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(b ∩ 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 probability of choosing a black ball first, \(P(B)=\frac{8}{12}=\frac{2}{3}\).

Step2: Calculate P(R|B)

If a black ball is chosen first, then there are \(12 - 1 = 11\) balls left, and 4 red balls. So \(P(R|B)=\frac{4}{11}\).

Step3: Calculate P(B ∩ R)

By the formula \(P(B\cap R)=P(B)\times P(R|B)\). Substitute \(P(B)=\frac{2}{3}\) and \(P(R|B)=\frac{4}{11}\), we get \(P(B\cap R)=\frac{2}{3}\times\frac{4}{11}=\frac{8}{33}\).

Step4: Convert P(B ∩ R) to percentage

\(\frac{8}{33}\approx 0.2424\), and \(0.2424\times100 = 24.24\%\)

Answer:

\(P(B)=\frac{2}{3}\)
\(P(R|B)=\frac{4}{11}\)
\(P(B\cap R)=\frac{8}{33}\)
The probability that the first ball chosen is black and the second ball chosen is red is about \(24.24\) percent.