Probability Test

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Probability Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

Alice, Bob and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice, Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is 1/6, independent of the outcome of any other toss.)

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

B
$$\dfrac{2}{9}$$

C
$$\dfrac{5}{18}$$

D
$$\dfrac{25}{91}$$

Solution

If Carol wins in the first round, then she must have rolled a six after two non-sixes have occurred. This happens with the probability
$$\left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{1}{6}\right)=\left(\dfrac{1}{6}\right)\left(\dfrac{5}{6}\right)^2$$
If Carol wins in the second round, then five non-sixes preceded her six. This happens with probability
$$\left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{5}{6}\right) \left(\dfrac{1}{6}\right) =\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{5}$$
This pattern continues. Carol wins in the third round with probability $$\left(\dfrac{1}{6}\right)\left(\dfrac{5}{6}\right)^{8}$$ and in the fourth with probability $$\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{11}$$.
It is possible (though unlikely) that the game could continue forever. The probability that Carol wins the game is equal to the sum of the probabilities that she wins in any given round:
$$=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}+\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{5}+\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{8}+....=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}\left[1+\left(\dfrac{5}{6}\right)^{3}+\left(\dfrac{5}{6}\right)^{6}+.....\right]$$
$$=\displaystyle \left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}\sum_{n=0}^{\infty} \left(\dfrac{5}{6}\right)^{3n}=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2}\sum_{n=0}^{\infty} \left(\dfrac{5^3}{6^3}\right)^{n}$$
$$=\left(\dfrac{1}{6}\right) \left(\dfrac{5}{6}\right)^{2} \left[\dfrac{1}{1-\dfrac{5^3}{6^3}}\right]$$
$$=\dfrac{25}{91}$$
Question 2
1 / -0

A candidate takes three tests in succession and the probability of passing the first test is P. The probability of passing each succeeding test is P or $$\frac{P}{2}$$ according as he passes or fails in the preceding one. The candidate is selected if he passes at least two tests. The probability that the cndidate is selected is

A
$$P^2(2-P)$$

B
$$P(2-P)$$

C
$$P+P^2+P^3$$

D
$$P^2(1-P)$$

Solution

Candidate will be selected in following cases
i) P P P
ii) P P F
iii) P F P
iv) F P P

$$P(I)=P \cdot P\cdot P=P^3$$
$$P(II) = P\cdot P\cdot (1-P)=P^2-P^3$$
$$P(III)=P\cdot (1-P)\cdot \dfrac{P}{2}=\dfrac{P^2}{2}-\dfrac{P^3}{2}$$
$$P(IV)=(1-P)\dfrac{P}{2}\cdot P=\dfrac{P^2}{2}-\dfrac{P^3}{2}$$
$$\therefore$$ Probability of sucess $$=P^3+(P^2-P^3)+\left (\dfrac{P^2-P_3}{2}\right)+\left (\dfrac{P^2-P^3}{2}\right)$$
$$=2P^2-P^3$$
$$=P^2(2-P)$$
Question 3
1 / -0

A certain party consists of four different group of people - 30 students, 35 politicians, 20 actors and 27 leaders. On a particular function day, the total cost spent on party members was Rs. 9000. It was found that 6 students spent as much as 7 politicians, 15 politicians spent as much as 12 actors and 10 actors spent as much as 9 leaders. How much did students spent ?

