Question

Three balls are randomly drawn from an urn that contains four white and seven red balls. (a) What is the probability of drawing a red ball on the third draw? (Round your answer to 3 decimal places.) (b) What is the probability of drawing a red ball on the third draw given that at least one red ball was drawn on the first two draws? (Round your answer to 3 decimal places.)

Answer 1

P(red ball on third draw)=P(no red balls in first two draw and red on 3rd draw)+P(1 red balls in first two draw and red on 3rd draw)+P(2 red balls in first two draw and red on 3rd draw)

= $\frac{\binom{4}{2}\binom{7}{0}}{\binom{11}{2}}*(7/9)+\frac{\binom{4}{1}\binom{7}{1}}{\binom{11}{2}}*(6/9)+\frac{\binom{4}{0}\binom{7}{2}}{\binom{11}{2}}*(5/9)$

=(6/55)*(7/9)+(28/55)*(6/9)+(21/55)*(5/9)=7/11

2)

P(at least one red ball on first two draws)= $\frac{\binom{4}{1}\binom{7}{1}}{\binom{11}{2}}+\frac{\binom{4}{0}\binom{7}{2}}{\binom{11}{2}}$ =49/55

hence P(drawing a red ball on the third draw and at least one red ball on first two draw)

=P(1 red balls in first two draw and red on 3rd draw)+P(2 red balls in first two draw and red on 3rd draw)

= $\frac{\binom{4}{1}\binom{7}{1}}{\binom{11}{2}}*(6/9)+\frac{\binom{4}{0}\binom{7}{2}}{\binom{11}{2}}*(5/9)$ =(28/55)*(6/9)+(21/55)*(5/9)=91/165

hence P(rawing a red ball on the third draw given at least one red ball on first two draw)

=(91/165)/(49/55)=91/147=13/21