Probability Test

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Probability Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

In throwing of a die, let $$A$$ be the event 'an odd humber turns up', $$B$$ be the event 'a number divisible by $$3$$ turns up' and $$C$$ be the event 'a number $$\leq 4$$ turns up', Then find the probability that exactly two of $$A, B$$ and $$C$$ occur.

A
$$\dfrac23$$

B
$$\dfrac16$$

C
$$\dfrac13$$

D
$$\dfrac56$$

Solution

Event $$A=\left \{ 1, 3, 5 \right \},$$ event $$B=\left \{ 3, 6 \right \}$$ and event $$C=\left \{ 1, 2, 3, 4 \right \}$$
$$\therefore $$ $$P$$(exactly two of $$A, B$$ and $$C$$ occur)$$=P\left ( A\cap B \right )+P\left ( B\cap C \right )+P\left ( C\cap A \right )-3P\left ( A\cap B\cap C \right )$$
$$\displaystyle =\frac{1}{6}+\frac{1}{6}+\frac{2}{6}-3\times \frac{1}{6}=\frac{1}{6}$$
Question 2
1 / -0

For an event E, P(E) + P(E') =

A
0

B
1

C
0.5

D
2

Solution

$$Let\quad E\quad be\quad an\quad event\quad and\quad probability\quad of\quad happening\quad of\quad event\quad be\quad p(E)\\ where\quad E'\quad is\quad the\quad complement\quad of\quad event\quad \quad \\ Hence,\quad p(E')=1-p(E)\\ \Rightarrow p(E)+p(E')=1\\ $$
Question 3
1 / -0

If $$n(A)=18$$, $$n(B)=12$$, and $$A\cap B=\emptyset$$, then $$n(A\cup B)=$$

A
$$6$$

B
$$12$$

C
$$30$$

D
$$20$$

Solution

$$n(A\cup B)=n(A)+n(B)-n(A\cap B)=18+12-0=30$$
Question 4
1 / -0

In a group of $$15$$ people $$7$$ read French $$8$$ read English while $$3$$ of them read none of them. How many read French and English both?

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

E
$$7$$

Solution

Since 3 peoples read none of the two subjects there are only 12 people who are reading either one of the two or both It is clear from the diagram that 3 people are studying both French and English
Question 5
1 / -0

If A is any event in a sample space then P(A') is____

A
P(A)

B
1+P(A)

C
1-P(A)

D
1-2P(A)

Solution

if A is any event in a sample space, then $$ P(A') $$ will have the elements which are not in $$ P(A) $$

Also, Total elements will be $$ 1 $$

So, $$ P(A') = 1 - P(A) $$
Question 6
1 / -0

A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if ti is known that the elder child is a female.

A
$$0.55,0.38$$

B
$$0.33,0.50$$

C
$$0.67,0.78,$$

D
$$0.56,0.67$$

Solution

If a couple has two children, then the sample space is
$$S=\left\{ \left( b,b \right) ,\left( b,g \right) ,\left( g,b \right) ,\left( g,g \right) \right\} $$
(i) Let $$E$$ and $$F$$ respectively denote the events that both children are males and at least one of the children is a male.
$$\therefore E\cap F=\left\{ \left( b,b \right) \right\} \Rightarrow P\left( E\cap F \right) =\displaystyle\frac { 1 }{ 4 } $$
$$P\left( E \right) =\displaystyle\frac { 1 }{ 4 } $$
$$P\left( F \right) =\displaystyle\frac { 3 }{ 4 } $$
$$\Rightarrow P\left( E|F \right) =\displaystyle\frac { P\left( E\cap F \right) }{ P\left( F \right) } =\displaystyle\frac { \displaystyle\frac { 1 }{ 4 } }{ \displaystyle\frac { 3 }{ 4 } } =\displaystyle\frac { 1 }{ 3 }=0.33 $$
(ii) Let $$A$$ and $$B$$ respectively denote the events that both children are females and the elder child is a female.
$$A=\left\{ \left( g,g \right) \right\} \Rightarrow P\left( A \right) =\displaystyle\frac { 1 }{ 4 } $$
$$B=\left\{ \left( g,b \right) ,\left( g,g \right) \right\} \Rightarrow P\left( B \right) =\displaystyle\frac { 2 }{ 4 } $$
$$A\cap B=\left\{ \left( g,g \right) \right\} \Rightarrow P\left( A\cap B \right) =\displaystyle\frac { 1 }{ 4 } $$
$$P\left( A|B \right) =\displaystyle\frac { P\left( A\cap B \right) }{ P\left( B \right) } =\displaystyle\frac { \displaystyle\frac { 1 }{ 4 } }{ \displaystyle\frac { 2 }{ 4 } } =\displaystyle\frac { 1 }{ 2 }=0.50 $$
Question 7
1 / -0

Three letters, to each of which corresponds an envelope, are placed in the envelopes at random. The probability that all the letters are not placed in the right envelopes, is

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

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

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

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

Solution

Three letters can be placed in 3 envelopes in $$3!$$ ways, whereas there is only one way of placing them in their right envelopes.
So, probability that all the letters are placed in the right envelopes$$=\dfrac{1}{3!}$$
Hence, required probability$$=1-\dfrac{1}{3!}=\dfrac{5}{6}$$
Question 8
1 / -0

The chance of throwing a total of $$3$$ or $$5$$ or $$11$$ with two dice is.

