The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is
The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is
- A
$1$
- B
Wrong
- C
Could be either
- D
Do not know
Similar Questions
Fill in the blanks in following table :
$P(A)$
$P(B)$
$P(A \cap B)$
$P (A \cup B)$
$0.5$
$0.35$
.........
$0.7$
Fill in the blanks in following table :
| $P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
| $0.5$ | $0.35$ | ......... | $0.7$ |
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to
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.
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.
If $A, B, C$ are three events associated with a random experiment, prove that
$P ( A \cup B \cup C ) $ $= P ( A )+ P ( B )+ P ( C )- $ $P ( A \cap B )- P ( A \cap C ) $ $- P ( B \cap C )+ $ $P ( A \cap B \cap C )$
If $A, B, C$ are three events associated with a random experiment, prove that
$P ( A \cup B \cup C ) $ $= P ( A )+ P ( B )+ P ( C )- $ $P ( A \cap B )- P ( A \cap C ) $ $- P ( B \cap C )+ $ $P ( A \cap B \cap C )$
Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$
Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$