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

Explanation:

Step1: Calculate \(P(B)\)

Total balls \(=4 + 8=12\).
\(P(B)=\frac{\text{Number of black balls}}{\text{Total number of balls}}=\frac{8}{12}=\frac{2}{3}\)

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

After choosing a black ball first, number of black balls left \(=8 - 1 = 7\), total balls left \(=12- 1=11\), number of red balls \(=4\).
\(P(R|B)=\frac{\text{Number of red balls}}{\text{Total number of balls left}}=\frac{4}{11}\)

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

By the formula \(P(B\cap R)=P(B)\times P(R|B)\)
\(P(B\cap R)=\frac{8}{12}\times\frac{4}{11}=\frac{8\times4}{12\times11}=\frac{32}{132}=\frac{8}{33}\)

Step4: Calculate the percentage for \(P(B\cap R)\)

\(P(B\cap R)=\frac{8}{33}\approx0.2424\)
To convert to percentage: \(0.2424\times100 = 24.24\%\approx24\%\)

Answer:

\(P(B)=\frac{2}{3}\), \(P(R|B)=\frac{4}{11}\), \(P(B\cap R)=\frac{8}{33}\), The probability is about \(24\) percent.