Probability Tes...

For any two independent events $${ E }_{ 1 }$$ and $${ E }_{ 2 }$$, $$P\left\{ \left( { E }_{ 1 }\cup { E }_{ 2 } \right) \cap \left( \overline { { E }_{ 1 } }  \right) \cap \left( \overline { { E }_{ 2 } }  \right)  \right\} $$ is

Let $$A$$ and $$B$$ be two events such that $$P(\displaystyle \overline{A\cup B})=\frac{1}{6}, P(A\displaystyle \cap B)=\frac{1}{4}$$ and $$P(\displaystyle \overline{A})=\frac{1}{4}$$, then events $$A$$ and $$B$$ are

If $$A$$ and $$B$$ are two events such that $$P\left ( A\cup B \right )= 0.65$$ and $$P\left ( A\cap B \right )= 0.15$$, then $$P\left ( \bar{A} \right )+P\left ( \bar{B} \right )=$$

An urn contains $$6$$ white and $$4$$ black balls. A fair die whose faces are numbered from $$1$$ to $$6$$ is rolled and number of balls equal to that of the number appearing on the die is drawn from the urn at random. The probability that all those are white is

Let $$A,B,C$$ be three events such that $$P\left( A \right) =0.3,P\left( B \right) =0.4,P\left( C \right) =0.8,P\left( A\cap B \right) =0.08,$$
$$P\left( A\cap C \right) =0.28,P\left( A\cap B\cap C \right)  =0.09$$. If $$P\left( A\cup B\cup C \right) \ge 0.75$$, then

A coin is tossed until a head appears or it has been tossed 3 times. Given that head does not appear on the first toss, the probability that the coin is tossed 3 times is

A and B be two independent events of a sample space such that $$P(A) = 0.2, P(B) = 0.5$$
LIST-I                  LIST-II
A) $$P(B/A)$$             1$$. 0.2$$
B) $$P(A/B)$$            2 $$. 0.1$$
C) $$P(A\cap B)$$          3$$  . 0.3$$
D) $$P(A\cup B)$$         4 $$. 0.6$$
                                5$$. 0.5$$
The correct match for List-I from List-II
        A B C D

A letter is taken out at random from the word ASSISTANT and an other from STATISTICS. The probability that they are the same letters is

If $${ A }_{ 1 },{ A }_{ 2 },{ A }_{ 3 },...,{ A }_{ n }$$ are any $$n$$ events, then

$$A,B,C$$ are any three events. If $$P\left( S \right) $$ denotes the probability of $$S$$ happening, then $$P\left( A\cap \left( B\cup C \right)  \right) =$$