A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not
replaced. Let event B be choosing a black ball first and event R be choosing a red ball second.
What are the following probabilities?
P(B) =
P(
RB) =
P(B n R) =
The probability that the first ball chosen is black and the second ball chosen is red is about percent.

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

Solution:

The probability is given as:

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

The probability that the first ball chosen is black and the second ball chosen is red is about percent

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

45

step-by-step explanation:

to find a percentage, add a decimal in front and multiply it by the other number.

.25 times 180= 45