3 P(A) = P(B) = frac{5}{13} and P(A mid B) = | vedclass

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

Question

(B) Given: $3 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{39}$.$P(B) = \frac{5}{13} = \frac{15}{39}$.$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{

Vedclass · Accepted Answer

$\frac{11}{39}$

Vedclass · Answer

(B) Given: $3 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{39}$.$P(B) = \frac{5}{13} = \frac{15}{39}$.$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{5}$.Therefore,$P(A \cap B) = P(A \mid B) 	imes P(B) = \frac{3}{5} 	imes \frac{5}{13} = \frac{3}{13} = \frac{9}{39}$.We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = \frac{5}{39} + \frac{15}{39} - \frac{9}{39} = \frac{11}{39}$.

$3 P(A) = P(B) = \frac{5}{13}$ and $P(A \mid B) = \frac{3}{5}$,then $P(A \cup B) = $ . . . . . . .

Similar Questions

The general solution of the differential equation $x^2 dy - (xy - y^2) dx = 0$ is

If the solution of the differential equation $\frac{dy}{dx} = \frac{2x+3y}{3x-2y}$ is $y = x \tan(f(x)) + c$,then $f(x) =$

The solution of $x \frac{dy}{dx} = y + x e^{y/x}$ with $y(1) = 0$ is

The solution of the differential equation $\frac{dy}{dx} = -\left(\frac{x^2+3y^2}{3x^2+y^2}\right)$,$y(1)=0$ is

Which of the following is not a homogeneous function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module