1. **Problem statement:** An oil company has a 20% chance (probability $p=0.2$) of striking oil on each well drilled. We need to find: (i) The probability that the first strike comes on the third well drilled. (ii) The probability that the third strike comes on the seventh well drilled. (iii) The mean and variance of the number of wells drilled to get three producing wells. --- 2. **Formulas and concepts:** - For (i), this is a geometric distribution problem where the first success occurs on the $k$-th trial: $$P(X=k) = (1-p)^{k-1} p$$ - For (ii), this is a negative binomial distribution problem where the $r$-th success occurs on the $k$-th trial: $$P(X=k) = \binom{k-1}{r-1} p^r (1-p)^{k-r}$$ - For (iii), the mean and variance of the negative binomial distribution with parameters $r$ and $p$ are: $$\text{Mean} = \frac{r}{p}$$ $$\text{Variance} = \frac{r(1-p)}{p^2}$$ --- 3. **Calculations:** (i) Probability first strike on third well: $$P(X=3) = (1-0.2)^{3-1} \times 0.2 = 0.8^2 \times 0.2 = 0.64 \times 0.2 = 0.128$$ (ii) Probability third strike on seventh well: $$P(X=7) = \binom{7-1}{3-1} (0.2)^3 (1-0.2)^{7-3} = \binom{6}{2} (0.2)^3 (0.8)^4$$ Calculate binomial coefficient: $$\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$$ Calculate powers: $$(0.2)^3 = 0.008$$ $$(0.8)^4 = 0.4096$$ Multiply all: $$15 \times 0.008 \times 0.4096 = 15 \times 0.0032768 = 0.049152$$ (iii) Mean and variance for $r=3$, $p=0.2$: $$\text{Mean} = \frac{3}{0.2} = 15$$ $$\text{Variance} = \frac{3 \times (1-0.2)}{0.2^2} = \frac{3 \times 0.8}{0.04} = \frac{2.4}{0.04} = 60$$ --- 4. **Summary:** - (i) Probability first strike on third well = $0.128$ - (ii) Probability third strike on seventh well = $0.049152$ - (iii) Mean number of wells to get 3 producing wells = $15$ - (iii) Variance of number of wells to get 3 producing wells = $60$