Solution :
P(H) = 0.7 ; P(T) = 0.3
If heads, then Urn H, 1 blue and 4 red marbles.
If tails, then Urn T , 3 blue and 1 red marbles.
a).
P ( choosing a Red marble )
= P (H) x P( Red from Urn H) + P (T) x P( Red from Urn T)
[tex]$=0.7 \times \frac{4}{5} + 0.3 \times \frac{1}{4}$[/tex]
= 0.56 + 0.075
= 0.635
b). If P (B, if coin showed heads)
If heads, then marble is picked from Urn H.
Therefore,
P (Blue) [tex]$=\frac{1}{5}$[/tex]
= 0.2
c). P (Tails, if marble was red)
[tex]$=P (T/R) = \frac{P(R/T)}{P(R)} \ P(T)$[/tex]
Where P (R/T) = P ( red, if coin showed tails)
[tex]$=\frac{1}{4}$[/tex]
= 0.25 (As Urn T is chosen)
P (R) = P (Red) = 0.635 (from part (a) )
P (T) = P (Tails) = 0.3
∴ [tex]$P(T/R) = \frac{0.25 \times 0.3}{0.635}$[/tex]
= 0.118
