Probability Test

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Probability Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

$$A, B$$ and $$C$$ are three mutually exclusive and exhaustive events such that $$P(A) = 2 P(B) = 3P(C)$$. What is $$P(B)$$?

A
$$\dfrac {6}{ 11}$$

B
$$\dfrac {6 }{ 22}$$

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

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

Solution

Given $$A,B,C$$ are three mutually exclusive $$\xi$$ exhaustive events.
$$\Rightarrow P(A)=2P(B)=3P(C)$$
$$\Rightarrow P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) $$
$$=2P\left( B \right) +P\left( B \right) +\dfrac { 11P\left( B \right) }{ 3 } $$
$$=\dfrac { 6P\left( B \right) +3P\left( B \right) +2P\left( B \right) }{ 3 } =\dfrac { 11P\left( B \right) }{ 3 } $$
$$\Rightarrow P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) =1$$
$$\Rightarrow 1=\dfrac { 11P\left( B \right) }{ 3 }$$
$$\Rightarrow P\left( B \right)=\dfrac{3}{11}=\dfrac{6}{22}$$

Hence, the answer is $$\dfrac{6}{22}.$$
Question 2
1 / -0

Consider the following in respect of two events $$A$$ and $$B$$:
(1) $$P(A$$ occurs but not $$B)=P(A)-P(B)$$ if $$B\subset A$$
(2) $$P(A$$ alone or $$B$$ alone occurs) $$=P(A)+P(B)-P(A\cap B)$$
(3) $$P(A\cup B)=P(A)+P(B)$$ if $$A$$ and $$B$$ are mutually exclusive
Which of the above is/are correct?

A
1 only

B
1 and 3 only

C
2 and 3 only

D
1 and 2 only

Solution

If $$B\subset A$$, then $$P(A - B) = P(A) - P(A\cap B) = P(A) - P(B)$$. Statement 1 is correct.
$$P (A \text{ alone or } B \text{ alone}) = P(A) - P(A\cap B) + P(B) - P(A\cap B) = P(A) + P(B) - 2P (A\cap B)$$Statement 2 is false.
If $$A$$ and $$B$$ are mutually exclusive, then $$P(A\cap B)=0\implies P(A\cup B) = P(A) + P(B)$$. Statement 3 is correct.
Hence, answer is 1 and 3 only.
Question 3
1 / -0

Consider the following statements:
(1) $$P(\bar { A } \cup B)=P(\bar { A } )+P(B)-P(\bar { A } \cap B)$$
(2) $$P(A\cap \bar B) = P(\bar B)-P(A\cap B)$$
(3) $$P(A\cap B)=P(B)P(A|B)\quad $$
Which of the above statements are correct?

A
1 and 2 only

B
1 and 3 only

C
2 and 3 only

D
1, 2 and 3

Solution

Statement 1: $$P(\overline{A}\cup B) = P(\overline{A}) + P(B) - P(\overline{A} \cap B)$$ is true by definition.
Statement 2: $$P(A\cup \overline{B}) = P(B) - P(A \cap B)$$ is false by definition.
Statement 3: $$P(A \cap B)=P(B)P(A|B)$$ is true by definition of conditional probability.
Hence, only statements 1 and 3 are true.
Question 4
1 / -0

For two events $$A$$ and $$B$$, let $$P(A) =\dfrac {1}{2}, P(A\cup B) = \dfrac {2}{3}$$ and $$P(A\cap B) = \dfrac {1}{6}$$, What is $$P(\overline {A} \cap B)$$ equal to?

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

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

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

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

Solution

$$P(A\cap B)=P(B)-P(\overline{A}\cap B)$$

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

$$\dfrac 23=\dfrac 12+P(B)-\dfrac 16$$

$$ \implies P(B)=\dfrac 56-\dfrac 36$$

$$ \implies P(B)=\dfrac{1}{3}$$

$$\Rightarrow P(\overline{A}\cap B)=P(B)-P(A \cap B)=\dfrac 13-\dfrac 16$$

$$\therefore P(\overline{A}\cap B)=\dfrac{1}{6}$$
Question 5
1 / -0

If $$P(B) = \dfrac {3}{4}, P(A\cap B\cap \overline {C}) = \dfrac {1}{3}$$ and $$P(\overline {A}\cap B\cap \overline {C}) = \dfrac {1}{3}$$, then what is $$P(B\cap C)$$ equal to?

A
$$\dfrac {1}{12}$$

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

C
$$\dfrac {1}{15}$$

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

Solution

From the above drawn Venn diagram,
$$P(B) = P(A\cap B\cap \bar{C}) + P(\bar{A} \cap B \cap \bar{C}) + P(B\cap C)$$
$$\therefore P(B\cap C) = P(B) - P(A\cap B\cap \bar{C}) - P(\bar{A} \cap B \cap \bar{C}) = \dfrac{3}{4} - \dfrac{1}{3} - \dfrac{1}{3} = \dfrac{1}{12}$$
Question 6
1 / -0

A survey of $$850$$ students in a University yields that $$680$$ students like music and $$215$$ like dance. What is the least number of students who like both music and dance?