A
Rs . $$2291$$

B
Rs . $$2292$$

C
Rs . $$2293$$

D
Rs . $$2294$$

Solution

Let $$S$$ be the amount spent for all students.
Let $$P$$ be the amount spent for all politicians.
Let $$A$$ be the amount spent for all actors.
Let $$L$$ be the amount spent for all leaders.
Given that 6 students spent as much as 7 politicians
i.e Amount spent per student $$\times 6=$$ Amount spent per politician $$\times 7$$
OR, (Total amount spent on students / Total students)$$\times$$ 6 = (Total amount spent on politicians / Total politicians) $$\times $$7
Or, $$(S/30 \times 6 = (P/35) \times 7$$
$$\Rightarrow S/5 = P/5$$
$$\Rightarrow S = P$$..........(i)
15 politicians spent as much as 12 actors:
Therefore, $$(P/35) \times 15 = (A/20) \times 12$$
$$ \Rightarrow 3P/7 = 3A/5$$
$$5P = 7P$$
$$5S = 7A$$ (since $$S = P$$)
$$A = 5S / 7$$............(ii)
and 10 actors spent as much as 9 leaders:
Therefore, $$(A/20) \times 10 = (L/27) \times 9$$
$$3A = 2L$$
$$A= 2L/3$$
$$5S/7 = 2L/3$$ (since $$A = 5S/7$$)
$$L= 15S/14$$...........(iii)
Total amount spent for the party is Rs. $$9000$$.
i.e. $$S + P + A + L = 9000$$
Using eq (i), (ii) and (iii), we get
$$S+S+5S/7+15S/14 = 9000$$
$$53 S / 14 = 9000$$
$$S = 9000 \times 14 / 53$$
$$S = 2290.90$$
Hence, the answer is Rs. $$2291$$.
Question 4
1 / -0

Two events $$A$$ and $$B$$ are such that $$P(A)=\cfrac{1}{4},P(B)=\cfrac{1}{2}$$ and $$P(B| A)=\cfrac{1}{2}$$
Consider the following statements
$$(I)$$ $$P(\overline { A } |\overline { B } )=\cfrac { 3 }{ 4 } $$
$$(II)$$ $$A$$ and $$B$$ are mutually exclusive
$$(III)$$ $$P(\overline { A } |\overline { B } )+P(A|\overline { B } )=1$$
Then

A
Only $$I$$ is correct

B
Only $$I$$ and $$II$$ are correct

C
Only $$I$$ and $$III$$ are correct

D
Only $$II$$ and $$III$$ are correct

Solution

$$P(A)=\cfrac{1}{4},P(B)=\cfrac{1}{2},P(B|A)=\cfrac { 1 }{ 2 } $$
$$P(B|A) = \cfrac{P(B\cap A)}{P(A)}\Rightarrow \cfrac12 = \cfrac{P(B\cap A)}{\dfrac14}\Rightarrow P(B\cap A) = \cfrac18 = P(A)P(B)$$
Hence, $$A$$ and $$B$$ are mutually independent events.
$$(I)\, P\left({ \overline { A } }|{\overline{ B} } \right) = \cfrac{P(\overline{A} \cap \overline{B})}{P(\overline{B})}=\cfrac{P(\overline{A})P(\overline{B})}{1-P(B)}=\cfrac{\left(1-\dfrac14\right)\left(1-\dfrac12\right)}{1-\dfrac12}=\cfrac { 3 }{ 4 } $$
Hence, $$I$$ is correct.
$$(II)\, P(A\cap B)\neq 0$$
Hence, $$A$$ and $$B$$ are not mutually exclusive. $$II$$ is not correct.
$$(III)\, P\left({\overline A}|{\overline B} \right) +P\left( { A }|{ \overline { B } } \right) = \cfrac{P(\overline{A} \cap \overline{B})}{P(\overline{B})} + \cfrac{P(A\cap \overline{B})}{P(\overline{B})} = \dfrac 34 + \cfrac{\dfrac14 \times \left(1 - \dfrac12\right)}{1-\dfrac12}=\cfrac { 3 }{ 4 } +\cfrac { 1 }{ 4 } =1$$
Hence $$III$$ is also correct.

$$\therefore$$ Opt:$$[C]$$ is correct
Question 5
1 / -0

A screw factory has two machines, the M1, which is old, and does $$75 \%$$ of all the screws, and the M2, newer but small, that does $$25 \%$$ of the screws. The M1 does $$4 \%$$ of defective screws, while the M2 just does $$2 \%$$ of defective screws. If we choose a screw at random: what is the probability that it turns out to be defective?

