Probability Test 10

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Probability Test 10

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Weekly Quiz Competition

Question 1
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A coin is tossed $$40$$ times and it showed tail $$24$$ times.The probability of getting a head was:

A
$$\displaystyle\frac{2}{5}$$

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

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

D
$$\displaystyle\frac{17}{40}$$

Solution

Coin tossed $$=40$$ times
showed tail $$=24$$ times
Probability (getting head) $$=\dfrac{40-24}{40}$$
$$=\dfrac{16}{40}$$
$$=\dfrac{2}{5}$$
Question 2
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A, B are two inaccurate arithmeticians whose chances of solving a given question correctly are (1/8) and (1/12) respectively. They solve a problem and obtained the same result. If it is 1000 to 1 against their making the same mistake, find the chance that the result is correct

A
$$\displaystyle \frac {13}{14}$$

B
$$\displaystyle \frac {2}{7}$$

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

D
$$\displaystyle \frac {11}{12}$$

Solution

$$A$$ : $$A$$ solves correctly $$B : B$$ solves correctly
$$E$$ : Commit same mistake
$$F$$ : same result
$$\displaystyle P(AB/F)=\frac {P(AB)}{P(AB)+P(\overline A\overline BE)}=\displaystyle \frac {\frac {1}{8}.\frac {1}{12}}{\frac {1}{8}.\frac {1}{12}+\frac {7}{8}.\frac {11}{12}.\frac {1}{1001}}$$
$$\displaystyle =\frac {1001}{1078}=\frac {13}{14}$$
Question 3
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In a test, an examine either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is $$\displaystyle \frac { 1 }{ 3 } $$ and the probability that he copies the answer is $$\displaystyle \frac { 1 }{ 6 } $$. The probability that his answer is correct given that he copied it is $$\displaystyle \frac { 1 }{ 8 }$$. The probability that he knew the answer to the question given that he correctly answered it, is

A
$$\displaystyle \frac { 24 }{ 29 } $$

B
$$\displaystyle \frac { 1 }{ 4 } $$

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

D
None of these

Solution

Let $${ E }_{ 1 }$$ be the event that the answer is guessed, $${ E }_{ 2 }$$ be the event that the answer is copied, $${ E }_{ 3 }$$ be the event that the examine knows the answer and $$E$$ be thye event that the examine answer correctly.
Given $$\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 3 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 6 } $$,
Assume that events $${ E }_{ 1 },{ E }_{ 2 }$$ and $${ E }_{ 3 }$$ are exhaustive.
$$\therefore P\left( { E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) +P\left( { E }_{ 3 } \right) =1$$
$$\displaystyle \therefore P\left( { E }_{ 3 } \right) =1-P\left( { E }_{ 1 } \right) -P\left( { E }_{ 2 } \right) =1-\frac { 1 }{ 3 } -\frac { 1 }{ 6 } =\frac { 1 }{ 2 } $$
Now $$\displaystyle P\left( \frac { E }{ { E }_{ 1 } } \right) \equiv $$ Probability of getting correct answer by guessing $$\displaystyle =\frac { 1 }{ 4 } $$ (Since $$4$$ alternatives)
$$\displaystyle P\left( \frac { E }{ { E }_{ 2 } } \right) \equiv $$ Probability of answering correctly by copying $$\displaystyle =\frac { 1 }{ 8 } $$
And $$\displaystyle P\left( \frac { E }{ { E }_{ 3 } } \right) \equiv $$ Probability of answering correctly by knowing $$=1$$
Clearly, $$\displaystyle \left( \frac { { E }_{ 3 } }{ E } \right) $$ is the vent he knew the answer to the question given that he correctly answered it.
$$\displaystyle P\left( \frac { { E }_{ 3 } }{ E } \right) =\frac { P\left( { E }_{ 3 } \right) .P\left( \frac { E }{ { E }_{ 3 } } \right) }{ P\left( { E }_{ 1 } \right) .P\left( \frac { E }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) .P\left( \frac { E }{ { E }_{ 2 } } \right) +P\left( { E }_{ 3 } \right) .P\left( \frac { E }{ { E }_{ 3 } } \right) } $$
$$\displaystyle =\frac { \frac { 1 }{ 2 } \times 1 }{ \frac { 1 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 6 } \times \frac { 1 }{ 8 } +\frac { 1 }{ 2 } \times 1 } =\frac { 24 }{ 29 } $$
Question 4
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Suppose that of all used cars of a particular year 30% have bad brakes. You are considering buying a used car of that year. You take the car to a mechanic to have the brakes checked. The chance that the mechanic will give you the wrong report is 20%. Assuming that the car you take to the mechanic is selected at random from the population of cars of that year. The chance that the car's brakes are good, given that the mechanic says its brakes are good, is

