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 \(P(B)\)

Total number of balls \(=4 + 8=12\).
Probability of choosing a black ball first, \(P(B)=\frac{8}{12}=\frac{2}{3}\).

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

After choosing a black ball first, number of balls left \(=11\), number of red balls \( = 4\).
Probability of choosing a red ball given a black ball was chosen first, \(P(R|B)=\frac{4}{11}\).

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

Using 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 \(P(B\cap R)=\frac{2}{3}\times\frac{4}{11}=\frac{8}{33}\approx0.2424\).
To convert to percentage, multiply by \(100\), \(0.2424\times100 = 24.24\%\).

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.