Probability Test

Result

Probability Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

If $$E_{1}$$ and $$E_{2}$$ are mutually exclusice events, then

A
$$P(E_{1})+P(E_{2}) \le 1$$

B
$$P(E_{1})+P(E_{2}) \ge 1$$

C
$$P(E_{1})+P(E_{2}) = 1$$

D
$$None\ of\ these$$

Solution
Question 2
1 / -0

If $$P(A) = 0.45, P(B) = 0.20, P (A \cap B) = 0.15$$, then $$P(\bar{A} \cap \bar{B})$$ =

A
0.38

B
0.39

C
0.40

D
0.50

Solution
Question 3
1 / -0

Let $$P(E)$$ denotes probability of an event $$E$$, if $$P(B)=\dfrac{3}{4},P(A \cap B \cap \bar {C})=\dfrac{1}{3}, p(\bar {A} \cap B \cap \bar {C})=\dfrac{1}{3}$$, then $$p(B \cap C)=$$

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

B
$$\dfrac{5}{12}$$

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

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

Solution
Question 4
1 / -0

If P (E) =0.83 then P $$\left( \overline { E } \right) =----$$

A
0.17

B
0.38

C
0.27

D
1
Question 5
1 / -0

A college offers 6 courses in the morning and 4 in the evening. The possible numbers of choices with the student if he wants to study one course in the morning and one in the evening is

A
24

B
2

C
12

D
10
Question 6
1 / -0

If A and B are even such that $$P(A \cup B)=\frac{3}{4}, P(A\cap B)=\frac{1}{4},P(\bar{A})=\frac{2}{3}, then P(\bar{A} \cap B)$$ is:

A
$$\frac{5}{12}$$

B
$$\frac{3}{8}$$

C
$$\frac{5}{8}$$

D
$$\frac{1}{4}$$

Solution
Question 7
1 / -0

If the probability of X to fail in the examination is 0.3 and that for Y is 0.2, then the probability that either X or Y fails in the examination is :

A
0.55

B
0.44

C
0.66

D
0.45

Solution
Question 8
1 / -0

If $$P\left( A \right) = 0.8,P\left( B \right) = 0.5,$$ then $$P\left( {A \cap B} \right)$$ lies in interval.

A
$$\left[ {0.2,\,0.5} \right]$$

B
$$\left[ {0.2,\,0.3} \right]$$

C
$$\left[ {0.3,\,0.5} \right]$$

D
$$\left[ {0.1,\,0.5} \right]$$

Solution
Question 9
1 / -0

Let A, B, C be three events such that :
$$P\left(A\right) = 0.3, P\left(B\right) = 0.4, P\left(C\right) = 0.5,P\left(A\cap { B }^{ ' }\right)=0.2,P\left(B\cap C\right)=0.3,P\left({ A }^{ ' }\cap { B }^{ ' }\cap { C }^{ ' }\right)=0.3$$ and $$P\left({ A\cap B }|{ C }^{ ' }\right)=0.1$$. Find the value of $$P\left({ B }^{ ' }|{ C }^{ ' }\right)$$

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

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

C
$$\dfrac { 3 }{ 5 } $$

D
$$\dfrac { 4 }{ 5 } $$

Solution

$$ (\frac{B^{1}}{C^{1}}) = \frac{P(B^{1}\cap C^{1})}{P(C^{1})} = \frac{1-0.3}{1-0.5} 1- P\frac{B\cap C}{P(C)}$$

$$ P(B^{1}/C^{1}) = 1 - 0.3/0.5 = \frac{0.2}{0.5} = \frac{2}{5}$$
Question 10
1 / -0

If $$A$$ and $$B$$ are two events such that $$P ( A ) = 1 / 2$$ and $$P ( B ) = 2 / 3 ,$$ then $$P \left( A ^ { \prime } \cap B \right)$$ may be

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

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

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

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