Suppose A and B are two events such that P(A) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that for any two events $A$ and $B$,the probability of their union is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Given $P(A) = 0.9

Vedclass · Accepted Answer

Always true

Vedclass · Answer

(A) We know that for any two events $A$ and $B$,the probability of their union is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Given $P(A) = 0.9$ and $P(B) = 0.8$,we have $P(A \cup B) = 1.7 - P(A \cap B)$.Since $P(A \cap B) \geq 0.7$,the maximum value of $P(A \cap B)$ is $0.9$ (because $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$).Also,$P(A \cup B) \leq 1$ must hold for any probability space.Substituting the minimum value of $P(A \cap B) = 0.7$,we get $P(A \cup B) = 1.7 - 0.7 = 1.0$.Since $P(A \cup B)$ can be at most $1$,and the condition $P(A \cap B) \geq 0.7$ allows for valid probability values,this case is always true.

Suppose $A$ and $B$ are two events such that $P(A) = 0.9$,$P(B) = 0.8$,and $P(A \cap B) \geq 0.7$. Then,we can conclude that such a case is . . . . . .

Similar Questions

$A$ bag contains $5$ distinct Red,$4$ distinct Green,and $3$ distinct Black balls. If balls are drawn one by one without replacement,what is the probability of getting a particular red ball in the fourth draw?

The probability of a student qualifying for $IITJEE$ and $EAMCET$ are $\frac{1}{5}$ and $\frac{3}{5}$ respectively. The probability that the student qualifies for at least one of these tests is:

Two dice are rolled one after the other. The probability that the number on the first die is smaller than the number on the second die is:

Suppose $A$ and $B$ are two events such that $P(A \cap B) = \frac{3}{25}$ and $P(B - A) = \frac{8}{25}$. Then, $P(B)$ is equal to

Describe the sample space for the indicated experiment: $A$ die is thrown two times.

Vedclass Test Series

Exam Paper Generator

Online Exam Module