Probability Test

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Probability Test - 14

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Weekly Quiz Competition

Question 1
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Let $$\mathrm{E}^{\mathrm{c}}$$ denote the complement of an event $$\mathrm{E}$$. Let $$\mathrm{E},\ \mathrm{F},\ \mathrm{G}$$ be pairwise independent events with $$\mathrm{P}(\mathrm{G})>0$$ and $$\mathrm{P}(\mathrm{E}\,\mathrm{\cap}\,\mathrm{F}\,\mathrm{\cap}\,\mathrm{G}) =0$$. Then $$\mathrm{P}(\mathrm{E}^{\mathrm{c}}\,\mathrm{\cap}\,\mathrm{F}^{\mathrm{c}}|\mathrm{G})$$ equals

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

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

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

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

Solution
Question 2
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A signal which can be green or red with probability $$\displaystyle \frac{4}{5}$$ and $$\displaystyle \frac{1}{5}$$ respectively, is received by station $$\mathrm{A}$$ and then transmitted to station $$\mathrm{B}$$. The probability of each station receiving the signal correctly is $$\displaystyle \frac{3}{4}$$. If the signal received at station $$\mathrm{B}$$ is green, then the probability that the original signal was green is

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

B
$$\displaystyle \frac{6}{7}$$

C
$$\displaystyle \frac{20}{23}$$

D
$$\displaystyle \frac{9}{20}$$

Solution

Let Event $$\mathrm{G}=$$ original signal is green
$$\mathrm{E}_{1}=\mathrm{A}$$ receives the signal correct
$$\mathrm{E}_{2}=\mathrm{B}$$ receives the signal correct
$$\mathrm{E}=$$ signal received by $$\mathrm{B}$$ is green
$$\mathrm{P}$$(signal received by $$\mathrm{B}$$ is green)
$$=\mathrm{P}(\mathrm{G}\mathrm{E}_{1}\mathrm{E}_{2})+\mathrm{P}(\mathrm{G}\overline{\mathrm{E}}_{1}\overline{\mathrm{E}}_{2})+\mathrm{P}(\overline{\mathrm{G}}\mathrm{E}_{1}\overline{\mathrm{E}}_{2})+\mathrm{P}(\overline{\mathrm{G}}\
\overline{\mathrm{E}}_{1}\mathrm{E}_{2})$$

$$ = \displaystyle \left( \frac{4}{5}\times\frac34\times\frac34 \right) +\left( \frac45\times\frac14\times\frac14 \right) +\left( \frac15\times\frac34\times\frac14 \right )+\left(\frac15\times\frac14\times\frac34 \right)$$

$$\displaystyle \mathrm{P}(\mathrm{E})=\frac{46}{80}$$
$$\mathrm{P}(\mathrm{G}/\mathrm{E}) =\displaystyle \frac{\displaystyle \left( \frac{4}{5}\times\frac34\times\frac34 \right) +\left( \frac45\times\frac14\times\frac14 \right)}{\dfrac{46}{80}}=\frac{20}{23}$$
Question 3
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Directions For Questions

There are five students $$S_1, S_2, S_3, S_4$$ and $$S_5$$ in a music class and for them are five seats $$R_1, R_2, R_3, R_4$$ and $$R_5$$ arranged in a row, where initially the seat $$R_i$$ is allotted to the student $$S_i$$, $$i=1, 2, 3, 4, 5$$. But, on the examination day, the five students are randomly allotted the five seats.

...view full instructions

For $$i=1, 2, 3, 4$$ let $$T_i$$ denote the event that the students $$S_i$$ and $$S_{i+1}$$ do NOT sit adjacent to each other on the day of the examination. Then the probability of the event $$T_1 \cap T_2 \cap T_3 \cap T_4$$ is?

A
$$\displaystyle\frac{1}{15}$$

B
$$\displaystyle\frac{1}{10}$$

C
$$\displaystyle\frac{7}{60}$$

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

Solution

$$n(T_1 \cap T_2 \cap T_3 \cap T_4)=$$Total $$-n(\bar{T_1}\cup \bar{T_2}\cup \bar{T_3}\cup \bar{T_4})$$
$$=5!-\left({^4C_1}4!2!-\left({^3C_1}.3!2!+{^3C_1}3!2!2!\right)+\left({^2C_1}2!2!+{^4C_1}.2.2!\right)-2\right)$$
$$=14$$
Probability $$=\displaystyle\frac{14}{5!}=\frac{7}{60}$$.
Question 4
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Directions For Questions

There are five students $$S_1, S_2, S_3, S_4$$ and $$S_5$$ in a music class and for them are five seats $$R_1, R_2, R_3, R_4$$ and $$R_5$$ arranged in a row, where initially the seat $$R_i$$ is allotted to the student $$S_i$$, $$i=1, 2, 3, 4, 5$$. But, on the examination day, the five students are randomly allotted the five seats.

