Probability Test 1

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Probability Test 1

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Question 1
1 / -0

An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in a tail then a card from a well-shuffled pack of nine cards numbered $$1, 2, 3, .., 9$$ is randomly picked and the number on the card is noted. The probability that the noted number is either $$7$$ or $$8$$ is?

A
$$\dfrac{13}{36}$$

B
$$\dfrac{19}{36}$$

C
$$\dfrac{19}{72}$$

D
$$\dfrac{15}{72}$$

Solution

$$P(7\ or\ 8)$$

$$=P(H)P(7\ or\ 8)+P(T)P(7\ or\ 8)$$(head then sum on dice 7 or 8,tell then number on ticket 7 or 8)

$$=\dfrac{1}{2}\times \dfrac{11}{36}+\dfrac{1}{2}\times \dfrac{2}{9}=\dfrac{19}{72}$$.
Question 2
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Two different families $$A$$ and $$B$$ are blessed with equal number of children. There are $$3$$ tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family $$B$$ is $$\displaystyle\dfrac{1}{12}$$, then the number of children in each family is?

A
$$4$$

B
$$6$$

C
$$3$$

D
$$5$$

Solution

There are two different families $$A$$ and $$B$$ with equal number of children.
Let the children in each family be $$x$$.
Thus the total number of children in both the families are $$2x$$

Now, it is given that $$3$$ tickets are distributed amongst the children of the families.
And all the tickets are distributed to the children in family $$B$$.

Thus, the probability that all the three tickets go to the children in family $$B$$ is given by

$$\dfrac{1}{12} = \dfrac{^x C_3}{^{2x} C_3}$$
On solving the above equation, we get,

Thus, $$\dfrac{1}{12} = \dfrac{x(x-1)(x-2)}{2x(2x-1)(2x-2)}$$
Thus, $$\dfrac{1}{3} = \dfrac{x-2}{2x-1}$$

$$\rightarrow 3x-6 = 2x-1$$
$$\rightarrow x = 5$$

Thus, the number of children in each family are $$5$$.
Question 3
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A box $$'A'$$ contains $$2$$ white, $$3$$ red and $$2$$ black balls. Another box $$'B'$$ contains $$4$$ white, $$2$$ red and $$3$$ black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box $$'B'$$ is

A
$$\dfrac {7}{16}$$

B
$$\dfrac {9}{32}$$

C
$$\dfrac {7}{8}$$

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

Solution

For the bag $$A$$ we can see that there are $$2$$ white, $$3$$ red and $$2$$ black balls. Similarly, from bag $$B$$ we have $$4$$ white, $$2$$ red and $$3$$ black balls.

Probability of choosing a white and then a red ball from bag $$B$$ is given by $$=\dfrac{^4C_1 \times \ ^2C_1}{^9C_2}$$

Probability of choosing a white ball then a red ball from bag $$A$$ is given by $$=\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}$$

So, the probability of getting a white ball and then a red ball from bag $$B$$ is given by
$$\dfrac{\dfrac{^4C_1 \times \ ^2C_1 }{^9C_2}}{\dfrac{^4C_1 \times ^2C_1}{^9C_2}+\dfrac{^2C_1 \times \ ^3C_1}{^7C_2}}$$ $$=\dfrac{\dfrac{2}{9}}{\dfrac{2}{7}+\dfrac{2}{9}} = \dfrac{2 \times 7}{18+14}=\dfrac{7}{16}$$
Question 4
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A box $$B_{1}$$ contains $$1$$ white ball, $$3$$ red balls and $$2$$ black balls. Another box $$B_{2}$$ contains $$2$$ white balls, $$3$$ red balls and $$4$$ black balls. A third box $$B_{3}$$ contains $$3$$ white balls, $$4$$ red balls and $$5$$ black balls.
If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $$B_{2}$$ is

