Answer:
a) Y = 3 trials : [SSS]
b) Y = 4 trials : [FSSS]
c) Y = 5 trials: [FFSSS; SFSSS]
d) Y = 6 trials: [FFFSSS; SSFSSS; SFFSSS; FSFSSS]
e) Y = 7 trials: [FFFFSSS; FFSFSSS; FSSFSSS; SSFFSSS; SFSFSSS; SFFFSSS; FSFFSSS]
Explanation:
Here the outcome of a trial is S(success) or F(failure). We are told the component is tested repeatedly until a success occurs on three consecutive trials, which means an outcome of S must be gotten 3 time consecutively (SSS).
Let's take Y as the number of trials, therefore
a) Y = 3 trials : [SSS]
This outcome is the only possibility here because we are limited to just 3 trials and S must happen 3 times consecutively.
b) 4 trials : [FSSS]
Similar to option A, S should be gotten an 3 consecutive trials.
c) 5 trials: [FFSSS; SFSSS]
In this case FSSSS is also possible, but if we get FSSS there would be no need for a fifth trial since we have gotten 3 consecutive S.
d) 6 trials: [FFFSSS; SSFSSS; SFFSSS; FSFSSS]
In this case the 4 last outcomes must be FSSS. For the outcomes before the last 4, we just had to make sure it doesn't have S 3 consecutive times
e) 7 trials: [FFFFSSS; FFSFSSS; FSSFSSS; SSFFSSS; SFSFSSS; SFFFSSS; FSFFSSS]
This is similar to option d