Hey!

the question is of baye's theorem.

Let us say that "E" is the event that the chosen coin always shows tail.

now, let E1 , E2 & E3 are the coins which shows tail on both sides, biased coin (70%) and unbiased coin respectively.

So, p (E1)=p(E2)=p(E3)=1/3 (as the chosen coin can be any of the three, so the probability of each coin to be chosen is 1/3).

Now, p(E/E1)= 1, p(E/E2) = 70/100 and p(E/E3)= 1/2    -----(1)

By Baye's theorem

p (E1/E) =   [p(E/E1)p(E1)]  /  [ {p(E/E1)p(E1) } + {p(E/E2)p(E2)} + {p(E/E3)p(E3)} ]

by putting the values from equation (1)

and solving, we get,

p(E/E1) = 5/11

thankyou