A
$$0.035$$

B
$$0.045$$

C
$$0.015$$

D
None of these

Solution

The diagrammatic representation of the problem is shown above.
We consider the following events;
$$M1 =$$ being produced by machine $$1$$.
Therefore, $$\overline{M1}=M2$$ being produced by machine $$2$$.
$$D=$$ defective screw.
In the diagram we can see the red branch in which we are interested.
The top branch being produced by machine $$1$$ and being defective has probability $$\dfrac{75}{100} \times \dfrac{4}{100}=0.75 \times 0.04=0.03$$.
The bottom branch being produced by machine $$2$$ and being defective has probability $$\dfrac{25}{100} \times \dfrac{2}{100}=0.25 \times 0.02=0.005$$.
Therefore, the probability of being defective is $$0.03+0.005=0.035$$.
Question 6
1 / -0

We roll a fair four-sided die. If the result is $$1$$ or $$2$$, we roll once more but otherwise, we stop. What is the probability that the sum total of our rolls is at least $$4$$?

A
$$\dfrac{1}{16}$$

B
$$\dfrac{4}{16}$$

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

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

Solution

$$𝐴_𝑖$$: the result of the first roll is 𝑖
$$𝑃(𝐴_𝑖 )=\dfrac 14, 𝑖=1,2,3,4.$$
𝐵: the sum total is at least 4.
$$𝑃(𝐵)=sum _{ i=1 }^{ 4 }{ 𝑃(𝐴_𝑖 )𝑃(B/𝐴_𝑖 ) }$$
Given:
$$𝐴_1$$: the sum total will be ≥4 if the second roll results in 3 or 4, which happens with probability $$\dfrac 12$$.
Thus $$𝑃(𝐵/𝐴_1)=\dfrac 12$$,
Similarly $$𝑃(𝐵/𝐴_2) =\dfrac 34$$.
Given $$𝐴_2$$, the sum total will be ≥4 if the second roll results in 2, 3, or 4, which happens with probability $$\dfrac 34$$
Given $$𝐴_3$$: you stop and the sum total remains below 4.
Thus $$𝑃(𝐵/𝐴_3)=0, $$
Given $$𝐴_4$$: you stop but the sum total is already 4.
Thus $$𝑃(𝐵/𝐴_4)=1$$.
By the total probability theorem
$$𝑃(𝐵)=\dfrac 14\times\dfrac 12+\dfrac 14\times\dfrac 34+\dfrac 14\times 0+\dfrac 14\times1=\dfrac 9{16}$$
Question 7
1 / -0

A business man is expecting two telephone calls. Mr Walia may call any time between $$2$$ p.m and $$4$$ p.m. while Mr Sharma is equally likely to call any time between $$2.30$$ p.m. and $$3.15$$ p.m. The probability that Mr Walia calls before Mr Sharma is:

A
$$\dfrac{1}{18}$$

B
$$\dfrac{1}{9}$$

C
$$\dfrac{1}{6}$$

D
None of these

Solution

The probability of getting a telephonic call $$\dfrac{1}{2}$$
Let $$E_1$$ be the event of getting a call from Mr. Walia between $$2$$ pm and $$4$$ pm.
Let $$E_2$$ be the event of getting a call from Mr. Sharma between $$2:30$$ pm and $$3:15$$ pm.
$$\Rightarrow P(E_1)=\dfrac{1}{3}$$         $$P(E_2)=\dfrac{2}{3}$$
$$\Rightarrow P\left( \dfrac { E }{ E_1 } \right)=P\left( \dfrac { E }{ E_2 } \right)=\dfrac{1}{2}$$
$$=\dfrac{1}{4}$$
$$\Rightarrow P\left( \dfrac { E }{ E_1 } \right)=\dfrac{P(E_1)P\left( \dfrac { A }{ E_1 } \right)}{P(E_1)P\left( \dfrac { A }{ E_1 } \right)+P(E_2)P\left( \dfrac { A }{ E_2 } \right)}$$
                       $$=\dfrac{\dfrac{1}{4}\times\dfrac{1}{2}}{\dfrac{1}{3}\times\dfrac{1}{2}+\dfrac{2}{3}\times\dfrac{1}{2}}$$
                 $$=\dfrac{1}{6}$$
Hence, the answer is $$\dfrac{1}{6}.$$
Question 8
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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 C?

