S2 EXAM QUESTION HELP

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

a) find the first 5 will occur on the sixth throw.
b)in the first eight throws there will be exactly three 5s
(8 marks)

I know the probably of getting a 5 is 2/7.. but stuck from there, please help I'm sure I'm missing something obvious

probability of getting 5 = 2/7
probability of getting x (x ≠ 5) = 1/7
probability of getting any number but 5 = (5/7)

a) First 5 will occur on the sixth throw.

First five throws are numbers other than 5
Sixth throw is a 5

p = (5/7)^5 * 2/7 = 6,250/117,649 = 0.0531 = 5.31%

b) In first eight throws there will be exactly three 5s (and five numbers other than 5)

p = (8 C 3) * (2/7)^3 * (5/7)^5 = 200,000/823,543 = 0.2429 = 24.29%

a) Denote a five by F and a non-five by F' then you want P(F'F'F'F'F) =
b) The P(F) is the same for each throw and throws are independent
with a fixed number of them so if the number of F in 8 throws is X
then X has a Bin(8,2/7) distribution) and you want P(X=3).

a) [(5/7)^5] * (2/7)

b) (8C5) * [(5/7)^5] * [(2/7)^3]