...view full instructions

The probability that, on the examination day, the student $$S_1$$ gets the previously allotted seat $$R_1$$ and None of the remaining students gets the seat previously allotted to him/her is?

A
$$\displaystyle\frac{3}{40}$$

B
$$\displaystyle\frac{1}{8}$$

C
$$\displaystyle\frac{7}{40}$$

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

Solution

Required probability $$=\displaystyle\frac{4!\left(\displaystyle\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right)}{5!}=\frac{9}{120}=\frac{3}{40}$$.
Question 5
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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 6
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A pack of playing cards was found to contain only $$51$$ cards. If the first $$13$$ cards which are examined are all red, then the probability thatthe missing card is black, is

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

B
$$\displaystyle \frac{2}{3}$$

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

D
$$\displaystyle \frac{^{25} C_{13}}{^{51} C_{13}}$$

Solution

$$Total\quad number\quad of\quad cards=52\\ Number\quad of\quad lost\quad cards=1\\ 13\quad cards\quad are\quad surely\quad red\quad therfore,\quad from\quad the\quad remaining\quad 39\quad cards\quad 26\quad are\quad black\quad and\quad 13\quad are\quad red\\ So\quad probability\quad of\quad lost\quad card\quad being\quad black\quad =\frac { \left( 26\\ 1 \right) }{ \left( 39\\ 1 \right) } =\frac { 26 }{ 39 } =\frac { 2 }{ 3 } $$
Question 7
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If $$P(A) + P(B) = 1$$; then which of the following option explains the event $$A$$ and $$B$$ correctly ?

A
Event $$A$$ and $$B$$ are mutually exclusive, exhaustive and complementary events.

B
Event $$A$$ and $$B$$ are mutually exclusive and exhaustive events.

C
Event $$A$$ and $$B$$ are mutually exclusive and complementary events.

D
Event $$A$$ and $$B$$ are exhaustive and complementary events.

Solution

Since $$P(A) +P(B) = 1$$
$$\therefore A \cap B = 0$$.
Thus, event $$A$$ and $$B$$ are mutually exclusive, exhaustive and complementary events.
Hence, option $$A$$ is correct.
Question 8
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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
Question 9
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There are 50 marbles of 3 colors: blue yellow and black The probability of picking up a blue marble is 3/10 and that of picking up a yellow marble is 1/2 The probability of picking up a black ball is

A
1/5

B
1/10

C
1/4

D
4/5

Solution

P(blue)=$$\displaystyle \frac{3}{10}=\frac{15}{50},$$i.e., there are 15 blue marbles
P(yellow)=$$\displaystyle \frac{1}{2}=\frac{25}{50},$$i.e., there are 25 yellow marbles
$$\displaystyle \therefore $$ Number of black marbles=10
$$\displaystyle \therefore $$ P(black)=$$\displaystyle \frac{10}{50}=\frac{1}{5}$$
Question 10
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A bag contains six red four green and eight white balls If a ball is picked at random the probability that it is not white is

A
1/3

B
4/9

C
5/9

D
2/3

Solution

Number of balls that are not white=10
Total=18
$$\displaystyle \therefore $$ P(not white)=$$\displaystyle \frac{10}{18}=\frac{5}{9}$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 14

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Probability Test - 14

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Question 1

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

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

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

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

Solution

Question 2

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

$$\displaystyle \frac{6}{7}$$

$$\displaystyle \frac{20}{23}$$

$$\displaystyle \frac{9}{20}$$

Solution

Question 3

$$\displaystyle\frac{1}{15}$$

$$\displaystyle\frac{1}{10}$$

$$\displaystyle\frac{7}{60}$$

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

Solution

Question 4

$$\displaystyle\frac{3}{40}$$

$$\displaystyle\frac{1}{8}$$

$$\displaystyle\frac{7}{40}$$

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

Solution

Question 5

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

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

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

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

Solution

Question 6

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

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

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

$$\displaystyle \frac{^{25} C_{13}}{^{51} C_{13}}$$

Solution

Question 7

Event $$A$$ and $$B$$ are mutually exclusive, exhaustive and complementary events.

Event $$A$$ and $$B$$ are mutually exclusive and exhaustive events.

Event $$A$$ and $$B$$ are mutually exclusive and complementary events.

Event $$A$$ and $$B$$ are exhaustive and complementary events.

Solution

Question 8

$$5$$

$$25$$

$$10$$

$$30$$

Solution

Question 9

1/5

1/10

1/4

4/5

Solution

Question 10

1/3

4/9

5/9

2/3

Solution

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