A
$$0.232$$

B
$$0.862$$

C
$$0.123$$

D
$$0.286$$

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 C if it is defective.
i.e., $$P(C/D)=\dfrac {P(C).P(D/C)}{P(A).P(D/A)+P(B).P(D/B)+P(C).P(D/C)}\\\implies = \dfrac {0.4\times 0.02}{0.25\times 0.05+0.35\times 0.04+0.4\times 0.02}\\\implies = \dfrac {0.008}{0.0345}\\\implies P(C/D)=0.232$$
Question 9
1 / -0

In a group $$14$$ males and $$6$$ females, $$8$$ and $$3$$ of the males and females respectively are aged above $$40$$ years. The probability that a person selected at random from the group is aged above $$40$$ years, given that the selected person is female, is

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

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

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

D
$$\dfrac{5}{6}$$

Solution

This is a question of conditional probability, the condition being given here that the selected person is a female.
Since the selected person is a female, the sample space becomes $$6$$ since there were $$6$$ females in total.
Next, out of those $$6$$ only $$3$$ females are above $$40$$ years of age and so the probability that the selected female is aged above $$40$$ becomes $$\cfrac{3}{6} = \cfrac{1}{2}$$
Question 10
1 / -0

A student answers a multiple choice question with $$5$$ alternatives, of which exactly one is correct. The probability that he knows the correct answer is $$p,0< p< 1$$. If he does not know the correct answer, he randomly ticks one answer. Given that he has answered the question correctly, the probability that he did not tick the answer randomly, is

A
$$\cfrac{3p}{4p+3}$$

B
$$\cfrac{5p}{3p+2}$$

C
$$\cfrac{5p}{4p+1}$$

D
$$\cfrac{4p}{3p+1}$$

Solution

Let the no. of alternatives be denoted as $$'p'$$
The question contains $$5$$ alternatives.
$$\Rightarrow$$ Let event $$E_1=$$ the answer that he ticked randomly.
$$\Rightarrow$$ Let event $$E_2=$$ the answer that he knows
$$\Rightarrow A=$$ answered correctly.
$$\Rightarrow p(E_1)=\dfrac{5\; possible\;answers}{5}$$
$$=\dfrac{5}{5}$$
$$=1$$
$$\Rightarrow p(E_2)=\dfrac{1}{5}$$
Probability that the ticked answer is wrong $$=\dfrac{4}{5}$$
$$\therefore \dfrac { p\left( { E }_{ 1 } \right) p\left( \dfrac{A}{{ E }_{ 1 }} \right) }{ p\left( { E }_{ 2 } \right) p\left( \dfrac{A}{ E }_{ 2 } \right) +p\left( { E }_{ 1 } \right) p\left(\dfrac{ A}{ E }_{ 1 } \right) } $$
$$=\dfrac{p\times1}{\dfrac{4}{5}\times p+\dfrac{1}{5}\times1}$$
$$=\dfrac{5p}{4p+1}$$
Hence, the answer is $$\dfrac{5p}{4p+1}.$$

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Re-Attempt Weekly Quiz Competition

Probability Test - 52

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Probability Test - 52

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SHARING IS CARING

Question 1

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

$$\dfrac{2}{9}$$

$$\dfrac{5}{18}$$

$$\dfrac{25}{91}$$

Solution

Question 2

$$P^2(2-P)$$

$$P(2-P)$$

$$P+P^2+P^3$$

$$P^2(1-P)$$

Solution

Question 3

Rs . $$2291$$

Rs . $$2292$$

Rs . $$2293$$

Rs . $$2294$$

Solution

Question 4

Only $$I$$ is correct

Only $$I$$ and $$II$$ are correct

Only $$I$$ and $$III$$ are correct

Only $$II$$ and $$III$$ are correct

Solution

Question 5

$$0.035$$

$$0.045$$

$$0.015$$

None of these

Solution

Question 6

$$\dfrac{1}{16}$$

$$\dfrac{4}{16}$$

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

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

Solution

Question 7

$$\dfrac{1}{18}$$

$$\dfrac{1}{9}$$

$$\dfrac{1}{6}$$

None of these

Solution

Question 8

$$0.232$$

$$0.862$$

$$0.123$$

$$0.286$$

Solution

Question 9

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

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

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

$$\dfrac{5}{6}$$

Solution

Question 10

$$\cfrac{3p}{4p+3}$$

$$\cfrac{5p}{3p+2}$$

$$\cfrac{5p}{4p+1}$$

$$\cfrac{4p}{3p+1}$$

Solution

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