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)
![$=0.7 \times \frac{4}{5} + 0.3 \times \frac{1}{4}$](https://tex.z-dn.net/?f=%24%3D0.7%20%5Ctimes%20%20%5Cfrac%7B4%7D%7B5%7D%20%2B%200.3%20%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D%24)
= 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) ![$=\frac{1}{5}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B5%7D%24)
= 0.2
c). P (Tails, if marble was red)
![$=P (T/R) = \frac{P(R/T)}{P(R)} \ P(T)$](https://tex.z-dn.net/?f=%24%3DP%20%28T%2FR%29%20%3D%20%5Cfrac%7BP%28R%2FT%29%7D%7BP%28R%29%7D%20%20%5C%20P%28T%29%24)
Where P (R/T) = P ( red, if coin showed tails)
![$=\frac{1}{4}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B4%7D%24)
= 0.25 (As Urn T is chosen)
P (R) = P (Red) = 0.635 (from part (a) )
P (T) = P (Tails) = 0.3
∴ ![$P(T/R) = \frac{0.25 \times 0.3}{0.635}$](https://tex.z-dn.net/?f=%24P%28T%2FR%29%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%200.3%7D%7B0.635%7D%24)
= 0.118