Probability Test

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Probability Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

In a construction job, following are some probabilities given:
Probability that there will be strike is $$0.65$$, probability that the job will be completed on time if there is no strike is $$0.80$$, probability that the job will be completed on time if there is strike is $$0.32$$. Determine probability that the construction job will get complete on time.

A
$$0.438$$

B
$$0.538$$

C
$$0.488$$

D
None of these

Solution

Let $$A$$ be the event that the construction job is completed on time and $$B$$ be the event that there is strike.

$$\Rightarrow P(B)=0.65$$

Hence probability that there will be no strike $$=P(B')$$
$$=1-P(B)$$
$$=1-0.65$$
$$=0.35$$

$$\therefore P(B')=0.35$$

By the Law of Total Probability we have $$P(A)=P(B) \times P(A|B)+P(B') \times P(A|B')$$

Given, $$P(construction \: job\: is\: completed\: with\: no\: strike)=P(A|B)=0.80$$
and $$P(construction \: job\: is\: completed\: with\: strike)=P(A|B')=0.32$$

$$\therefore P(A)=0.65 \times 0.80+0.35 \times 0.32=0.488$$

Hence the probability that the construction job will get complete on time is $$0.488$$
Question 2
1 / -0

The events $$E_1, E_2, ........$$ represents the partition of the sample space $$S$$, if they are:

A
pairwise disjoint

B
exhaustive

C
have non-zero probabilities

D
All are correct

Solution

A set of events $$E_1 , E_2 ,...$$ is said to represent a partition of a sample space $$S$$, if

$$(a)$$ $$E_i \cap E_j = \phi, i\neq j; i, j = 1, 2, 3,..., n$$ (pairwise disjoint)

$$(b)$$ $$E_i \cup E_2 \cup ... \cup E_n = S$$ (exhaustive)

$$(c)$$ Each $$E_i \neq \phi, i.e, P(E_i) > 0$$ for all $$i = 1, 2, ..., n$$ (have non-zero probabilities)

That is the events should be pairwise disjoint, exhaustive and should have non zero probabilities.

Hence, option D is correct.
Question 3
1 / -0

From a city population, the probability of selecting a male or smoker is $$\dfrac {7}{10}$$, a male smoker is $$\dfrac {2}{5}$$ and a male, if a smoker is already selected, is $$\dfrac {2}{3}$$. Then, Probability of selecting a smoker, if a male is first selected

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

B
$$\dfrac {1}{5}$$

C
$$\dfrac {4}{5}$$

D
$$\dfrac {8}{5}$$

Solution

Suppose $$A:$$ a male is selected
$$B:$$ a smoker is selected
Given:
$$P(A\cup B) = \dfrac {7}{10}, P(A\cap B) = \dfrac {2}{5}$$ and $$P\left (\dfrac {A}{B}\right ) = \dfrac {2}{3}$$
The probability of selecting a smoker.
$$P(B) = \dfrac {P(A\cap B)}{P\left (\dfrac {A}{B}\right )}$$
$$= \dfrac {2\times 3}{5\times 2} = \dfrac {3}{5}$$
The probability of selecting a non-smoker
So, $$P(B) = 1 - P(B)$$ ... [using conditional probability]
$$= 1 - \dfrac {3}{5} = \dfrac {2}{5}$$
The probability of selecting a male
$$P(A) = P(A\cup B) + P(A\cap B) - P(B)$$
$$= \dfrac {7}{10} + \dfrac {2}{5} - \dfrac {3}{5}$$
$$= \dfrac {7 + 4 - 6}{10} = \dfrac {1}{2}$$
Probability of selecting a smoker, if a male is first selected, is given by
$$P\left (\dfrac {B}{A}\right ) = \dfrac {P(A\cap B)}{P(A)}$$ ..... [Using conditional probability]
$$= \dfrac {2}{5}\times \dfrac {2}{1} = \dfrac {4}{5}$$
Hence, only option C is correct.
Question 4
1 / -0

If $$A, B, C$$ are three events associated with a random experiment, then $$P(A) P\left (\dfrac {B}{A}\right )P\left (\dfrac {C}{A} \cap B\right )$$ is

