Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.5$ | $0.35$ | ......... | $0.7$ |
Here, $P ( A )=0.5$, $P ( B )=0.35$, $P (A \cup B)=0.7$
We know that $P (A \cup B)= P ( A )+ P ( B )- P (A \cap B)$
$\therefore 0.7=0.5+0.35- P (A \cap B)$
$\Rightarrow P (A \cap B)=0.5+0.35-0.7$
$\Rightarrow P (A \cap B)=0.15$
A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
A die is tossed thrice. Find the probability of getting an odd number at least once.
Given $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$. Find $P(A $ or $B),$ if $A$ and $B$ are mutually exclusive events.
The probability that a student will pass the final examination in both English and Hindi is $0.5$ and the probability of passing neither is $0.1$. If the probability of passing the English examination is $0.75$, what is the probability of passing the Hindi examination?