A
$$40$$

B
$$45$$

C
$$50$$

D
$$55$$

Solution

$$680$$ students like music
$$215$$ like dance
$$Total \rightarrow 680 + 215 = 895$$
Least no. of students who like both dance and music
$$= 895 - 850$$
$$= 45$$.
Question 7
1 / -0

In a class, $$54$$ students are good in Hindi only, $$63$$ students are good in Mathematics only and $$41$$ students are good in English only. There are $$18$$ students who are good in both Hindi and Mathematics. $$10$$ students are good in all three subjects. What is the number of students who are good in either Hindi or Mathematics but not in English?

A
$$99$$

B
$$107$$

C
$$125$$

D
$$130$$

Solution

This is a question of venn diagram
No. of students who are good in either Hindi or Mathematics but not in English
$$= 54 + 18 + 63$$
$$= 125$$
Let $$H_{1}M_{1}E_{1}$$ denote the set of students studying Hindi, Mathematics and English.
No. of students of English only $$= 41$$
No. of students of Hindi only $$= 63$$
No. of students of Maths only $$= 54$$
$$n(H\cap M\cap E) = 10$$
$$n(H\cap M) = 18$$
No. of students who study $$'H'$$ or $$'M'$$ but not $$'E'$$
$$= 63 + 54 + 18 = 125$$.
Question 8
1 / -0

If $$A$$ and $$B$$ are independent events, $$P(A)=0.1$$ and $$P(B)=0.9$$, then $$P\left( A\cup B \right) =$$ ____

A
$$0.91$$

B
$$0.09$$

C
$$0.99$$

D
$$0.90$$

Solution

Since A and B are independent events so $$P(A \cap B)=P(A)\times P(B)=0.1\times 0.9=0.09$$
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A)=0.1$$
$$P(B)=0.9$$
$$P(A\cap B)=0.09$$
So
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$P(A\cup B)=0.1+0.9-0.09$$
$$P(A\cup B)=1-0.09=0.91$$
Question 9
1 / -0

In a survey of $$400$$ students in a school, $$100$$ are listed as taking apple juice, $$150$$ as taking orange juice and $$75$$ were listed as taking both apple as well as orange juice. Find how many students were taking "Neither apple nor orange juice"

A
$$175$$

B
$$225$$

C
$$275$$

D
$$none\ of\ these$$

Solution

$$There\quad are\quad 400\quad students.So,n(s)=400.$$
$$ 100\quad students\quad take\quad apple\quad juice,n(A)=100$$
$$ 150\quad take\quad orange\quad juice,n(o)=150$$
$$ 75\quad students\quad take\quad both\quad apple\quad abd\quad orange\quad juice,n(A\cap o)=75$$
$$ So,n(A\cup o)=n(A)+n(o)-n(A\cap o)$$
$$ \quad \quad \quad \quad \quad \quad \quad \quad =100+150-75$$
$$ \quad \quad \quad \quad \quad \quad \quad \quad =175$$
$$ Number\quad of\quad students\quad taking\quad neither\quad apple\quad nor\quad orange$$
$$ juice,n(\overline { A\cup o } )=n(s)-n(A\cup o)$$
$$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =400-175$$
$$ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =225$$
Question 10
1 / -0

Tickets numbered from $$1$$ to $$30$$ are mixed up and then a ticket is drawn at random. What is the probability that the drawn ticket has a number which is divisible by both $$2$$ and $$6$$?

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

B
$$\dfrac{2}{5}$$

C
$$\dfrac{8}{15}$$

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

Solution

Here, $$S=\{1,2,3,4,5,6,.....29,30\}$$

Let $$E$$ be the event of number divisible by both $$2$$ and $$6$$.

$$E=\{6,12,18,24,30\}$$

$$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{5}{30}=\dfrac{1}{6}$$

$$\therefore$$ the probability that the drawn ticket has a number which is divisible by both $$2$$ and $$6$$ is $$\dfrac{1}{6}$$

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Probability Test - 48

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Probability Test - 48

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SHARING IS CARING

Question 1

$$\dfrac {6}{ 11}$$

$$\dfrac {6 }{ 22}$$

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

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

Solution

Question 2

1 only

1 and 3 only

2 and 3 only

1 and 2 only

Solution

Question 3

1 and 2 only

1 and 3 only

2 and 3 only

1, 2 and 3

Solution

Question 4

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

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

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

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

Solution

Question 5

$$\dfrac {1}{12}$$

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

$$\dfrac {1}{15}$$

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

Solution

Question 6

$$40$$

$$45$$

$$50$$

$$55$$

Solution

Question 7

$$99$$

$$107$$

$$125$$

$$130$$

Solution

Question 8

$$0.91$$

$$0.09$$

$$0.99$$

$$0.90$$

Solution

Question 9

$$175$$

$$225$$

$$275$$

$$none\ of\ these$$

Solution

Question 10

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

$$\dfrac{2}{5}$$

$$\dfrac{8}{15}$$

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

Solution

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