a box contains four red balls and eight black balls. two balls are randomly chosen from the box, and are not…

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.

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.

Answer

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)

After choosing a black ball first (without replacement), there are 11 balls left, 4 of which are red. So (P(R|B)=\frac{4}{11}).

Step3: Calculate P(B ∩ R)

By the multiplication - rule (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: Calculate the percentage

To convert (\frac{8}{33}) to a percentage, we calculate (\frac{8}{33}\times100\approx 24.24%).

Answer:

(P(B)=\frac{2}{3}), (P(R|B)=\frac{4}{11}), (P(B\cap R)=\frac{8}{33}), about (24.24%)