Probability Test

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Probability Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

A biased coin (with probability of obtaining a Head equal to $$p > 0$$) is tossed repeatedly and independently until the first head is observed. Compute the probability that the first head appears at an even numbered toss.

A
$$\dfrac{2-p}{3-p}$$

B
$$\dfrac{1-p}{3-p}$$

C
$$\dfrac{1-p}{2-p}$$

D
none of these

Solution

Let event $$H_1$$ - head on the first toss;
• event E - first head on an even-numbered toss.
We want P(E).
Using the Theorem of Total Probability
$$P(E/H_1) = 0$$ (given $$H_1$$, that a head appears on the first toss, E cannot occur)
$$P(E/H_1')=P(E')=1-P(E)$$
(given that a head does not appear on the first toss, the required conditional
probability is merely the probability that the sequence concludes after a
further odd number of tosses, that is, the probability of E'). Hence P(E) satisfies
$$P(E)=P(E/H_1)P(H_1)+P(E/H_1')P(H_1')\\\implies P(E)=0\times p+(1-P(E))\times (1-p)=(1-p)(1-P(E))\\\implies P(E)=\dfrac {1-p}{2-p}$$
Question 2
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A factory production line is manufacturing bolts using three machines, A, B and C. Of the total output, machine A is responsible for $$25$$%, machine B for $$35$$% and machine C for the rest. It is known from previous experience with the machines that $$5$$% of the output from machine A is defective, $$4$$% from machine B and $$2$$% from machine C. A bolt is chosen at random from the production line and found to be defective. What is the probability that it came from machine B

A
$$0.557$$

B
$$0.652$$

C
$$0.473$$

D
none of these

Solution

Let
A be bolt manufactured from machine A
B be bolt manufactured from machine B
C be bolt manufactured from machine C
D be bolt is defective
Given:
$$P(A)=25\%=0.25, P(B)=35\%=0.35, P(C)=40\%=0.4$$
$$P(D/A)=5\%=0.05, P(D/B)=4\%=0.04, P(D/C)=2\%=0.02$$
To find:
The probability that the bolt is manufactured by machine B if it is defective.
i.e., $$P(B/D)=\dfrac {P(B).P(D/B)}{P(A).P(D/A)+P(B).P(D/B)+P(C).P(D/C)}\\\implies = \dfrac {0.35\times 0.04}{0.25\times 0.05+0.35\times 0.04+0.4\times 0.02}\\\implies = \dfrac {0.014}{0.0345}\\\implies P(B/D)=0.406$$
Question 3
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From a batch of $$100$$ items of which $$20$$ are defective, exactly two items are chosen, one at a time, without replacement. Calculate the probabilities that both items chosen are defective

A
$$\dfrac{20}{495}$$

B
$$\dfrac{19}{495}$$

C
$$\dfrac{99}{495}$$

D
none of these

Solution

Let A={first item chosen is defective}, B ={second item chosen is defective}
Given:
Total number of items = 100.
Number of defective items = 20
To find:
the probabilities that both items chosen are defective
$$P(A\cap B)=P(A/B)P(A)=\dfrac {20}{100}\times \dfrac {19}{99}=\dfrac {19}{495}$$
Question 4
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While watching a game of Champions League football in a cafe, you observe someone who is clearly supporting Manchester United in the game. What is the probability that they were actually born within 25 miles of Manchester? Assume that:
$$\implies $$the probability that a randomly selected person in a typical local bar environment is born within $$25$$ miles of Manchester is $$\dfrac{1}{20}$$, and;
$$\implies$$the chance that a person born within $$25$$ miles of Manchester actually supports United is $$\dfrac{7}{10}$$;
$$\implies$$the probability that a person not born within $$25$$ miles of Manchester supports United with probability $$\dfrac{1}{10}$$