A
$$P (A\cup B\cap C)$$

B
$$P(A \cap B\cap C)$$

C
$$P\left (\dfrac {C}{A}\cap B\right )$$

D
$$P\left (\dfrac {B}{A}\right )$$

Solution

$$P(A) P\left (\dfrac {B}{A}\right )P\left (\dfrac {C}{A} \cap B\right )$$

$$= P(A\cap B) P\left (\dfrac {C}{A}\cap B\right )$$ .......... (using conditional probability)

$$= \dfrac {P(A\cap B)P[C\cap (A\cap B)]}{P(A\cap B)}$$

$$= P(A\cap B\cap C)$$ ............. (using conditional probability)

Hence, option B is correct.
Question 5
1 / -0

If $$x$$ is a continuous random variable then $$P(x \geq a) = $$

A
$$P (x < a)$$

B
$$1 - P(x > a)$$

C
$$P(x > a)$$

D
$$1 - P (x \leq a - 1)$$

Solution

Given $$x$$ is continuous random variable , Let us consider that there are $$n$$ points in total, Since $$x$$ is continuous, $$n\rightarrow \infty $$
The probability of $$x=a$$ is $$\dfrac{1}{n}\rightarrow 0$$ , therefore $$P(x=a)=0$$
Which implies $$P(x\ge a) = P(x>a)+P(x=a) = P(x>a)$$
Therefore the correct option is $$C$$
Question 6
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Directions For Questions

For the next four (04) items that follow :
Number $$X$$ is randomly selected from the set of odd numbers and $$Y$$ is randomly selected from the set of even numbers of the set $$\{1, 2, 3, 4, 5, 6, 7\}$$. Let $$Z = (X + Y)$$.

...view full instructions

What is $$P(Z = 5)$$ equal to ?

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

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

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

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

Solution

Given set $$=\{1,2,3,4,5,6,7\}$$
$$Z=X+Y$$
Where $$X$$ is randomly selected from set of odd numbers and
$$Y$$ is randomly selected from set of even numbers
That gives $$Z$$ is odd and greater than one
Now, for $$Z=5$$
Possible ordered pairs are $$\{(1,4),(2,3)\}$$
Probability of getting these 2 pairs $$=P(1)\times P(4)+P(2)\times P(3)$$
$$=\cfrac{1}{4}\times \cfrac{1}{3}+\cfrac{1}{4}\times \cfrac{1}{3}$$
$$=\cfrac{1}{6}$$
Question 7
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Directions For Questions

For the next four (04) items that follow :
Number $$X$$ is randomly selected from the set of odd numbers and $$Y$$ is randomly selected from the set of even numbers of the set $$\{1, 2, 3, 4, 5, 6, 7\}$$. Let $$Z = (X + Y)$$.

...view full instructions

What is $$P(Z = 10)$$ equal to ?

A
$$0$$

B
$$1/2$$

C
$$1/3$$

D
$$1/5$$

Solution

Given set $$=\{1,2,3,4,5,6,7\}$$
$$Z=X+Y$$
Where $$X$$ is randomly selected from set of odd numbers and $$Y$$ is randomly selected from set of even numbers.
Which gives $$Z$$ as odd and greater than one
Given : $$Z=10$$
But $$10$$ is even
So, $$P(Z=10)=0$$
Question 8
1 / -0

For two independent events $$A$$ and $$B$$, which of the following pair of events need not be independent?

A
$$A', B'$$

B
$$A, B'$$

C
$$ A', B$$

D
$$A - B, B - A$$

Solution

$$A$$ and $$B$$ are independent event.
$$\Rightarrow P(A\cap B) = P(A).P(B)$$

Option A:
$$P(A'\cap B') $$ $$= P((A\cup B)')\\=1-P(A\cup B)\\=1-[P(A)+P(B)-P(A\cap B)]\\=1-P(A)-P(B)+P(A)P(B)\\=(1-P(A))(1-P(B))\\=P(A')P(B')$$
Hence, $$A',B'$$ are also independent events.

