Answer:
0.0004
Step-by-step explanation:
A fair die has 6 faces.
The probability of obtaining a 2 [tex]=\dfrac{1}{6}[/tex]
The probability of a number other than 2 [tex]=\dfrac{5}{6}[/tex]
The experiment consists of tossing a fair die until 2 occurs 7 times.
Therefore, the probability that the process ends after exactly ten tosses with 2 occurring on the ninth and tenth tosses
[tex]=^{10}C_4 \times \left(\dfrac{5}{6}\right)^3 \times \left(\dfrac{1}{6}\right)^7\\\\=0.00043412701 \\ \approx 0.0004 $(correct to 4 decimal places)[/tex]
