Probability Test 8

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Probability Test 8

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Weekly Quiz Competition

Question 1
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A box contain $$N$$ coins, $$m$$ of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is $$\dfrac{1}{2}$$, while it is $$\dfrac{2}3{}$$. when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is

A
$$\displaystyle \frac{8\mathrm{m}}{8\mathrm{N}+\mathrm{m}}$$

B
$$\displaystyle \frac{\mathrm{m}}{8\mathrm{N}+\mathrm{m}}$$

C
$$\displaystyle \frac{9\mathrm{m}}{8\mathrm{N}+\mathrm{m}}$$

D
$$\displaystyle \frac{9\mathrm{N}}{8\mathrm{N}+\mathrm{m}}$$

Solution

Let $$E$$ be the event that coin tossed twice , show head at first
time and tail at second time and $$F$$ be the event that coin dawn is fair.
$$ \displaystyle P\left( \frac { F }{ E } \right) =\frac { P\left( \frac { E }{ F } \right) .P\left( F \right) }{ P\left( \frac { E }{ F } \right) .P\left( F \right) +P\left( \frac { E }{ F' } \right) .P\left( F' \right) } $$

$$ \displaystyle =\dfrac { \dfrac { 1 }{ 2 } .\dfrac { 1 }{ 2 } .\dfrac { m }{ N } }{ \dfrac { 1 }{ 2 } .\dfrac { 1 }{ 2 } .\dfrac { m }{ N } +\dfrac { 2 }{ 3 } .\dfrac { 1 }{ 3 } .\dfrac { N-m }{ N } } $$

$$ \displaystyle =\dfrac { \dfrac { m }{ 4 } }{ \dfrac { m }{ 4 } +\dfrac { 2\left( N-m \right) }{ 9 } } =\dfrac { 9m }{ 8N+m } $$
Question 2
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It is given that event A & B are such that $$\displaystyle P(A)=\frac{1}{4},P\left(\frac{A}{B}\right)=\frac{1}{2},P\left(\frac{B}{A}\right)=\frac{2}{3},$$ then $$\displaystyle P(B)=$$

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

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

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

D
$$\displaystyle \frac{1}{6}$$

Solution

We have that $$\displaystyle P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}$$

$$\quad \displaystyle =\frac{P(A\cap B)}{P(A)}\times \frac{P(A)}{P(B)}$$

$$\Rightarrow \displaystyle P\left(\frac{A}{B}\right)= P\left(\frac{B}{A}\right)\cdot \frac{P(A)}{P(B)}$$

$$\displaystyle \therefore P(B)=\frac{2/3\times 1/4}{1/2}=\frac{1}{3}$$
Question 3
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Form two digit numbers using the digit $$0,1,2,3,4,5$$ without repeating the digits.
$$P$$ is the event that the number so formed is even.
$$Q$$ is the event that the number so formed is divisible by $$3$$.
$$R$$ is the event that the number so formed is greater than $$50$$.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 5$$

B
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 2$$

C
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 4$$

D
$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 8$$

Solution

The given digits are $$0,1,2,3,4,5$$
Number of two digit numbers that can be formed, as first place can be filed with 5 numbers and second place can be filed with another 5 numbers; $$5 \times 5 = 25$$
Now, number is an even number, with 0 at the end there can be $$5$$ numbers, with $$2$$ at the end there can be $$4$$ numbers and with $$4$$ at the end there can be $$4$$ numbers. Hence, total numbers which are even $$= 4 + 4 + 5 = 13$$
Numbers divisible by $$3$$ are, $$12, 15, 21, 24, 30, 42, 45, 51, 54$$
Thus, total numbers are, $$9$$
There are only $$4$$ number, viz. $$51, 52,53,54$$ which are greater than $$50$$
Hence, $$n(S) = 25, n(P) = 13, n(Q) = 9, n (R) = 4$$
Question 4
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A die is thrown :
$$P$$ is the event of getting an odd number.
$$Q$$ is the event of getting an even number.
$$R$$ is the event of getting a prime number.
Which of the following pairs is mutually exclusive?

