Probability Test

Result

Probability Test - 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A researcher conducted a survey to determine whether people in a certain town prefer watching sports on television to attending the sporting event. The researcher asked 117 people who visited a local restaurant on a Saturday, and 7 people refused to respond. Which of the following factors makes it least likely that a reliable conclusion can be drawn about the sports-watching preferences of all people in the town?

A
Sample size

B
Population size

C
The number of people who refused to respond

D
Where the survey was given.

Solution

Considering the population of a town, the number $$117$$ is very much low.
We can't take a survey of $$117$$ people and conclude the result for the whole town.
So, the $$sample\ size$$ of the survey is very low compared to the population. $$[A]$$
Question 2
1 / -0

If $$ n(A) = 1000, n (B)=500 $$ and if $$ n ( A \cap B ) \ge 1 $$ and $$ n( A \cup B ) = p , $$ then

A
$$ 500 \le p \le 1000$$

B
$$ 1001 \le p \le 1498 $$

C
$$ 1000 \le p \le 1498 $$

D
$$ 999 \le p \le 1499 $$

E
$$ 1000 \le p \le 1499 $$

Solution

Given, $$ n(A) = 1000, n (B) = 500 $$

and $$ n ( A \cap B ) \ge 1 $$ and $$ n ( A \cup B ) = p $$

Therefore, $$n ( A \cap B ) = n (A) + n (B) - n (A \cup B) $$
$$ = 1000 + 500 - p $$
$$ = 1500 - p.$$ ..... [Since $$ n ( A \cap B ) \ge 1 $$]
$$ \Rightarrow 1500 - p \ge 1 $$
$$ \Rightarrow p \le 1499$$ .....(i)

Also, $$ n ( A \cup B) \ge n (A) $$

$$ \Rightarrow p \ge 1000$$ ...(ii)
From Equations. (i) and (ii), we get

$$ 1000 \le p \le 1499 $$
Question 3
1 / -0

Sita and Geta are friends, what is the probability that both will have different birthdays (ignoring a leap year)

A
$$\dfrac { 1 }{ 365 } $$

B
$$\dfrac { 1 }{ 364 } $$

C
$$\dfrac { 364 }{ 365 } $$

D
None of these

Solution

$$P=\dfrac { 365\times 364 }{ 365\times 365 } =\dfrac { 364 }{ 365 } $$
Question 4
1 / -0

If $$A$$ and $$B$$ are events with $$P(A\cup B) = \dfrac {3}{4}, P(A') = \dfrac {2}{3}$$ and $$P(A\cap B) = \dfrac {1}{4}$$ then $$P(B)$$ is

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

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

C
$$\dfrac {3}{4}$$

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

Solution

Given $$P(A\cup B)=\dfrac{3}{4}, P(A')=\dfrac{2}{3}, P(A\cap B)=\dfrac{1}{4}$$.

We have to find $$P(B)$$.

Since $$P(A')=\dfrac{2}{3}, P(A)=1-P(A')=1-\dfrac{2}{3}=\dfrac{1}{3}$$

Hence $$P(A)=\dfrac{1}{3}$$

We know that, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

Substituting the values we get,

$$\dfrac{3}{4}=\dfrac{1}{3}+P(B)-\dfrac{1}{4}$$

$$\Rightarrow P(B)=\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{3}$$

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

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

$$\therefore P(B)=\dfrac{2}{3}$$
Question 5
1 / -0

A die is thrown .The probability that the number comes up even is ______ .

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

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

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

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

Solution

Possible Outcomes of a Dice $$= 1, 2, 3, 4, 5, 6$$
Even Outcomes $$= 2, 4, 6$$
Probability of an event $$P(E) = \dfrac{No.\ of\ Favourable\ Outcomes}{Total\ No.\ of\ Outcomes}$$

Here,
Favourable Outcomes $$=$$ Getting an Even Number on Dice
Number of Favourable Outcomes $$=$$ Number of Possible Even Outcomes $$= 3$$
Total Number of Outcomes $$= 6$$

$$\Rightarrow P(E) = \dfrac{3}{6} = \dfrac{1}{2}$$
Question 6
1 / -0

The probability that an event does not happens in one trial is 0.8.The probability that the event happens atmost once in three trails is

A
$$0.896$$

B
$$0.791$$

C
$$0.642$$

D
$$0.592$$

Solution
Question 7
1 / -0

If $$A$$ and $$B$$ are events having probabilities, $$P(A)=0.6,P(B)=0.4,P(A\cap B)=0$$, then the probability that neither $$A$$ nor $$B$$ occurs is

A
$$\cfrac{1}{4}$$

B
$$1$$

C
$$\cfrac{1}{2}$$

D
$$0$$

Solution

P(A u B) $$=$$ P(A) + P(B) - P(A n B)

P(A u B) $$=$$ probability that either A or B occurs
P(A n B) $$=$$ probability that both A & B occur
Plugging the values of P(A), P(B), P(A n B)

we get,
P(A u B) $$= 0.6 + 0.4 - 0$$
P(A u B) $$= 1$$

This means, there is $$100 \%$$ chance that either event A or B occurs all the time.