A
$$\displaystyle\frac{5}{36}$$

B
$$\displaystyle\frac{1}{9}$$

C
$$\displaystyle\frac{2}{9}$$

D
$$\displaystyle\frac{19}{36}$$

Solution

The chance of throwing a total of $$3$$ is $$2$$, i.e., $$\left\{(1, 2), (2, 1)\right\}$$.
The chance of throwing a total of $$5$$ is $$4$$, i.e., $$\left\{(1, 4), (4, 1), (2, 3), (3, 2)\right\}$$.
The chance of throwing a total of $$11$$ is $$2$$, i.e., $$\left\{(5, 6), (6, 5)\right\}$$.
$$\therefore$$ Required probability$$=\displaystyle\frac{2}{36}+\frac{4}{36}+\frac{2}{36}$$
$$=\displaystyle\frac{8}{36}=\frac{2}{9}$$
Question 9
1 / -0

Given that the events $$A$$ and $$B$$ are such that $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 } , P\left( A\cup B \right) = \displaystyle\frac { 3 }{ 5 } $$ and $$P\left( B \right) =p$$. Find $$p$$ if they are (i) mutually exclusive (ii) independent.

A
$$0.5,0.6$$

B
$$0.1,0.2$$

C
$$0,2,0.4$$

D
$$0.1,0.6$$

Solution

It is given that $$P\left( A \right) = \displaystyle\frac { 1 }{ 2 } , P\left( A\cup B \right) = \displaystyle\frac { 3 }{ 5 } $$ and $$P\left( B \right) = p$$
(i) When $$A$$ and $$B$$ are mutually exclusive, $$A\cap B=\phi $$
$$\therefore P\left( A\cap B \right) =0$$
It is known that, $$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow \displaystyle\frac { 3 }{ 5 } = \displaystyle\frac { 1 }{ 2 } + p - 0$$
$$\Rightarrow p = \displaystyle\frac { 3 }{ 5 } - \displaystyle\frac { 1 }{ 2 } = \displaystyle\frac { 1 }{ 10 }=0.1 $$
(ii) When $$A$$ and $$B$$ are independent, $$P\left( A\cap B \right) =P\left( A \right) \cdot P\left( B \right) = \displaystyle\frac { 1 }{ 2 } p$$
It is known that, $$P\left( A\cup B \right) =P\left( A \right) + P\left( B \right) -P\left( A\cap B \right) $$
$$\Rightarrow \displaystyle\frac { 3 }{ 5 } = \displaystyle\frac { 1 }{ 2 } + p - \displaystyle\frac { 1 }{ 2 } p$$
$$\Rightarrow \displaystyle\frac { 3 }{ 5 } = \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { p }{ 2 } $$
$$\Rightarrow \displaystyle\frac { p }{ 2 } = \displaystyle\frac { 3 }{ 5 } -\displaystyle\frac { 1 }{ 2 } = \displaystyle\frac { 1 }{ 10 } $$
$$\Rightarrow p = \displaystyle\frac { 2 }{ 10 } = \displaystyle\frac { 1 }{ 5 }=0.2 $$
Question 10
1 / -0

If $$A$$ and $$B$$ are any two events such that $$P\left( A \right) +P\left( B \right) -P\left( A and B \right) =P\left( A \right) $$, then

A
$$P\left( B|A \right) =1$$

B
$$P\left( A|B \right) =1$$

C
$$P\left( B|A \right) =0$$

D
$$P\left( A|B \right) =0$$

Solution

$$P\left( A \right) +P\left( B \right) -P\left( A and B \right) =P\left( A \right) $$
$$\Rightarrow P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) =P\left( A \right) $$
$$\Rightarrow P\left( B \right) -P\left( A\cap B \right) =0$$
$$\Rightarrow P\left( A\cap B \right) =P\left( B \right) $$
$$\therefore P\left( A|B \right) =\displaystyle\frac { P\left( A\cap B \right) }{ P\left( B \right) } =\displaystyle\frac { P\left( B \right) }{ P\left( B \right) } =1$$
Thus, the correct answer is B.

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Probability Test - 36

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Probability Test - 36

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SHARING IS CARING

Question 1

$$\dfrac23$$

$$\dfrac16$$

$$\dfrac13$$

$$\dfrac56$$

Solution

Question 2

0

1

0.5

2

Solution

Question 3

$$6$$

$$12$$

$$30$$

$$20$$

Solution

Question 4

$$3$$

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 5

P(A)

1+P(A)

1-P(A)

1-2P(A)

Solution

Question 6

$$0.55,0.38$$

$$0.33,0.50$$

$$0.67,0.78,$$

$$0.56,0.67$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$$\displaystyle\frac{5}{36}$$

$$\displaystyle\frac{1}{9}$$

$$\displaystyle\frac{2}{9}$$

$$\displaystyle\frac{19}{36}$$

Solution

Question 9

$$0.5,0.6$$

$$0.1,0.2$$

$$0,2,0.4$$

$$0.1,0.6$$

Solution

Question 10

$$P\left( B|A \right) =1$$

$$P\left( A|B \right) =1$$

$$P\left( B|A \right) =0$$

$$P\left( A|B \right) =0$$

Solution

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