Probability Test

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Probability Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

If $$\dfrac{1+3p}{3},\dfrac{1-2p}{2} $$ are probabilities of two mutually exclusive event, then $$p$$ lies in the interval

A
$$\left [-\dfrac{1}{3},\dfrac{1}{2}\right]$$

B
$$\left (-\dfrac{1}{2},\dfrac{1}{2}\right)$$

C
$$\left [-\dfrac{1}{2},\dfrac{2}{3}\right]$$

D
$$\left (-\dfrac{1}{3},\dfrac{2}{3}\right)$$

Solution

$$Probabilities\quad are\quad \frac { 1+3p }{ 3 } \quad and\frac { 1-2p }{ 2 } \\ so,\quad 0\le \frac { 1+3p }{ 3 } \le 1\\ \Rightarrow 0\le 1+3p\le 3\\ \Rightarrow -1/3\le p\le 2/3\\ Alse,\quad 0\le \frac { 1-2p }{ 2 } \le 1\\ \Rightarrow 0\le 1-2p\le 2\\ \Rightarrow -1\le -2p\le 1\\ \Rightarrow -1\le 2p\le 1\\ \Rightarrow -1/2\le p\le 1/2\\ Intersection\quad range\quad of\quad p\quad for\quad both\quad probabilities\quad to\quad be\quad valid\quad =-1/3\le p\le 1/2$$
Question 2
1 / -0

If P(A) = P(B), then

A
A and B are the same events

B
A and B must be same events

C
A and B may be different events

D
A and B are mutually exclusive events.

Solution

Given $$P(A) = P(B)$$
Then $$A$$ can be different event from $$B$$ because $$A\;\xi\;B$$ are Mutually exclusive.
i.e, $$A\;\xi\;B$$ may be different events.
Hence, the answer is $$A$$ and $$B$$ may be different events.
Question 3
1 / -0

$$A$$ and $$B$$ are two events such that $$P(A\cup B) = \dfrac {3}{4}, P(A\cap B) = \dfrac {1}{4}, P(\overline {A}) = \dfrac {2}{3}$$; then $$P(\overline {A} \cap B)$$ is

A
$$1/12$$

B
$$5/12$$

C
$$4/9$$

D
$$1/3$$

Solution

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

$$=\dfrac{2}{3}-\dfrac{1}{4}=\dfrac{5}{12}$$
Question 4
1 / -0

If $$A$$ and $$B$$ are mutually exclusive events, then

A
$$P(A) = P(A - B)$$

B
$$P(B) = P(A - B)$$

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

D
$$P(B) = P(A \cap B)$$

Solution

Given $$A$$ and $$B$$ are mutually exclusive
$$\Rightarrow P(A\cup B)=P(A)+(B)$$
$$\Rightarrow P(A-B)=P(A)-P(B)$$
When $$P(B)=0$$ i.e, $$P(A_B)+P(A)$$
$$\Rightarrow P(B)=0$$ is not a sure event.
Hence, the answer is $$P(A)=P(A-B).$$
Question 5
1 / -0

Two events $$A$$ and $$B$$ are such that
$$P(A)=\cfrac { 1 }{ 4 } ,P(A|B)=\cfrac { 1 }{ 4 } $$ and $$P(B|A)=\cfrac { 1 }{ 2 } $$
Consider the following statements:
(I) $$P(\overline { A } |\overline { B } )=\cfrac { 3 }{ 4 } $$
(II) $$A$$ and $$B$$ are mutually exclusive
(III) $$P(A|B)+P(A|\overline { B } )=1$$
Then

A
Only (I) is correct

B
Only (I) and (II) are correct

C
Only (I) and (III) are correct

D
Only (II) and (III) are correct

Solution

$$P(\overline{A}|\overline{B})=1-\dfrac{1}{4}=\dfrac{3}{4}$$

for A to B mutually Exclusive , $$P(A|B)$$x$$P(B|A)=1$$ but here $$P(A|B)$$x$$P(B|A)=\dfrac18$$

$$(III)$$ is also incorrect because $$(I)$$ is the correct property that is

$$P(\overline{A}|\overline{B})+P(A|B)=1$$
Question 6
1 / -0

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

A
$$6/11$$

B
$$6/22$$

C
$$1/6$$

D
$$1/4$$

Solution

Given: A, B, C are mutually exclusive and exhaustive events, so this means
$$P(A\cup B\cup C)=P(S)=1$$
And ,$$P(A\cap B\cap C)=P(A\cap B)=P(B\cap C)=P(A\cap C)=0$$
We know,
$$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\\\implies 1=P(A)+P(B)+P(C)-0-0-0+0\\\implies P(A)+P(B)+P(C)=1$$
Give, $$P(A)=2P(B)=2P(C)$$
Hence, $$P(A)+P(B)+P(C)=1 $$ becomes
$$2P(B)+P(B)+P(B)=1\\\implies 4P(B)=1\\\implies P(B)=\dfrac 14$$
Question 7
1 / -0

