Answer:
0
Step-by-step explanation:
given that we roll a fair die repeatedly until we see the number four appear and then we stop.
the number 4 can appear either in I throw, or II throw or .... indefinitely
So X = the no of throws can be from 1 to infinity
This is a discrete distribution countable.
Sample space= {1,2,.....}
b) Prob ( 4 never appears) = Prob (any other number appears in all throws)
= ![\frac{5}{6} *\frac{5}{6} *\frac{5}{6} *\frac{5}{6} *...\\=(\frac{5}{6} )^n](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B6%7D%20%2A%5Cfrac%7B5%7D%7B6%7D%20%2A%5Cfrac%7B5%7D%7B6%7D%20%2A%5Cfrac%7B5%7D%7B6%7D%20%2A...%5C%5C%3D%28%5Cfrac%7B5%7D%7B6%7D%20%29%5En)
where n is the number of throws
As n tends to infinity, this becomes 0 because 5/6 is less than 1.
Hence this probability is approximately 0
Or definitely 4 will appear atleast once.