Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.35$ | ........... | $0.25$ | $0.6$ |
Here, $P(A)=0.35$, $P(A \cap B)=0.25$, $P(A \cup B)=0.6$
We know that $P (A \cup B)= P ( B )+ P ( B )- P (A \cap B)$
$\therefore $ $0.6=0.35+ P ( B )-0.25$
$\Rightarrow $ $P ( B )=0.6-0.35+0.25$
$\Rightarrow $ $P ( B )=0.5$
If the odds in favour of an event be $3 : 5$, then the probability of non-occurrence of the event is
$A, B, C$ are any three events. If $P (S)$ denotes the probability of $S$ happening then $P\,(A \cap (B \cup C)) = $
The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $
If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P\,(F) < 1,$ then
Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that both balls are red.