**CONCEPT: **

[We are using the conditional probability, Bayes' theorem of P(B|A)= P(B) P(A|B) / P(A)

Where A has already occurred(we're getting 6 heads in a row) & B is the event we are interested in to calculate(coin is a 2 headed coin) when A has already occurred.

P(A/B)=Probability of getting 6 heads in a row, if coin is 2-headed= 1

P(B)=Probability that the coin is two- headed =1/65

P(B/A)=Probability of B, if A has already occurred

P(A)=Probability of getting 6 heads in a row=64/65)*(1/64)}+{(1/65)*1})

**SOLUTION:**

The Probability of selecting a fair coin= 64/65

The Probability of getting 6 heads in a row with the fair coin= (1/2)^6= 1/64

The Probability of selecting the unfair 2-headed coin= 1/65

Probability of getting 6 heads in a row with the 2-headed coin= 1

If a coin, chosen at random from the bag and then tossed, turns up heads 6 times in a row, the probability that it is the two-headed coin(P (B/A))

= [(1/65)*1] / [{(64/65)*(1/64)}+{(1/65)*1}]

= (1/65)/(2/65)

= 1/2

Hence **OPTION (E)**