You flip a coin that is not fair, the prbability of heads on each flip is 0.7. if the coin shows heads, you draw a marble from urn h with 1 blue and 4 red marbles. if the coin shows tails, you draw a marble from urn t with 3 blue and 1 red marble. Find the following probabilities: a. The probability of choosing a red marble.
b. The probability of choosing a blue marble, given that the coin showed heads.
c. The probability that the coin showed tails, given that the marble was red.

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)

= 0.56 + 0.075

= 0.635

b).  If P (B, if coin showed heads)

If heads, then marble is picked from Urn H.

Therefore,

P (Blue)

             = 0.2

c). P (Tails, if marble was red)

Where P (R/T) = P ( red, if coin showed tails)

                        = 0.25 (As Urn T is chosen)

P (R) =  P (Red) = 0.635 (from part (a) )

P (T) = P (Tails) = 0.3

∴

                = 0.118

what are the options? ?

step-by-step explanation: