In the given problem we have 3 black ball and 5 white ball, the probability tree be constructed as per ball picks from bag.
(a) Refer the attached figure for the probability tree.
(b)
(i) The probability that Paul picks two black balls is [tex]\dfrac{9}{64}[/tex].
(ii) The probability that Paul picks a black ball in his second draw is [tex]\dfrac{3}{8}[/tex].
Given:
The bag contain 3 black and 5 white balls.
(a)
Refer the attached figure for the construction of probability tree.
(b)
(i)
Since the getting two black ball are independent event so multiply the branch B ( refer attached figure)
[tex]P(\rm two\: black)=\dfrac{3}{8}\times\dfrac {3}{8}\\P(\rm two\: black)=\dfrac{9}{24}[/tex]
Thus, the probability that Paul picks two black balls is [tex]\dfrac{9}{64}[/tex].
(ii)
There should be two outcomes either (B,B) or (W, B).
From the attached figure,
[tex]P(\rm B,B)=\dfrac{9}{64}[/tex]
[tex]P(\rm W, B)=\dfrac{15}{64}[/tex]
Calculate the probability of second ball black.
[tex]P(\rm second\: ball\: black)=P(B, B) + P(W, B)\\P(\rm second\: ball\: black)=\dfrac{9}{64}+\dfrac{15}{64}\\P(\rm second\: ball\: black)=\dfrac{3}{8}[/tex]
Thus, the probability that Paul picks a black ball in his second draw is [tex]\dfrac{3}{8}[/tex].
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