A
$$P,Q$$

B
$$Q,R$$

C
$$P,R$$

D
None of these

Solution

Let $$S$$ be the sample space
$$S=\{1,2,3,4,5,6\}$$
$$n(S)=6$$
$$P$$ is the event of getting an odd number
$$\therefore P=\{1,3,5\}\implies n(P)=3$$
$$Q$$ is the event of getting an even number
$$\therefore Q=\{2,4,6\}\implies n(Q)=3$$
$$R$$ is the vent of getting a prime number
$$\therefore R=\{2,3,5\}\implies n(R)=3$$
The set $$S$$ contains two types of numbers even and odd
We have, $$P=S-Q$$ and $$Q=S-P$$
$$\therefore n(P)$$ and $$n(Q)$$ both are complementary.
Two events are mutually exclusive if they are disjoint
Now, $$P\cap Q=\phi$$
$$\implies P$$ and $$Q$$ are disjoint i.e. they don't have common elements.
Hence, $$P$$ and $$Q$$ are mutually exclusive.
Question 5
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There are $$3$$ red, $$3$$ white and $$3$$ green balls in a bag. One ball is drawn at random from a bag:
$$P$$ is the event that ball is red.
$$Q$$ is the event that ball is not green.
$$R$$ is the event that ball is red or white.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 9, n(P) =3, n(Q) = 6, n(R) = 6$$

B
$$n(S) = 4, n(P) =3, n(Q) = 6, n(R) = 6$$

C
$$n(S) = 6, n(P) =3, n(Q) = 6, n(R) = 6$$

D
$$n(S) = 7, n(P) =3, n(Q) = 6, n(R) = 6$$

Solution

There are $$3$$ red balls, $$3$$ white balls and $$3$$ green balls.
Thus, total number of balls $$= 9$$, $$n (S) = 9$$
$$P$$, the ball is red, $$n(R) = 3$$
$$Q$$, the ball is not green, $$n(Q) = 9 - 3 = 6$$
$$R$$, the ball is red or white, $$n(R) = 3 + 3 = 6$$
Question 6
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There are $$3$$ men and $$2$$ women. a 'Gramswachhatta Abhiyan' committee of two is to be formed :
$$P$$ is event that the committee should contain at least one woman.
$$Q$$ is event that the committee should contain one man and one women.
$$R$$ is the event there should not be a women in the committee.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 10, n(P) = 7, n(Q) = 1, n(R) = 3$$

B
$$n(S) = 10, n(P) = 7, n(Q) = 3, n(R) = 3$$

C
$$n(S) = 10, n(P) = 7, n(Q) = 2, n(R) = 3$$

D
$$n(S) = 10, n(P) = 7, n(Q) = 6, n(R) = 3$$

Solution

$$n(S) = 10, n(P) = 7, n(Q) = 6, n(R) = 3$$
$$P$$ and $$ R$$ are complementary, mutually exclusive events and exhaustive events.
$$Q$$ and $$R$$ are complementary and mutually exclusive events.
Question 7
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A coin is tossed and a die is thrown simultaneously :
$$P$$ is the event of getting head and a odd number.
$$Q$$ is the event of getting either $$H$$ or $$T$$ and an even number.
$$R$$ is the event of getting a number on die greater than $$7$$ and a tail.
$$S$$ is the sample space.
Which of the following options is correct?

A
$$n(S) = 12, n(P) = 3, n(Q) = 2, n(R) = 0$$

B
$$n(S) = 12, n(P) = 3, n(Q) = 3, n(R) = 0$$

C
$$n(S) = 12, n(P) = 3, n(Q) = 6, n(R) = 0$$

D
$$n(S) = 12, n(P) = 3, n(Q) = 5, n(R) = 0$$

Solution

A coin is tossed and a die is thrown simultaneously.
$$ S = \{ H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 \} $$
Thus $$ n (S) = 12$$
$$P$$ is the event of getting head and an odd number.
$$P = \{ H1, H3, H5 \}$$
Thus $$ n (P) = 3$$
$$Q$$ is the event of getting either $$H$$ or $$T$$ and an even number.
$$Q = \{ H2, H4, H6, T2, T4, T6 \}$$
Thus $$ n (Q) = 6$$
$$R$$ is the event of getting a number greater than $$7$$ and tail $$1$$.
$$R = \{ \}$$
$$\therefore n(R)=0$$
Hence, option C is correct.
Question 8
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Two dice are thrown :
$$P$$ is the event that the sum of the scores on the uppermost faces is a multiple of $$6$$.
$$Q$$ is the event that the sum of the scores on the uppermost faces is at least $$10$$.
$$R$$ is the event that same scores on both dice.
Which of the following pairs is mutually exclusive?

A
$$P,Q$$

B
$$P,R$$

C
$$Q,R$$

D
None of these

Solution

Possibilities of$$ P$$, $$(3,3), (6,6), (1,5), (5,1), (4,2),(2,4)$$
Possibilities of $$Q: (5,5), (5,6),(6,5), (6, 6)$$
Possibilities of $$R: (1,1), (2,2),(3, 3), (4,4), (5,5), (6,6)$$
Thus, the possibilities are neither exhaustive, nor mutually exclusive nor these are complementary probabilities.
Question 9
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There are two bags, one of which contains three black and four white balls while the other contains four black and three white balls. A die is cast: if the face 1 or 3 turns up, a ball is taken from the first bag; and if any other face turns up, a ball is chosen from the second bag. Find the probability of choosing a black ball.

