Answer:
[tex](a)\ \{SSS\}[/tex]
[tex](b)\ \{FSSS\}[/tex]
[tex](c)\ \{FFSSS, SFSSS\}[/tex]
[tex](d)\ \{FFFSSS, SSFSSS, SFFSSS, FSFSSS\}[/tex]
[tex](e)\ \{FFFFSSS, FFSFSSS, FSSFSSS, SSFFSSS, SFSFSSS, SFFFSSS, FSFFSSS\}[/tex]
Step-by-step explanation:
Given
[tex]S\to[/tex] Success
[tex]F\to[/tex] Failure
[tex]n \to 3[/tex] Consecutive trials until S happens
[tex]Y \to[/tex] Number of trials
For every possible outcome, the last 3 must be SSS. So, we have:
[tex](a)\ Y = 3[/tex]
The only possibility here is: [tex]\{SSS\}[/tex]
because 3 trials implies that all outcomes must be S.
[tex](b)\ Y = 4[/tex]
The only possibility here is:
[tex]\{FSSS\}[/tex]
because 4 trials implies that the first outcome must be F
[tex](c)\ Y = 5[/tex]
The possibilities are:
[tex]\{FFSSS, SFSSS\}[/tex]
because 5 trials implies that the first and the second outcomes must be FS or SF.
[tex](d)\ Y = 6[/tex]
The possibilities are:
[tex]\{FFFSSS, SSFSSS, SFFSSS, FSFSSS\}[/tex]
because 6 trials implies that the first three outcomes are: FFF, SSF and SFF
[tex](e)\ Y = 7[/tex]
The possibilities are:
[tex]\{FFFFSSS, FFSFSSS, FSSFSSS, SSFFSSS, SFSFSSS, SFFFSSS, FSFFSSS\}[/tex]
because 7 trials implies that the first three outcomes are: FFFF, FSSF and SSFF and FFSF (similar to (e))