Probability Test

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Probability Test - 9

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Question 1
1 / -0

If $$\displaystyle \frac { 1+4p }{ 4 } ,\frac { 1-p }{ 2 } ,\frac { 1-2p }{ 2 } $$ are probabilities of three mutually exclusive events, then

A
$$\displaystyle \frac { 1 }{ 3 } \le p\le \frac { 1 }{ 2 } $$

B
$$\displaystyle \frac { 1 }{ 3 } \le p\le \frac { 2 }{ 3 } $$

C
$$\displaystyle \frac { 1 }{ 6 } \le p\le \frac { 1 }{ 2 } $$

D
None of these

Solution

As $$\displaystyle \frac { 1+4p }{ 4 } ,\frac { 1-p }{ 2 } ,\frac { 1-2p }{ 2 } $$ are probabilities of three mutually exclusive events, we must have
$$\displaystyle 0\le \frac { 1+4p }{ 4 } \le 1,0\le \frac { 1-p }{ 2 } \le 1,0\le \frac { 1-2p }{ 2 } \le 1$$
and $$\displaystyle 0\le \frac { 1+4p }{ 4 } +\frac { 1-p }{ 2 } +\frac { 1-2p }{ 2 } \le 1$$
$$\displaystyle \Rightarrow -\frac { 1 }{ 4 } \le p\le \frac { 3 }{ 4 } ,-1\le p\le 1,-\frac { 1 }{ 2 } \le p\le \frac { 1 }{ 2 } $$ and $$\displaystyle \frac { 1 }{ 2 } \le p\le \frac { 5 }{ 2 } $$
$$\displaystyle \Rightarrow \max { \left\{ -\frac { 1 }{ 4 } ,-1,-\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right\} } <p\le \min { \left\{ \frac { 3 }{ 4 } ,1,\frac { 1 }{ 2 } ,\frac { 5 }{ 2 } \right\} } $$
$$\displaystyle \Rightarrow \frac { 1 }{ 2 } \le p\le \frac { 1 }{ 2 } \Rightarrow p=\frac { 1 }{ 2 } $$
Question 2
1 / -0

Probability of an event $$E\, +$$ Probability of the event 'not $$E$$'$$=$$_____

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Let the probability of an event ($$E$$) $$=P(E)$$
Then, probability of not an event ($$E$$) $$=1 - P(E)$$
therefore, probability of an event ($$E$$) $$+$$ probability of not an event ($$E$$) $$=P(E) + 1 - P(E) = 1$$
Question 3
1 / -0

The probability of an event that cannot happen is ____?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

We know that. all probabilities are between $$0$$ and $$1$$ inclusive.
A probability of $$0$$ means an event is impossible, it cannot happen.
A probability of $$1$$ means an event is certain to happen, it must happen.
Thus the probability of an event that cannot happen is $$0$$.
Question 4
1 / -0

If S is a sample space $$P\left ( A \right )= \dfrac{1}{3} P\left ( B \right )$$ and $$S= A\cup B$$ where A and B are two mutually exclusive events, then $$P\left ( A \right )= $$

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

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

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

D
$$\dfrac{3}{8}$$

Solution

A and B are two mutually exclusive events .So, $$P(A\cap B)=0$$
Because $$S=A\cup B$$ so: $$P(A\cup B)=1$$. It is a case of an Exhaustive Event too.
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$1=P(A)+3P(A)-0$$
$$P(A)=\dfrac{1}{4}$$
Question 5
1 / -0

A student appears for tests, I, II and III. The student is successful if he passes in tests I, II or I, III. The probabilities of the student passing in tests I, II and III are respectively p, q and $$\dfrac{1}{2}$$. If the probability of the student to be successful is $$\dfrac{1}{2}$$ then

A
p(1 + q) $$=$$ 1

B
q(1 + p) $$=$$ 1

C
pq $$=$$ 1

D
$$\dfrac{1}{p} + \dfrac{1}{q} = 1$$

Solution

$$P(I, II), P (I, III)$$
$$P(I)=P$$ $$(P(I, II) \cup P(I)(II)) =\dfrac{1}{2}$$
$$P(II)=q$$ $$\dfrac{1}{2} =P(I)^2 P(II)P(III)$$
$$P(III)=\dfrac{1}{2}$$
$$\displaystyle \dfrac{1}{2} =pq + \dfrac{p}{2} -\dfrac{pq}{2}$$
$$P(1+q)=1$$
Question 6
1 / -0

The probability of an event that is certain to happen is ____?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

The event that is sure to happen is called a certain event and probability of such an event is $$1$$ as this event is bound to happen.
Question 7
1 / -0

In a sample study of $$642$$ people, it was found that $$514$$ people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is :

A
$$0.5$$

B
$$0.6$$

C
$$0.7$$

D
$$0.8$$

Solution

Given that:
Total number of people $$n(S)=642$$
No. of peoples having high school certificate $$n(E)=514$$
$$P(E)=$$Probability that the person has a school certificate
$$\therefore P(E)=\dfrac{n(E)}{n(S)}=\dfrac{514}{642}=0.8006$$
Hence, D is the correct option.
Question 8
1 / -0

If A and B are two mutually exclusive events then $$P\left ( A\cap B \right )= $$

A
$$P(A). P(B)$$

B
$$P(A)$$

C
$$0$$

D
$$1$$

Solution

The basic meaning of exclusive event is the events are unique and there will be no set of common elements between them, so the probability $$P(A \cap B) \ = 0 $$ .
Question 9
1 / -0

If A and B are two Mutually Exclusive events in a sample space S such that $$P\left ( B \right )= 2P\left ( A \right )$$ and $$A\cup B=S$$ then $$P\left ( A \right )=$$

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

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

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

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

Solution

A and B are two Mutually Exclusive events so $$P(A\cap B)=0$$
$$A\cup B=S \Rightarrow P(A\cup B)=1$$
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$1=P(A)+2P(A)-0$$
$$P(A)=\dfrac{1}{3}$$
Question 10
1 / -0

If A and B are two mutually exclusive events then $$P\left ( A\cup B \right )= $$

A
$$P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$$

B
$$P\left ( A \right )-2P\left( A\cap B \right )$$

C
$$P(A) + P(B)$$

D
$$P(A).P(B)$$

Solution

Two exclusive events signify that they have nothing in common between them. so $$ P(A \cap B) = \ 0 $$
$$P(A \cup B) \ = P(A) + \ P(B) \ - P(A \cap B) \\so \ P(A \cup B) = P(A) \ + P(B)$$

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Probability Test - 9

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Probability Test - 9

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Question 1

$$\displaystyle \frac { 1 }{ 3 } \le p\le \frac { 1 }{ 2 } $$

$$\displaystyle \frac { 1 }{ 3 } \le p\le \frac { 2 }{ 3 } $$

$$\displaystyle \frac { 1 }{ 6 } \le p\le \frac { 1 }{ 2 } $$

None of these

Solution

Question 2

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 4

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

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

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

$$\dfrac{3}{8}$$

Solution

Question 5

p(1 + q) $$=$$ 1

q(1 + p) $$=$$ 1

pq $$=$$ 1

$$\dfrac{1}{p} + \dfrac{1}{q} = 1$$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 7

$$0.5$$

$$0.6$$

$$0.7$$

$$0.8$$

Solution

Question 8

$$P(A). P(B)$$

$$P(A)$$

$$0$$

$$1$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$$

$$P\left ( A \right )-2P\left( A\cap B \right )$$

$$P(A) + P(B)$$

$$P(A).P(B)$$

Solution

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