be the event $$\geq 4$$ (when a die is cast) $$'E'$$ be the event drawn ball is Black.
$$\therefore P(A_{1}) = \dfrac {1}{2} = P(A_{2})$$
$$P\left (\dfrac {E}{A_{1}}\right ) = \dfrac {3}{7}, P\left (\dfrac {E}{A_{2}}\right ) = \dfrac {4}{7}$$
$$\therefore P\left (\dfrac {A_{1}}{E}\right ) = \dfrac {P(A_{1})P\left (\dfrac {E}{A_{1}}\right )}{P(A_{1}) P\left (\dfrac {E}{A_{1}}\right ) + P(A_{2}) P\left (\dfrac {E}{A_{2}}\right )}$$
$$= \dfrac {\dfrac {1}{2}\times \dfrac {3}{7}}{\dfrac {1}{2} \times \dfrac {3}{7} + \dfrac {1}{2}\times \dfrac {4}{7}} = \dfrac {3}{7}$$
Hence choice (d) is correct.
Question 5
1 / -0

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

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

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

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 6
1 / -0

$$X$$ has three children in his family. What is the probability that all the three children are boys?

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

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

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

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 all the three chilren being boy, $$n(E)=1$$
Hence, the probability that all the three children are boys is $$\dfrac {n(E)}{n(S)}=\dfrac 18$$
Question 7
1 / -0

A committee of three persons is to be randomly selected from a group of three men and two women and the chair person will be randomly selected from the committee.The probability that the committee will have exactly two women and one man, and that the chair person will be a women, is / are

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

B
$$\dfrac{8}{15}$$

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

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

Solution

No of ways of selecting 3 persons from the group of 3 men and two woman = $$^{5}C_{3}$$
No. of ways of selecting exactly 2 women and 1 man = $$^{2}C_{2}\times ^{3}C_{1}= 3$$
$$\therefore $$ Probability of selecting exactly 2 women and 1 man = $$\dfrac{3}{^{5}C_{3}}$$
$$=\dfrac{3}{\frac{5!}{3! 2!}}=\dfrac{3}{10}$$
probability that the chair person selected from the committee is a woman = $$\dfrac{2}{3}$$
$$\therefore $$ Probability of selecting exactly 2 women and 1 man for the committee and chairman selected i.e a woman = $$\dfrac{3}{10}\times \dfrac{2}{3}$$
$$=\dfrac{1}{5}$$
Question 8
1 / -0

$$X$$ has three children in his family. Probability of atleast two girls in the family is.....

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

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

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

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

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 at-least two girls are, $$n(E)=4$$
Hence, the probability that there are at-least two girls is $$\dfrac {n(E)}{n(S)}=\dfrac 48=\dfrac 12$$
Question 9
1 / -0

A box contains $$2$$ silver coins and 4 gold coins and the second box contains $$4$$ silver coins and $$3$$ gold coins. If a coin is selected from one of the box, what is the probability that it is a silver coin.

A
$$0.3$$

B
$$0.4$$

C
$$0.5$$

D
$$0.6$$

Solution

Box $$I:2$$ Silver coins $$+\;{4}$$ gold coins
Box $$II:4$$ Silver coins $$+\;{3}$$ gold coins
Let $$E_1$$ be the event of selecting a gold coin from box $$I$$
and $$E_2$$ be the event of selecting a gold coin from box $$II.$$
$$A:$$ Probability of gold coin
$$\Rightarrow P(E_1)={^6C_2}$$            $$P(E_2)={^7C_4}$$
$$\Rightarrow P\left( \dfrac { A }{ E_1} \right) =P\left( \dfrac { A }{ E_2} \right) =\dfrac{1}{2}$$
$$\Rightarrow P\left( \dfrac { E_1 }{ E_2} \right) =\dfrac{P(E_1)P\left( \dfrac { A }{ E_1} \right) }{P(E_1)P\left( \dfrac { A }{ E_1} \right) P(E_2)P\left( \dfrac { A }{ E_2} \right) }$$
                        $$=\dfrac{{^6C_2}\times\dfrac{1}{2}}{{^6C_2}\times\dfrac{1}{2}+{^7C_4}\times\dfrac{1}{2}}$$
                        $$=\dfrac{7.5}{7.5+17.5}$$
                        $$=0.3$$
Hence, the answer is $$0.3$$
Question 10
1 / -0

$$X$$ has three children in his family. What is the probability of two or more boys in the family?

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 two or more boys are, $$n(E)=4$$
Hence, the probability that there are two or more boys in the family is $$\dfrac {n(E)}{n(S)}=\dfrac 48=\dfrac 12$$

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Probability Test - 34

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Probability Test - 34

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SHARING IS CARING

Question 1

$$\frac{14}{3}$$

$$\frac{13}{3}$$

$$\frac{12}{3}$$

$$\frac{11}{3}$$

Solution

Question 2

$$\cfrac { 37 }{ 44 } $$

$$\cfrac { 31 }{ 44 } $$

$$\cfrac { 21 }{ 44 } $$

$$\cfrac { 41 }{ 44 } $$

Solution

Question 3

$$\dfrac {8}{256}$$

$$\dfrac {9}{16}$$

$$\dfrac {27}{256}$$

$$\dfrac {9}{64}$$

Solution

Question 4

$$\dfrac {3}{14}$$

$$\dfrac {4}{7}$$

$$\dfrac {2}{7}$$

$$\dfrac {3}{7}$$

Solution

Question 5

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

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

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

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

Solution

Question 6

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

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

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

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

Solution

Question 7

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

$$\dfrac{8}{15}$$

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

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

Solution

Question 8

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

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

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

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

Solution

Question 9

$$0.3$$

$$0.4$$

$$0.5$$

$$0.6$$

Solution

Question 10

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

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

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

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

Solution

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