A
$$\dfrac{7}{26}$$

B
$$\dfrac{8}{26}$$

C
$$\dfrac{9}{26}$$

D
$$\dfrac{10}{26}$$

Solution

Probability that they support Manchest or not $$=\dfrac{1}{2}$$
Let $$E_1$$ and $$E_2$$ be the events of a person who was born $$25$$ miles of Manchester and $$E_2$$ be event of a person not born with in $$25$$ miles of Manchester
$$\therefore P(E_1)=\dfrac{1}{20}$$            $$P(E_2)=\dfrac{1}{10}$$
$$A=$$ Actually born with in $$25$$ miles
$$\Rightarrow P\left( \dfrac { E }{ E_1 } \right)=\dfrac { P\left( { E }_{ 1 } \right) P\left( { \dfrac{A}E }_{ 1 } \right) }{ 2\left[ P\left( { E }_{ 1 } \right) P\left( { \dfrac{A}E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) P\left( { \dfrac{A}E }_{ 2 } \right) \right] } $$
                        $$=\dfrac { \dfrac { 1 }{ 20 } \times \dfrac { 7 }{ 10 } }{ 2\left[ \dfrac { 1 }{ 20 } \times \dfrac { 7 }{ 10 } +\dfrac { 1 }{ 10 } \times \dfrac { 3 }{ 10 } \right] } $$
                       $$=\dfrac { 7 }{ 26 }.$$
Hence, the answer is $$\dfrac { 7 }{ 26 }.$$
Question 5
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A garage mechanic keeps a box of good springs to use as replacements on customers cars. The box contains $$5$$ springs. A colleague, thinking that the springs are for scrap, tosses three faulty springs into the box. The mechanic picks two springs out of the box while servicing a car. Find the probability that the second spring drawn is faulty.

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

B
$$\dfrac{2}{8}$$

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

D
$$\dfrac{4}{8}$$

Solution

Let A ={first spring chosen is faulty}, B ={second spring chosen is faulty}
Given:
Total number of springs in box = 8
Number of faulty springs = 3
To find:
The probability that the second spring drawn is faulty
$$P(B) = P(B/A)P(A)+P(B/A')P(A')=\dfrac 27\times \dfrac 38+\dfrac 37\times \dfrac 58=\dfrac {21}{56}=\dfrac 38$$
Question 6
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Tossing a coin is an example of .........

A
Infinite discrete sample space

B
Finite sample space

C
Continuous sample space

D
None of these

Solution

A coin is tossed.
The possible outcomes are "$$\text{head,tail}$$"
Thus $$ S=\{H,T\}$$
We get a finite sample space.
Thus tossing a coin is an example of finite sample space.
Question 7
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For a random experiment, all possible outcomes are called

A
numerical space.

B
event space.

C
sample space.

D
both b and c.

Solution

We know that,
An outcome is a result of a random experiment.
The set of all possible outcomes is called the sample space.
The subset of the sample space is called event space.
Thus, for a random experiment, all possible outcomes are called sample space and event space.
Thus option $$(D)$$ is correct.
Question 8
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Two players A and B are competing at a trivia quiz game involving a series of questions. On any individual question, the probabilities that A and B give the correct answer are $$\alpha$$ and $$\beta$$ respectively, for all questions, with outcomes for different questions being independent. The game finishes when a player wins by answering a question correctly.
Compute the probability that A wins if A answers the first question

A
$$\dfrac{\alpha}{1-(1-\alpha)(1-\beta)}$$

B
$$\dfrac{\alpha}{1-(\alpha)(\beta)}$$

C
$$\dfrac{\beta}{1-(1-\alpha)(1-\beta)}$$

D
none of these

Solution

Let event A - A answers the first question;
event F - game ends after the first question;
event W - A wins.
To find:
$$P(W/A)$$
Now, clearly
$$P(F/A) = P$$ [A answers first question correctly] = $$\alpha$$,
$$P(F'/A)= 1 − α,$$ and
$$P(W/A\cap F)=1$$, but $$P(W/A\cap F')=P(W/A')$$, so that
$$P(W/A)=P(W/A\cap F)P(F/A)+P(W/A\cap F')P(F'/A)$$
$$P(W/A) = (1 \times \alpha) + (P(W/A')\times (1-\alpha))=\alpha +P(W/A')(1-\alpha)....(i)$$
We have,
$$P(F/A')=P$$ [B answers first question correctly] = $$\beta$$,
$$ P(F'/A)=1-\beta$$
but $$PW/A'\cap F)=0$$. Finally $$P(W/A'\cap F')=P(W/A)$$, so that
$$P(W/A')=(0\times \beta)+(P(W/A)\times (1-\beta))=P(W/A)(1-\beta).......(ii)$$
Solving (i) and (ii) simultaneously gives, for A answers the first question,
$$P(W/A)=\dfrac {\alpha}{1-(1-\alpha)(1-\beta)}$$
Question 9
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Two players A and B are competing at a trivia quiz game involving a series of questions. On any individual question, the probabilities that A and B give the correct answer are $$\alpha$$ and $$\beta$$ respectively, for all questions, with outcomes for different questions being independent. The game finishes when a player wins by answering a question correctly.
Compute the probability that A wins if B answers the first question.

