Answer:
The answer is "[tex]\bold{\frac{2}{5}\ \ and \ \ \frac{6}{13}}[/tex]".
Step-by-step explanation:
You have 4/10 opportunities to choose a white ball, and there'll be 7 white balls and 6 black balls out of 13, and so the second time they choose a white one is 7/13, as well as the second time they choose a black, 6/13. people will also have a 4/10 chance.
There are 6/10 chances which users pick its black ball and 4 white balls would still be picked, but 9 black balls and out 13 balls and thus, its second and third time you select the white one is 4/13 but you are likely to pick a black for the second time is 9/13.
Taking the diagram of the next tree. The very first draw is marked with a and the second draw is marked with b.
[tex]\to P(a) = \frac{4}{10}\ \ \ \ \ \ \ \ \ P(b) = \frac{6}{10}\\\\\to P(\frac{a2}{a1}) = \frac{7}{13} \ \ \ \ \ \ \ \ \ \ P(\frac{a}{b}) = \frac{4}{13}\\\\\to P(\frac{b2}{a1}) = \frac{6}{13} \ \ \ \ \ \ \ \ \ \ P(\frac{b2}{b1}) = \frac{9}{13}[/tex]
Calculating the second drawn ball is white:
[tex]\to P(b2)=P(a)P(\frac{a2}{b1})+P(b)P(\frac{a}{b})\\[/tex]
[tex]=\frac{4}{10}\frac{7}{13}+\frac{6}{10}\frac{4}{13}\\\\=\frac{28}{130}+\frac{24}{130}\\\\=\frac{28+24}{130}\\\\=\frac{52}{130}\\\\=\frac{2}{5}\\\\[/tex]
In point b:
[tex]\to P(\frac{b}{a1})= \frac{P(B)P(\frac{a}{b})}{P(a)P(\frac{a2}{b1})+P(b)P(\frac{a}{b})\\}[/tex]
[tex]=\frac{\frac{6}{10} \frac{4}{13}}{\frac{52}{130}}\\\\=\frac{\frac{24}{130}}{\frac{52}{130}}\\\\=\frac{24}{130} \times \frac{130}{52}\\\\=\frac{24}{52}\\\\=\frac{6}{13}\\[/tex]
