Question

) A jar contains 4 white balls and 6 black balls. A ball is chosen at random, and its color noted. The ball is then replaced, along with 3 more balls of the same color. Then, another ball is drawn at random from the jar. (a) Find the chance that the second ball drawn is white. (b) Given that the second ball drawn is white, what is the probability that the first ball drawn is black

Answer 1

Answer:

The answer is "$\bold{\frac{2}{5}\ \ and \ \ \frac{6}{13}}$".

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.

$\to P(a) = \frac{4}{10}\ \ \ \ \ \ \ \ \ P(b) = \frac{6}{10}\\\\\to P(\frac{a2}{a1}) = \frac{7}{13} \ \ \ \ \ \ \ \ \ \ P(\frac{a}{b}) = \frac{4}{13}\\\\\to P(\frac{b2}{a1}) = \frac{6}{13} \ \ \ \ \ \ \ \ \ \ P(\frac{b2}{b1}) = \frac{9}{13}$

Calculating the second drawn ball is white:

$\to P(b2)=P(a)P(\frac{a2}{b1})+P(b)P(\frac{a}{b})$

              $=\frac{4}{10}\frac{7}{13}+\frac{6}{10}\frac{4}{13}\\\\=\frac{28}{130}+\frac{24}{130}\\\\=\frac{28+24}{130}\\\\=\frac{52}{130}\\\\=\frac{2}{5}$

In point b:

$\to P(\frac{b}{a1})= \frac{P(B)P(\frac{a}{b})}{P(a)P(\frac{a2}{b1})+P(b)P(\frac{a}{b})\\}$

              $=\frac{\frac{6}{10} \frac{4}{13}}{\frac{52}{130}}\\\\=\frac{\frac{24}{130}}{\frac{52}{130}}\\\\=\frac{24}{130} \times \frac{130}{52}\\\\=\frac{24}{52}\\\\=\frac{6}{13}$