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 1^{st} 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 2^{nd} part similarly

reference: Intro to Algorithms: CHAPTER 12: HASH TABLES (ustc.edu.cn)