Question

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(bcap r) = ) the probability that the first ball chosen is black and the second ball chosen is red is about percent.

Explanation:

Step1: Calculate \(P(B)\)

The total number of balls is \(4 + 8=12\). The number of black balls is \(8\).
Using 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 total number of balls left is \(12 - 1 = 11\), and the number of red balls is still \(4\).
So \(P(R|B)=\frac{4}{11}\).

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

By the formula \(P(B\cap R)=P(B)\times P(R|B)\).
Substitute \(P(B)=\frac{2}{3}\) and \(P(R|B)=\frac{4}{11}\) into the formula: \(P(B\cap R)=\frac{2}{3}\times\frac{4}{11}=\frac{8}{33}\approx0.2424\).
To convert to a percentage, multiply by \(100\): \(0.2424\times100 = 24.24\%\approx24\%\).

Answer:

\(P(B)=\frac{2}{3}\), \(P(R|B)=\frac{4}{11}\), \(P(B\cap R)\approx24\%\)