A
$$\displaystyle \frac{116}{181}$$

B
$$\displaystyle \frac{126}{181}$$

C
$$\displaystyle \frac{65}{181}$$

D
$$\displaystyle \frac{55}{181}$$

Solution

Let A: one ball is white and other is red

$$\mathrm{E}_{1}$$ : both balls are from box $$\mathrm{B}_{1}$$
$$\mathrm{E}_{2}$$ : both balls are from box $$\mathrm{B}_{2}$$
$$\mathrm{E}_{3}$$ : both balls are from box $$\mathrm{B}_{3}$$

Required Probability $$=\displaystyle \mathrm{P}(\frac{\mathrm{E}_{2}}{\mathrm{A}})$$
$$=\frac{\displaystyle
\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{2}})\cdot
\mathrm{P}(\mathrm{E}_{2})}{\displaystyle
\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{1}})\cdot
\mathrm{P}(\mathrm{E}_{1})+\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{2}})\cdot
\mathrm{P}(\mathrm{E}_{2})+\mathrm{P}(\frac{\mathrm{A}}{\mathrm{E}_{3}})\cdot
\mathrm{P}(\mathrm{E}_{3})}$$

$$=\displaystyle
\dfrac{\dfrac{^{2}\mathrm{C}_{1}\times ^{3}\mathrm{C}_{1}}{^{9}\mathrm{C}_{2}}\times\frac{1}{3}}{\dfrac{^{1}\mathrm{C}_{1}\times ^{3}\mathrm{C}_{1}}{^{6}\mathrm{C}_{2}}\times\dfrac{1}{3}+\dfrac{^{2}\mathrm{C}_{1}\times ^{3}\mathrm{C}_{1}}{^{9}\mathrm{C}_{2}}\times\dfrac{1}{3}+\dfrac{^{3}\mathrm{C}_{1}\times ^{4}\mathrm{C}_{1}}{^{12}\mathrm{C}_{2}}\times\dfrac{1}{3}}=\dfrac{\dfrac{1}{6}}{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{2}{11}}=\dfrac{55}{181}$$
Question 5
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A computer producing factory has only two plants $${ T }_{ 1 }$$ and $$ { T }_{ 2 }$$. Plant $$ { T }_{ 1 }$$ produces $$20\%$$ and plant $${ T }_{ 2 }$$ produces $$80\%$$ of total computers produced. $$7\%$$ of computers produced in the factory turn out to be defective. It is known that $$P$$ (computer turns out to be defective given that it is produced in plant $${ T }_{ 1 }$$) $$=10P$$ (computer turns out to be defective given that it is produced in plant $$\displaystyle { T }_{ 2 }$$).
where $$P(E)$$ denotes the probability of an event $$E$$. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $$ { T }_{ 2 }$$ is

A
$$\displaystyle \frac { 36 }{ 73 } $$

B
$$\displaystyle \frac { 47 }{ 79 } $$

C
$$\displaystyle \frac { 78 }{ 93 } $$

D
$$\displaystyle \frac { 75 }{ 83 } $$

Solution

Let $$x = P$$ (computer turns out to be defective given that it is produced in plant $$T_2$$),

$$\dfrac{7}{100}=\dfrac{1}{5}\times (10x)+\dfrac{4}{5}\times x \Rightarrow 7 = 200x + 80x \Rightarrow x = \dfrac{7}{280}$$
Event A = computer is produced in $$T_2$$
Event B = computer is not defective
$$P$$ (produced in $$T_2$$ not defective) $$= \dfrac{P(A \cap B)}{P(B)}$$

$$=\displaystyle \dfrac{4/5(1-x)}{\dfrac{1}{5}(1-10x)+\dfrac{4}{5}(1-x)}=\dfrac{\dfrac{4}{5}\left(\dfrac{273}{280}\right)}{\dfrac{1}{5}\left(\dfrac{280-70}{280}\right)+\dfrac{4}{5}\left(\dfrac{273}{280}\right)}=\dfrac{4\times 273}{210+4\times 273} = \dfrac{2\times 273}{105+2\times 273}=\dfrac{546}{651} = \dfrac{78}{93}$$
Question 6
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A bag contains 12 balls out of which x are white.If one ball is drawn at random, what is the probability it will be a white ball?

