Question

A bag contains 2 black balls, 4 yellow balls and 4 white balls. Event A is defined as drawing a white ball on the first draw and event B is defined as drawing a black ball on the second draw. If two balls are drawn from the bag, one after the other and not replaced, what is P(B|A) expressed in simplest form? A. B. C. D.

Answer 1

Answer: $P(B|A)=\dfrac{4}{9}$

Step-by-step explanation:

Given: A bag contains 2 black balls, 4 yellow balls and 4 white balls.

Event A is defined as drawing a white ball on the first draw.

Event B is defined as drawing a black ball on the second draw.

P(B|A) is expressed as the conditional probability of occurring B given that A.

i.e. it is the probability of happening B where A is already happended.

If a white ball is drawn at first, then the number of black ball(4) remains unchanged but the total number of balls (10) will become 9.

$Probability=\dfrac{Number\ of \ favorable \ outcomes}{Total\ outcomes}$

Then, $P(B|A)=\dfrac{4}{9}$