Events $A$ and $B$ are such that $P(A)=\frac{1}{2}$,$P(B)=\frac{7}{12}$ and $P(\text{not } A \text{ or not } B)=\frac{1}{4}$. State whether $A$ and $B$ are independent?
(B) It is given that $P(A)=\frac{1}{2}$,$P(B)=\frac{7}{12}$ and $P(A' \cup B')=\frac{1}{4}$.
Using De Morgan's Law,$A' \cup B' = (A \cap B)'$.
So,$P((A \cap B)') = \frac{1}{4}$.
Since $P(E') = 1 - P(E)$,we have $1 - P(A \cap B) = \frac{1}{4}$.
Therefore,$P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$.
Since $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$,we observe that $P(A \cap B) \neq P(A) \times P(B)$.
Thus,events $A$ and $B$ are not independent.
Using De Morgan's Law,$A' \cup B' = (A \cap B)'$.
So,$P((A \cap B)') = \frac{1}{4}$.
Since $P(E') = 1 - P(E)$,we have $1 - P(A \cap B) = \frac{1}{4}$.
Therefore,$P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$.
Now,calculate $P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$.
Since $P(A \cap B) = \frac{3}{4}$ and $P(A) \times P(B) = \frac{7}{24}$,we observe that $P(A \cap B) \neq P(A) \times P(B)$.
Thus,events $A$ and $B$ are not independent.
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Similar Questions
Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:
List-$I$ List-$II$ $(A)$ $P(E_2)$ $(i)$ $1/4$ $(B)$ $P(E_1 \cup E_2)$ $(ii)$ $5/8$ $(C)$ $P(\bar{E}_1 / \bar{E}_2)$ $(iii)$ $1/8$ $(D)$ $P(E_1 / \bar{E}_2)$ $(iv)$ $1/2$ $(v)$ $3/8$ $(vi)$ $3/4$
| List-$I$ | List-$II$ |
| $(A)$ $P(E_2)$ | $(i)$ $1/4$ |
| $(B)$ $P(E_1 \cup E_2)$ | $(ii)$ $5/8$ |
| $(C)$ $P(\bar{E}_1 / \bar{E}_2)$ | $(iii)$ $1/8$ |
| $(D)$ $P(E_1 / \bar{E}_2)$ | $(iv)$ $1/2$ |
| $(v)$ $3/8$ | |
| $(vi)$ $3/4$ |
If $A$ and $B$ are two events such that $P(A | B) = 0.6$, $P(B | A) = 0.3$, and $P(A) = 0.1$, then $P(\bar{A} \cap \bar{B})$ equals:
Two dice are thrown. What is the probability that the sum of the numbers appearing on the two dice is $11$,given that $5$ appears on the first die?
Easy
View SolutionIf $E_1$ and $E_2$ are two events of a sample space such that $P(E_1) = \frac{1}{4}$,$P(E_2 \mid E_1) = \frac{1}{2}$,and $P(E_1 \mid E_2) = \frac{1}{4}$,then $P(\bar{E}_1 \mid E_2) = $
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