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

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

C
$$ \displaystyle \frac{x}{10}$$

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

Solution

Total number of balls = 12
Number of white balls = x
P (white ball) = $$= \displaystyle \frac{x}{12}$$
Question 7
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STATEMENT - 1 : Dependent events are those in which the outcome of one does not affect and is not affected by the other.
STATEMENT - 2 : Dependent events are those in which the outcome of one affects and is affected by the other.

A
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

B
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

C
Statement - 1 is True, Statement - 2 is False

D
Statement - 1 is False, Statement - 2 is True

Solution

Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.
Question 8
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All possible outcomes of a random experiment forms the -

A
Events

B
Sample space

C
Both

D
None of these

Solution

Sample Space is the set of all possible outcomes of an experiment. It is denoted by S.
Examples:
When a coin is tossed, S = {H, T} where H = Head and T = Tail
When a dice is thrown, S = {1, 2 , 3, 4, 5, 6}
When two coins are tossed, S = {HH, HT, TH, TT} where H = Head and T = Tail
Question 9
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The sum of the probabilities of all the elementary events of an experiment is ____?

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

The probability $$p$$ of any event must be greater than or equal to $$0$$.
In other words, $$0 \leq p \leq 1$$.
Since $$1$$ is the maximum limit, all probabilities must add to $$1$$.
Hence, the sum of the probabilities of all the elementary events of an experiment is $$1$$.
Question 10
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A bag contains 40 balls out of which some are red, some are blue and remaining are black. If the probability of drawing a red ball is $$\displaystyle \frac{11}{20}$$ and that of blue ball is $$\displaystyle \frac{1}{5}$$, then the number of black ball is?

A
$$5$$

B
$$25$$

C
$$10$$

D
$$30$$

Solution

Let no. of red balls = x, no. of blue balls =y and no.of black balls = z.
Probability of drawing a red ball = P (red ball) =$$\displaystyle \frac{11}{20} =\frac{x}{40}$$
$$\Rightarrow $$ x=22
Probability of drawing a blue ball = P (Blue ball) = $$\displaystyle \frac{1}{5} =\frac{y}{40}$$
$$\Rightarrow$$ y=8
$$\therefore$$ No of black balls = 40- 22 -8 =10

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Probability Test 1

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Probability Test 1

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SHARING IS CARING

Question 1

$$\dfrac{13}{36}$$

$$\dfrac{19}{36}$$

$$\dfrac{19}{72}$$

$$\dfrac{15}{72}$$

Solution

Question 2

$$4$$

$$6$$

$$3$$

$$5$$

Solution

Question 3

$$\dfrac {7}{16}$$

$$\dfrac {9}{32}$$

$$\dfrac {7}{8}$$

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

Solution

Question 4

$$\displaystyle \frac{116}{181}$$

$$\displaystyle \frac{126}{181}$$

$$\displaystyle \frac{65}{181}$$

$$\displaystyle \frac{55}{181}$$

Solution

Question 5

$$\displaystyle \frac { 36 }{ 73 } $$

$$\displaystyle \frac { 47 }{ 79 } $$

$$\displaystyle \frac { 78 }{ 93 } $$

$$\displaystyle \frac { 75 }{ 83 } $$

Solution

Question 6

$$ \displaystyle \frac{x}{2}$$

$$\displaystyle \frac{x}{12}$$

$$ \displaystyle \frac{x}{10}$$

$$ \displaystyle \frac{12}{x}$$

Solution

Question 7

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is False

Statement - 1 is False, Statement - 2 is True

Solution

Question 8

Events

Sample space

Both

None of these

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 10

$$5$$

$$25$$

$$10$$

$$30$$

Solution

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