Question

applying the general multiplication rule
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(r|b)
p(b
the pr y that the first ball chosen is black and the second ball chosen is red is about percent.

Explanation:

Step1: Calculate \(P(B)\)

Total number of balls \(=4 + 8=12\).
Number of black balls \(=8\).
By the probability formula \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\), we have \(P(B)=\frac{8}{12}=\frac{2}{3}\).

Step2: Calculate \(P(R|B)\)

After choosing a black ball first (without replacement), the number of balls left \(=12 - 1 = 11\).
Number of red balls \(=4\).
So \(P(R|B)=\frac{4}{11}\).

Step3: Calculate \(P(B\cap R)\)

By the general multiplication rule \(P(B\cap R)=P(B)\times P(R|B)\).
Substitute \(P(B)=\frac{8}{12}\) and \(P(R|B)=\frac{4}{11}\) into the formula:
\(P(B\cap R)=\frac{8}{12}\times\frac{4}{11}=\frac{8\times4}{12\times11}=\frac{32}{132}=\frac{8}{33}\approx0.2424\).
To convert to a percentage, multiply by \(100\): \(0.2424\times100 = 24.24\%\approx24\%\).

Answer:

\(P(B)=\frac{2}{3}\), \(P(R|B)=\frac{4}{11}\), \(P(B\cap R)=\frac{8}{33}\), and the probability that the first ball chosen is black and the second ball chosen is red is about \(24\) percent.