the expected number of probes in an unsuccessful search is at most 1/(1 - alpha)
where ɑ = load factor
also
Given an open-address hash table with load factor < 1, the expected number of probes in a successful search is at most
$1/\alpha \ln (1/1-\alpha )+1/\alpha$
when load factor is ¾ hence for 1st case max probe is = $1/1-3/4 = 4$ for unsuccessful
and for successful = $4/3\\ln (4) + 4/3 = 3.18 approx$
calculate for 2nd part similarly
reference: Intro to Algorithms: CHAPTER 12: HASH TABLES (ustc.edu.cn)