# For two events and , the probability that occurs is 0.8, the probability that occurs is 0.4, and the probability that both occurs is 0.2. Given that occurred, what is the probability that also occurred?

P(B|A)=0.25  , P(A|B) =0.5

Step-by-step explanation:

The question provides the following data:

P(A)= 0.8

P(B)= 0.4

P(A∩B) = 0.2

Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.

To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:

P(B|A) = P(A∩B)/P(A)

= (0.2) / (0.8)

P(B|A)=0.25

To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:

P(A|B) = P(A∩B)/P(B)

= (0.2)/(0.4)

P(A|B) =0.5

## Related Questions

The sum of my digits is 11 when rounded to the nearest hundred i am 500 rounding to the nearest ten makes me 530 what number am I?

The mystery number is 533. Rounds to 500. Rounds to 530 and 5+3+3=11

Which eqation is equivalent to 3x + 4y = 15 A) y=15-3x
-------
4

B) y=3x-15
------
4

C) y=15-3x

D) y=3x-15

Step-by-step explanation:

Here we are given an equation in x and y and we are required to solve it for y

we have

subtracting 3x from each sides we get

now we divide both sides by 4 we get

Hence A) is our right answer.

A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top. if water is poured into the cup at a rate of 2cm3/s, how fast is the water level rising when the water is 5 cm deep?

Using implicit differentiation, it is found that the water level is rising at a rate of 0.0764 centimetres per second.

The volume of a cone of radius r and height h is given by:

Applying implicit differentiation, the rate of change is given by:

In this problem:

• The radius is constant, thus .
• Height of 5 cm and radius of 3 cm, thus .
• Water poured at a rate of 2 cm³/s, thus

Then

The water level is rising at a rate of 0.0764 centimetres per second.

A similar problem is given at brainly.com/question/13461339

Let
h: height of the water
r: radius of the circular top of the water
V: the volume of water in the cup.
We have:
r/h = 3/10
So,
r = (3/10)*h
the volume of a cone is:
V = (1/3)*π*r^2*h
Rewriting:
V (t) = (1/3)*π*((3/10)*h(t))^2*h(t)
V (t) =(3π/100)*h(t)^3
Using implicit differentiation:
V'(t) = (9π/100)*h(t)^2*h'(t)
Clearing h'(t)
h'(t)=V'(t)/((9π/100)*h(t)^2)
the rate of change of volume is V'(t) = 2 cm3/s when h(t) = 5 cm.
substituting:
h'(t) = 8/(9π) cm/s