Option B:
$$P(A\cap B')$$ $$=P((A'\cup B)')\\=1-P(A'\cup B)\\=1-[P(A')+P(B)-P(A'\cap B)]\\=1-[1-P(A)+P(B)-P(B)+P(A\cap B)]\\=P(A)-P(A\cap B)\\=P(A)(1-P(B))\\=P(A)P(B')$$
Hence, $$A,B'$$ are also independent.

Option C:
Similar as above, $$A',B$$ can also be proven to be independent.

Option D:
Event $$A-B$$ implies the happenings in event $$A$$ which do not occur in event $$B$$.
Event $$B-A$$ implies the happenings in event $$B$$ which do not occur in event $$A$$.
From the above, it is clear that $$A-B$$ and $$B-A$$ are mutually exclusive events. Hence $$P((A-B)\cap (B-A)) = 0$$
Hence, $$A-B,B-A$$ are not independent events.
Option $$D$$ is the correct answer.
Question 9
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What is the probability that a leap year selected at random contains 53 Mondays ?

A
$$1/7$$

B
$$2/7$$

C
$$7/366$$

D
$$26/183$$

Solution

In a leap year there will be $$52$$ weeks and $$2$$ days.
So, there will be $$52$$ Mondays fix.
Now that pair of $$2$$ days can be
Monday and Tuesday
Tuesday and Wednesday
Wednesday and Thursday
Thursday and Friday
Friday and Saturday
Saturday and Sunday
Sunday and Monday
There are $$7$$ possibilities, out of which $$2$$ has Mondays.
So, the probability that a leap year has $$53$$ Mondays $$=\dfrac { 2 }{ 7 }$$
Hence, option B is correct.
Question 10
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Suppose A and B are two events. Event B has occured and it is known that $$P(B) <1$$. What is $$P(A|B^c)$$ equal to?

A
$$\displaystyle\frac{P(A)-P(B)}{1-P(B)}$$

B
$$\displaystyle\frac{P(A)-P(A\cap B)}{1-P(B)}$$

C
$$\displaystyle\frac{P(A)+P(B^c)}{1-P(B)}$$

D
None of the above

Solution

Given that $$A$$ and $$B$$ are two events and event $$B$$ has occured,
From the division rule of probability we know that $$P\left(\dfrac { A }{ B } \right)=\dfrac { P(A\cap B) }{ P(B) } $$,
similarly,
$$P\left(\dfrac { A }{ { B }^{ c } }\right )=\dfrac { P(A\cap { B }^{ c }) }{ P({ B }^{ c }) } $$

$$\\ \Longrightarrow \dfrac { P(A)-P(A\cap B) }{ 1-P(B) } $$

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Probability Test - 31

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Probability Test - 31

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SHARING IS CARING

Question 1

$$0.438$$

$$0.538$$

$$0.488$$

None of these

Solution

Question 2

pairwise disjoint

exhaustive

have non-zero probabilities

All are correct

Solution

Question 3

$$\dfrac {3}{2}$$

$$\dfrac {1}{5}$$

$$\dfrac {4}{5}$$

$$\dfrac {8}{5}$$

Solution

Question 4

$$P (A\cup B\cap C)$$

$$P(A \cap B\cap C)$$

$$P\left (\dfrac {C}{A}\cap B\right )$$

$$P\left (\dfrac {B}{A}\right )$$

Solution

Question 5

$$P (x < a)$$

$$1 - P(x > a)$$

$$P(x > a)$$

$$1 - P (x \leq a - 1)$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$0$$

$$1/2$$

$$1/3$$

$$1/5$$

Solution

Question 8

$$A', B'$$

$$A, B'$$

$$ A', B$$

$$A - B, B - A$$

Solution

Question 9

$$1/7$$

$$2/7$$

$$7/366$$

$$26/183$$

Solution

Question 10

$$\displaystyle\frac{P(A)-P(B)}{1-P(B)}$$

$$\displaystyle\frac{P(A)-P(A\cap B)}{1-P(B)}$$

$$\displaystyle\frac{P(A)+P(B^c)}{1-P(B)}$$

None of the above

Solution

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