Fill in the blanks in following table :
$P(A)$
$P(B)$
$P(A \cap B)$
$P (A \cup B)$
$\frac {1}{3}$
$\frac {1}{5}$
$\frac {1}{15}$
........
Fill in the blanks in following table :
| $P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
| $\frac {1}{3}$ | $\frac {1}{5}$ | $\frac {1}{15}$ | ........ |
$P ( A )=\frac{1}{3}$, $P ( B )=\frac{1}{5}$, $P ( A \cap B )=\frac{1}{15}$
Here,
We know that $P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )$
$\therefore P(A \cup B)$ $=\frac{1}{3}+\frac{1}{5}+\frac{1}{15}$ $=\frac{5+3-1}{15}$ $=\frac{7}{15}$
Similar Questions
Two events $A$ and $B$ will be independent, if
Two events $A$ and $B$ will be independent, if
If $P(A) = 2/3$, $P(B) = 1/2$ and ${\rm{ }}P(A \cup B) = 5/6$ then events $A$ and $B$ are
If $P(A) = 2/3$, $P(B) = 1/2$ and ${\rm{ }}P(A \cup B) = 5/6$ then events $A$ and $B$ are
In a certain population $10\%$ of the people are rich, $5\%$ are famous and $3\%$ are rich and famous. The probability that a person picked at random from the population is either famous or rich but not both, is equal to
In a certain population $10\%$ of the people are rich, $5\%$ are famous and $3\%$ are rich and famous. The probability that a person picked at random from the population is either famous or rich but not both, is equal to
One bag contains $5$ white and $4$ black balls. Another bag contains $7$ white and $9$ black balls. A ball is transferred from the first bag to the second and then a ball is drawn from second. The probability that the ball is white, is
One bag contains $5$ white and $4$ black balls. Another bag contains $7$ white and $9$ black balls. A ball is transferred from the first bag to the second and then a ball is drawn from second. The probability that the ball is white, is
Given two independent events $A$ and $B$ such that $P(A) $ $=0.3, \,P(B)=0.6$ Find $P(A$ and $B)$.
Given two independent events $A$ and $B$ such that $P(A) $ $=0.3, \,P(B)=0.6$ Find $P(A$ and $B)$.