A
$$\displaystyle \frac{28}{130}$$

B
$$\displaystyle \frac{29}{31}$$

C
$$\displaystyle \frac{37}{62}$$

D
$$\displaystyle \frac{29}{62}$$

Solution

Given $$30$$% of the cars of bad brakes
$$P(E_1)=70$$%$$=\dfrac7{10}$$ $$P(E_2)=30$$%$$=\dfrac3{10}$$
$$\Rightarrow P\left( \dfrac { { E } }{ { E }_{ 1 } } \right) =0.2\times0.2=0.04$$
$$\Rightarrow P\left( \dfrac { { E } }{ { E }_{ 2 } } \right) =0.8\times0.8=0.64$$
$$\therefore P\left( \dfrac { { E } }{ { E }_{ 1 } } \right) =\dfrac { \dfrac { 7 }{ 10 } \times \dfrac { 2 }{ 10 } \times \dfrac { 2 }{ 10 } }{ \dfrac { 7\times 4 }{ 1000 } +\dfrac { 3 }{ 10 } +\dfrac { 8 }{ 10 } +\dfrac { 8 }{ 10 } } =\dfrac { 28 }{ 102+28 } =\dfrac { 28 }{ 130 } $$
Hence, the answer is $$\dfrac { 28 }{ 130}.$$
Question 5
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If $$E$$ and $$F$$ are independent events such that $$P(E) = 0.7$$ and $$P(F) = 0.3$$, then P$$\displaystyle (E\cap F)$$

A
$$0.4$$

B
$$1$$

C
$$0.21$$

D
None

Solution

Since, $$E$$ and $$F$$ are independent events with $$P(E) = 0.7$$ and $$P(F) = 0.3 $$
$$\therefore $$ P$$\displaystyle (E\cap F) = P(E)\times P(F)$$
$$\Rightarrow $$ P$$\displaystyle (E\cap F) = 0.7\times 0.3 $$
$$\Rightarrow $$ P$$\displaystyle (E\cap F) = 0.21$$
$$\therefore $$ Option C is correct.
Question 6
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A pack of cards was found to contain only $$51\ cards$$. If first $$13\ cards$$, which are examined, are all red, then find the probability that the missing cards is black?

A
$$2/3$$

B
$$1/3$$

C
$$2/9$$

D
$$1/6$$

Solution

Let $$A=$$ Missing card is black.
$$B=$$ Missing card is red.
$$D=$$ First $$13$$ cards found are red.
$$\displaystyle P\left( \frac { A }{ D } \right) =\frac { P\left( \frac { D }{ A } \right) P\left( A \right) }{ P\left( \frac { D }{ A } \right) P\left( A \right) +P\left( \frac { D }{ B } \right) P\left( B \right) } $$
$$\displaystyle=\frac { \dfrac { 26 }{ 51 } \times \dfrac { 25 }{ 50 } \times ...\times \dfrac { 14 }{ 39 } \times \dfrac { 1 }{ 2 } }{ \dfrac { 26 }{ 51 } \times \dfrac { 25 }{ 50 } \times ...\times \dfrac { 14 }{ 39 } \times \dfrac { 1 }{ 2 } +\dfrac { 25 }{ 51 } \times \dfrac { 24 }{ 50 } \times ...\times \dfrac { 13 }{ 39 } \times \dfrac { 1 }{ 2 } }$$
$$\displaystyle=\frac { 26 }{ 26+13 } =\frac { 2 }{ 3 } $$
Question 7
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A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all the three occasions. The chance that the die chosen was a rigged die, is

A
$$\displaystyle \frac {216}{217}$$

B
$$\displaystyle \frac {215}{219}$$

C
$$\displaystyle \frac {216}{219}$$

D
none

Solution

Given, $$A$$ is the event which selects the rigged one and $$B$$ is the event which selects the fair one.
Let E is the event which shows 5 in all three times
Probability of event A for given event E, $$P(A/E)=\dfrac{1}{4}*(1)^3=\dfrac{1}{4}$$ ($$\dfrac{1}{4}$$ in the equation is probability of selecting one dice among 4)
Probability of event B for given event E, $$P(B/E)=\dfrac{3}{4}*(\dfrac{1}{6})^3=\dfrac{1}{1152}$$ (since probability of getting 5 in fair dice case=$$\dfrac{1}{5}$$)
By Baye's theorem,Probability of selecting the rigged case among both=$$\dfrac{P(A/E)}{P(A/E)+P(B/E)}=\dfrac{(\dfrac{1}{4})}{(\dfrac{1}{4})+(\dfrac{1}{1152})}=\dfrac{216}{219}$$
Question 8
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In a simultaneous throw of two dice, what is the number of exhaustive events?