If for two events $$A$$ and $$B, P(A\cap B)\ne P(A) \times P(B)$$, then the two events $$A$$ and $$B$$ are

A
Independent

B
Dependent

C
Not equally likely

D
Not exhaustive

Solution

For independent.events $$P\left( A\cap B \right) =P\left( A \right) .P\left( B \right) $$
So, $$P\left( A\cap B \right) \neq P\left( A \right) .P\left( B \right) $$ implies that A and B are independent.
Question 8
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: $$P(A)=\dfrac 35, P(B)=\dfrac 23$$, $$A$$ and $$B$$ are independent.
To find: $$P(A+B)=?$$
Sol: As $$A$$ and $$B$$ are independent, $$P(A\cap B)=P(A)P(B)$$
Now, $$P(A+B)=P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
$$\implies P(A+B)=P(A)+P(B)-P(A)P(B)\\\implies P(A+B)=\dfrac 35+\dfrac 23-\dfrac 35\times \dfrac 23\\\implies = \dfrac {10+9-6}{15}=\dfrac {13}{15}$$
Question 9
1 / -0

If $$A, B$$ and $$C$$ are mutually exclusive and exhaustive events, then $$P(A) + P(B) + P(C)$$ equals to -

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

B
$$1$$

C
$$0$$

D
Any value between $$0$$ and $$1$$

Solution

Given $$A,B,C$$ are mutually exclusive and exhaustive events
When $$A,B,C$$ are mutually exclusive events,
$$\Rightarrow P(A\cup B\cup C)=P(A)+P(B)+P(C)$$
When $$A,B,C$$ are exheustive event,
$$\Rightarrow P(A)+P(B)+P(C)=1$$
Hence, the answer is probability always $$1.$$
Question 10
1 / -0

Let $$A$$ and $$B$$ be subsets of $$X$$ and $$C=\left( A\cap B' \right) \cup \left( A'\cap B \right) $$ where $$A'$$ and $$B'$$ are complements of $$A$$ and $$B$$ respectively in $$X$$. What is $$C$$ equal to?

A
$$\left( A\cap B' \right) -\left( A\cap B' \right) $$

B
$$\left( A'\cup B \right) -\left( A'\cap B \right) $$

C
$$\left( A\cup B \right) -\left( A\cap B \right) $$

D
$$\left( A'\cup B' \right) -\left( A'\cap B' \right) $$

Solution

Given : $$C = (A\cap B') \cup (A'\cap B)$$.
Hence, $$C = (A\cup B) - (A\cap B)$$

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Probability Test - 47

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Probability Test - 47

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SHARING IS CARING

Question 1

$$\left [-\dfrac{1}{3},\dfrac{1}{2}\right]$$

$$\left (-\dfrac{1}{2},\dfrac{1}{2}\right)$$

$$\left [-\dfrac{1}{2},\dfrac{2}{3}\right]$$

$$\left (-\dfrac{1}{3},\dfrac{2}{3}\right)$$

Solution

Question 2

A and B are the same events

A and B must be same events

A and B may be different events

A and B are mutually exclusive events.

Solution

Question 3

$$1/12$$

$$5/12$$

$$4/9$$

$$1/3$$

Solution

Question 4

$$P(A) = P(A - B)$$

$$P(B) = P(A - B)$$

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

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

Solution

Question 5

Only (I) is correct

Only (I) and (II) are correct

Only (I) and (III) are correct

Only (II) and (III) are correct

Solution

Question 6

$$6/11$$

$$6/22$$

$$1/6$$

$$1/4$$

Solution

Question 7

Independent

Dependent

Not equally likely

Not exhaustive

Solution

Question 8

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

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

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

$$0.65$$

Solution

Question 9

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

$$1$$

$$0$$

Any value between $$0$$ and $$1$$

Solution

Question 10

$$\left( A\cap B' \right) -\left( A\cap B' \right) $$

$$\left( A'\cup B \right) -\left( A'\cap B \right) $$

$$\left( A\cup B \right) -\left( A\cap B \right) $$

$$\left( A'\cup B' \right) -\left( A'\cap B' \right) $$

Solution

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