joseph1506
contestada

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.

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tramserran

black ball   and    red ball

8 out of 12   x     4 out of 11

      [tex]\frac{8}{12}[/tex]            x             [tex]\frac{4}{11}[/tex]

= [tex]\frac{8(4)}{12(11)} = \frac{8}{3(11)} = \frac{8}{33}[/tex] = 0.24  = 24%

Answer: 24%

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.

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