A
$$6$$

B
$$12$$

C
$$36$$

D
$$18$$

Solution

A dice has $$6$$ faces.
So, for each face of the first dice, there will be $$6$$ combinations with the faces of the second dice.
$$\therefore $$ The number of exhaustive events
=$$6\times 6=36$$.
Question 9
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In figure points A, B, C and D are the centre of four circles that each have a radius of length one unit. If a point is selected at random from the interior of square ABCD. What is the probability that the point will be chosen from the shaded region?

A
$$ \displaystyle \left (1 - \frac{ \pi}{4} \right )$$

B
$$ \displaystyle \left (1 - \frac{ \pi}{2} \right )$$

C
$$ \displaystyle \left (1 - \frac{ \pi}{6} \right )$$

D
$$ \displaystyle \left (2 - \frac{ \pi}{4} \right )$$

Solution

Side of a square $$= 2$$ unit.
Area of square $$= (2)^2$$ sq. unit $$= 4$$ sq. unit
Area of shaded region $$=$$ Area of a square $$-$$ Area of circle
$$=4- 4 \times \frac{1}{4} \pi (1)^2 = 4 - \pi$$
$$P $$(point chosen from shaded region) $$= \displaystyle \frac{4 - \pi}{4}$$
$$= \displaystyle \left (1 - \frac{ \pi}{4} \right )$$
Question 10
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A & B are sharp shooters whose probabilities of hitting a target are $$\displaystyle \frac{9}{10}$$ & $$\displaystyle \frac{14}{15}$$ respectively. If it is knownthat exactly one of them has hit the target, then the probability that it was hit by A is equal to

A
$$\displaystyle \frac{24}{55}$$

B
$$\displaystyle \frac{27}{55}$$

C
$$\displaystyle \frac{9}{23}$$

D
$$\displaystyle \frac{10}{23}$$

Solution

$$E_1$$ : only A hits the target
$$E_2$$ : only B hits the target
$$E$$ : exactly one hits the target.
$$\therefore \displaystyle P(E_1 / E) = \frac{P(E_1). P (E / E_1)}{P (E_1). P (E/ E_1) + P (E_2). P (E/ E_2)}$$
$$=

\displaystyle \frac{\displaystyle \frac{9}{10} \times \frac{1}{15}}{

\displaystyle \frac{9}{10} \times \frac{1}{15} + \frac{14}{15} \times

\frac{1}{10}}\\ = \dfrac{9}{23}$$

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Re-Attempt Weekly Quiz Competition

Probability Test 10

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Probability Test 10

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SHARING IS CARING

Question 1

$$\displaystyle\frac{2}{5}$$

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

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

$$\displaystyle\frac{17}{40}$$

Solution

Question 2

$$\displaystyle \frac {13}{14}$$

$$\displaystyle \frac {2}{7}$$

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

$$\displaystyle \frac {11}{12}$$

Solution

Question 3

$$\displaystyle \frac { 24 }{ 29 } $$

$$\displaystyle \frac { 1 }{ 4 } $$

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

None of these

Solution

Question 4

$$\displaystyle \frac{28}{130}$$

$$\displaystyle \frac{29}{31}$$

$$\displaystyle \frac{37}{62}$$

$$\displaystyle \frac{29}{62}$$

Solution

Question 5

$$0.4$$

$$1$$

$$0.21$$

None

Solution

Question 6

$$2/3$$

$$1/3$$

$$2/9$$

$$1/6$$

Solution

Question 7

$$\displaystyle \frac {216}{217}$$

$$\displaystyle \frac {215}{219}$$

$$\displaystyle \frac {216}{219}$$

none

Solution

Question 8

$$6$$

$$12$$

$$36$$

$$18$$

Solution

Question 9

$$ \displaystyle \left (1 - \frac{ \pi}{4} \right )$$

$$ \displaystyle \left (1 - \frac{ \pi}{2} \right )$$

$$ \displaystyle \left (1 - \frac{ \pi}{6} \right )$$

$$ \displaystyle \left (2 - \frac{ \pi}{4} \right )$$

Solution

Question 10

$$\displaystyle \frac{24}{55}$$

$$\displaystyle \frac{27}{55}$$

$$\displaystyle \frac{9}{23}$$

$$\displaystyle \frac{10}{23}$$

Solution

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