S2 binomial distribution question

A six-sided die is biased. When the die is thrown the number 5 is twice as likely to appear as any other number. All the other faces are equally likely to appear. The die is thrown repeatedly.
Find the probability that (b) (i) the first 5 will occur on the sixth throw,

(ii) in the first eight throws there will be exactly three 5s.

I do not know how to do part b)i)

It is just a Binomial distribution question. The number of 5's in 8 throws will have the Binomial distribution B(8,2/7) so probability of 3 5's will; be $^8C_3 \times \left(\frac{2}{7}\right)^3 \times \left(\frac{5}{7} \right)^5$

It is just a Binomial distribution question. The number of 5's in 8 throws will have the Binomial distribution B(8,2/7) so probability of 3 5's will; be $^8C_3 \times \left(\frac{2}{7}\right)^3 \times \left(\frac{5}{7} \right)^5$

why is it 2/7 and not 2/6?