We have,
$\text{P(A)}\neq0$ and $\text{P(B)}\neq1$
Now,
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
Hence, the correct alternative is option (C).
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?
A and B are two events such that P(A) = 0.25 and P(B) = 0.50. The probability pf both happening together is 0.14. The probability of both A and B hot happening is.
Choose the correct answer from the given four options.
If $\text{P}(\text{A})=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$ is equal to:
A bag $A$ contains $4$ green and $6$ red balls. Another bag $B$ contains $3$ green and $4$ red balls. If one ball is drawn from each bag, find the probability that both are green:
Mark the correct alternative in the following question:The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is:
In each of the following choose the correct answer:$\text{If}\ \text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=0,\ \text{then}\ \text{P}(\text{A}|\text{B})\ \text{is}:$