An oil drilling company ventures into various locations, and

Step 1
Given Information:
Probability of success at any specific location i.e., ${\displaystyle p=0.30}$
i) To find the probability that a driller drills 10 locations and finds 1 success:
Let X be a random variable which denotes the number of successes among 10 drills. Because trials are independent, X follows a Binomial distribution with parameters ${\displaystyle n=10\text{and}p=0.30}$.
Probability mass function of Binomial variable X is given as:
${\displaystyle P(X=x){=}^n{C}_x{p}^x{(1-p)}^{n-x}}$
Required probability is obtained as follows:
${\displaystyle P(X=1){=}^{10}{C}_1{\left(0.30\right)}^1{(1-0.30)}^{10-1}}$
${\displaystyle =\frac{10!}{1!(10-1)!}\times 0.30\times {\left(0.70\right)}^9}$
${\displaystyle =\frac{10\times 9!}{9!}\times 0.30\times 0.040353607}$
${\displaystyle =10\times 0.0121060821}$
${\displaystyle =0.121060821}$
${\displaystyle \approx 0.1211}$
Thus, the probability that a driller drills 10 locations and finds 1 success is 0.1211
Step 2
ii) Let X represent the number of drills required to get first success.
Find the probability that the driller prospects for bankruptcy.
In other words, that there are no successes ${\displaystyle (X=0)}$ in 10 trials:
Required probability is obtained as follows:
${\displaystyle P(X=0){=}^{10}{C}_0{\left(0.30\right)}^0{(1-0.30)}^{10-0}}$
${\displaystyle =\frac{10!}{0!(10-0)!}\times 1\times {\left(0.70\right)}^{10}}$
${\displaystyle =\frac{10!}{10!}\times 1\times 0.0282475249}$
${\displaystyle =0.0282475249}$
${\displaystyle \approx 0.0282}$