Which means, the probability of neither A nor B occurring will be 0, because one of them will occur $$100 \%$$ of the time.
Question 8
1 / -0

$$A$$ and $$B$$ are two independent events such that $$P(A)=\cfrac { 1 }{ 2 } ;P(B)=\cfrac { 1 }{ 3 } $$. Then $$P$$(neither $$A$$ nor $$B$$) is equal to

A
$$\cfrac { 2 }{ 3 } $$

B
$$\cfrac { 1 }{ 6 } $$

C
$$\cfrac { 5 }{ 6 } $$

D
$$\cfrac { 1 }{ 3 } $$

Solution

Given $$P(A)=\dfrac{1}{2},P(B)=\dfrac{1}{3}$$

Since $$A,B$$ are independent events $$P(A\cap B)=P(A) \cdot P(B)$$

$$\therefore P(A\cap B)=\dfrac{1}{2} \cdot \dfrac{1}{3}=\dfrac{1}{6}$$

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}$$

$$\therefore P(A\cup B)=\dfrac{2}{3}$$

$$P(neither\: A\:nor\:B)=P(A'\cap B')$$

$$=P(A\cup B)'$$

$$=1-P(A\cup B)$$

$$=1-\dfrac{2}{3}$$

$$P(neither\: A\:nor\:B)=\dfrac{1}{3}$$
Question 9
1 / -0

For any two events $$A$$ and $$B$$ -

A
$$P(A) + P(B) > P(A \cap B)$$

B
$$P(A) + P(B) < P(A \cap B)$$

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

D
$$P(A) \times P(B) \leq P(A \cap B)$$

Solution

Consider the events $$A$$ $$\xi$$ $$B$$
$$P(A\cup B)\le P(A)+P(B)-P(A\cap B)$$
$$\Rightarrow$$ When $$P(A\cup B)=0$$
$$P(A)+P(B)\ge P(A\cap B).$$
Hence, the answer is $$P(A)+P(B)\ge P(A\cap B).$$
Question 10
1 / -0

For two independent events $$A$$ and $$B$$, what is $$P(A + B)$$, given $$P(A) = \dfrac{3}{5}$$ and $$P(B) = \dfrac{2}{3}$$?

A
$$\dfrac{11}{15}$$

B
$$\dfrac{13}{15}$$

C
$$\dfrac{7}{15}$$

D
$$0.65$$

Solution

Given: $$P(A)=\dfrac 35, P(B)=\dfrac 23$$, and A, B are independent.
To find: $$P(A+B)=?$$
Sol: As A, B are independent, $$\implies P(A\cap B)=P(A)P(B)$$
And also $$P(A+B)=P(A\cup B)$$
$$\therefore, P(A+B)=P(A)+P(B)-P(A\cap B)\\\implies P(A+B)=\dfrac 35+\dfrac 23-\left(\dfrac 35\times \dfrac 23\right)\\\implies P(A+B)=\dfrac {9+10-6}{15}=\dfrac {13}{15}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Probability Test - 46

Result

Probability Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

Sample size

Population size

The number of people who refused to respond

Where the survey was given.

Solution

Question 2

$$ 500 \le p \le 1000$$

$$ 1001 \le p \le 1498 $$

$$ 1000 \le p \le 1498 $$

$$ 999 \le p \le 1499 $$

$$ 1000 \le p \le 1499 $$

Solution

Question 3

$$\dfrac { 1 }{ 365 } $$

$$\dfrac { 1 }{ 364 } $$

$$\dfrac { 364 }{ 365 } $$

None of these

Solution

Question 4

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

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

$$\dfrac {3}{4}$$

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

Solution

Question 5

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

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

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

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

Solution

Question 6

$$0.896$$

$$0.791$$

$$0.642$$

$$0.592$$

Solution

Question 7

$$\cfrac{1}{4}$$

$$1$$

$$\cfrac{1}{2}$$

$$0$$

Solution

Question 8

$$\cfrac { 2 }{ 3 } $$

$$\cfrac { 1 }{ 6 } $$

$$\cfrac { 5 }{ 6 } $$

$$\cfrac { 1 }{ 3 } $$

Solution

Question 9

$$P(A) + P(B) > P(A \cap B)$$

$$P(A) + P(B) < P(A \cap B)$$

$$P(A) + P(B) \geq P(A \cap B)$$

$$P(A) \times P(B) \leq P(A \cap B)$$

Solution

Question 10

$$\dfrac{11}{15}$$

$$\dfrac{13}{15}$$

$$\dfrac{7}{15}$$

$$0.65$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.