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.

Answer 1

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent.

Answer 2

black ball   and    red ball

8 out of 12   x     4 out of 11

      $\frac{8}{12}$            x             $\frac{4}{11}$

= $\frac{8(4)}{12(11)} = \frac{8}{3(11)} = \frac{8}{33}$ = 0.24  = 24%

Answer: 24%