A
$$\dfrac{\alpha}{1-(1-\alpha)(1-\beta)}$$

B
$$\dfrac{(1-\beta)}{1-(\alpha)(\beta)}$$

C
$$\dfrac{(1-\beta)\alpha}{1-(1-\alpha)(1-\beta)}$$

D
none of these

Solution

Let event A - A answers the first question;
event F - game ends after the first question;
event W - A wins.
To find:
$$P(W/A')$$
Now, clearly
$$P(F/A) = P$$ [A answers first question correctly] = $$\alpha$$,
$$P(F'/A)= 1 − α,$$ and
$$P(W/A\cap F)=1$$, but $$P(W/A\cap F')=P(W/A')$$, so that
$$P(W/A)=P(W/A\cap F)P(F/A)+P(W/A\cap F')P(F'/A)$$
$$P(W/A) = (1 \times \alpha) + (P(W/A')\times (1-\alpha))=\alpha +P(W/A')(1-\alpha)....(i)$$
We have,
$$P(F/A')=P$$ [B answers first question correctly] = $$\beta$$,
$$ P(F'/A)=1-\beta$$
but $$P(W/A'\cap F)=0$$. Finally $$P(W/A'\cap F')=P(W/A)$$, so that
$$P(W/A')=(0\times \beta)+(P(W/A)\times (1-\beta))=P(W/A)(1-\beta).......(ii)$$
Solving (i) and (ii) simultaneously gives, for B answers the first question,
$$P(W/A')=\dfrac {(1-\beta)\alpha}{1-(1-\alpha)(1-\beta)}$$
Question 10
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Sample space for experiment in which a dice is rolled is

A
$$4$$

B
$$8$$

C
$$12$$

D
None of these

Solution

A dice is rolled.
Thus $$S=\{1,2,3,4,5,6\}$$
$$\Rightarrow n(S)=6$$.
That is, sample space for experiment in which a dice is rolled is $$6$$.
Since $$6$$ is not listed in the given options we choose D.

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Re-Attempt Weekly Quiz Competition

Probability Test - 29

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Probability Test - 29

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SHARING IS CARING

Question 1

$$\dfrac{2-p}{3-p}$$

$$\dfrac{1-p}{3-p}$$

$$\dfrac{1-p}{2-p}$$

none of these

Solution

Question 2

$$0.557$$

$$0.652$$

$$0.473$$

none of these

Solution

Question 3

$$\dfrac{20}{495}$$

$$\dfrac{19}{495}$$

$$\dfrac{99}{495}$$

none of these

Solution

Question 4

$$\dfrac{7}{26}$$

$$\dfrac{8}{26}$$

$$\dfrac{9}{26}$$

$$\dfrac{10}{26}$$

Solution

Question 5

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

$$\dfrac{2}{8}$$

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

$$\dfrac{4}{8}$$

Solution

Question 6

Infinite discrete sample space

Finite sample space

Continuous sample space

None of these

Solution

Question 7

numerical space.

event space.

sample space.

both b and c.

Solution

Question 8

$$\dfrac{\alpha}{1-(1-\alpha)(1-\beta)}$$

$$\dfrac{\alpha}{1-(\alpha)(\beta)}$$

$$\dfrac{\beta}{1-(1-\alpha)(1-\beta)}$$

none of these

Solution

Question 9

$$\dfrac{\alpha}{1-(1-\alpha)(1-\beta)}$$

$$\dfrac{(1-\beta)}{1-(\alpha)(\beta)}$$

$$\dfrac{(1-\beta)\alpha}{1-(1-\alpha)(1-\beta)}$$

none of these

Solution

Question 10

$$4$$

$$8$$

$$12$$

None of these

Solution

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