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

B
$$\displaystyle \frac{11}{21}$$

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

D
$$\displaystyle \frac{10}{21}$$

Solution

Let $$\displaystyle E_{1}$$ be the event that a ball is drawn from first bag.

$$\displaystyle E_{2}$$ be the event that a ball is drawn form the second bag.

$$E$$ be the event a black ball is chosen.

$$P(E_1)=\dfrac{2}{6}=\dfrac{1}{3}$$

$$P(E_2)=\dfrac{4}{6}=\dfrac{2}{3}$$

Now, $$\displaystyle P\left ( E \right )=P\left ( E_{1} \right )P\left (
E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{2} \right )$$

$$=\displaystyle \frac{1}{3}.\frac{3}{7}+\frac{2}{5}.\frac{4}{7}$$

$$=\dfrac{11}{21}.$$
Question 10
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A letter is known to have come eithe from London or Clifton; on the post only the consecutive letters ON are legible; what is the chance that it came from London?

A
$$\displaystyle \frac{12}{17}$$

B
$$\displaystyle \frac{5}{17}$$

C
$$\displaystyle \frac{5}{12}$$

D
$$\displaystyle \frac{7}{12}$$

Solution

If letter came from Clifton there are $$6$$ pairs of consecutive letters i.e., $$cl, li, if, ft, to$$, $$ON $$ in which $$ON$$ appears only once.
$$\displaystyle \therefore $$ the chance that this was the legible couple on the Clifton hypothesis $$\displaystyle = \frac{1}{6}$$
pairs of consecutive letters in the word London are $$lo, on, nd, do$$, $$ON $$ in which $$ON$$ occurs twice.
$$\displaystyle \therefore $$ the chance that this was the legible couple on the London hypothesis$$=2/5.$$
$$\displaystyle \therefore $$ The a posteriori chances that the letter was from Clifton or London are $$\displaystyle \frac{1/6}{\dfrac{1}{6}+\dfrac{2}{5}}$$ and $$\displaystyle \frac{2/5}{\dfrac{1}{6}+\dfrac{2}{5}}$$ respectively.
Thus the reqd.chance $$\displaystyle = \frac{12}{17}.$$

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Re-Attempt Weekly Quiz Competition

Probability Test 8

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Probability Test 8

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Question 1

$$\displaystyle \frac{8\mathrm{m}}{8\mathrm{N}+\mathrm{m}}$$

$$\displaystyle \frac{\mathrm{m}}{8\mathrm{N}+\mathrm{m}}$$

$$\displaystyle \frac{9\mathrm{m}}{8\mathrm{N}+\mathrm{m}}$$

$$\displaystyle \frac{9\mathrm{N}}{8\mathrm{N}+\mathrm{m}}$$

Solution

Question 2

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

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

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

$$\displaystyle \frac{1}{6}$$

Solution

Question 3

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 5$$

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 2$$

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 4$$

$$n(S) = 25, n(P) = 13, n(Q) = 9, n(R) = 8$$

Solution

Question 4

$$P,Q$$

$$Q,R$$

$$P,R$$

None of these

Solution

Question 5

$$n(S) = 9, n(P) =3, n(Q) = 6, n(R) = 6$$

$$n(S) = 4, n(P) =3, n(Q) = 6, n(R) = 6$$

$$n(S) = 6, n(P) =3, n(Q) = 6, n(R) = 6$$

$$n(S) = 7, n(P) =3, n(Q) = 6, n(R) = 6$$

Solution

Question 6

$$n(S) = 10, n(P) = 7, n(Q) = 1, n(R) = 3$$

$$n(S) = 10, n(P) = 7, n(Q) = 3, n(R) = 3$$

$$n(S) = 10, n(P) = 7, n(Q) = 2, n(R) = 3$$

$$n(S) = 10, n(P) = 7, n(Q) = 6, n(R) = 3$$

Solution

Question 7

$$n(S) = 12, n(P) = 3, n(Q) = 2, n(R) = 0$$

$$n(S) = 12, n(P) = 3, n(Q) = 3, n(R) = 0$$

$$n(S) = 12, n(P) = 3, n(Q) = 6, n(R) = 0$$

$$n(S) = 12, n(P) = 3, n(Q) = 5, n(R) = 0$$

Solution

Question 8

$$P,Q$$

$$P,R$$

$$Q,R$$

None of these

Solution

Question 9

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

$$\displaystyle \frac{11}{21}$$

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

$$\displaystyle \frac{10}{21}$$

Solution

Question 10

$$\displaystyle \frac{12}{17}$$

$$\displaystyle \frac{5}{17}$$

$$\displaystyle \frac{5}{12}$$

$$\displaystyle \frac{7}{12}$$

Solution

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