The Binomial Model in Four Practice Questions

a. There are eight fixed trials. Each result is green or not green, the green probability stays $2/5$, and independence is given. Thus

$$X\sim\operatorname{Bin}(8,2/5).$$

Here $n=8$, $p=2/5=0.4$, and $X$ can be $0,1,2,\ldots,8$. The five spinner regions still give two categories for the event being counted.

b. The second experiment has no fixed number of spins. Also, $Y$ counts trials until the first success rather than successes within a fixed set of trials. A binomial model does not describe $Y$, even though the spins remain independent and their green probability stays the same.

The fixed seven trials, two outcomes, common probability and stated independence give $X\sim\operatorname{Bin}(7,0.60)$. A specified pattern of five germinating seeds and two that do not germinate has probability $(0.60)^5(0.40)^2$. There are

$$\binom75=\frac{7!}{5!2!}=21$$

ways to choose the five successful positions. These patterns cannot occur together, so add their probabilities:

$$\begin{aligned}\Pr(X=5)&=21(0.60)^5(0.40)^2\\&=0.2612736\approx0.2613.\end{aligned}$$

The probability is about $26.13\%$. The expression $(0.60)^5$ alone leaves out both the two failures and the possible positions. In real planting, a shared condition such as inadequate watering could undermine the independence assumption; the answer uses the model stated in the question.

The assumptions give $X\sim\operatorname{Bin}(6,0.15)$. A delivery that does not fail has probability $0.85$.

a. At least two means $X\ge2$. Its complement is $X=0$ or $X=1$:

$$\begin{aligned}\Pr(X=0)&=(0.85)^6,\\\Pr(X=1)&=6(0.15)(0.85)^5.\end{aligned}$$

Therefore

$$\begin{aligned}\Pr(X\ge2)&=1-(0.85)^6\\&\quad-6(0.15)(0.85)^5\\&=0.223515703125\approx0.2235.\end{aligned}$$

b. More than two means $X\ge3$. Remove exactly two failures from the previous event:

$$\begin{aligned}\Pr(X=2)&=\binom62(0.15)^2(0.85)^4\\&=0.176177109375,\\\Pr(X>2)&=\Pr(X\ge2)-\Pr(X=2)\\&=0.04733859375\approx0.0473.\end{aligned}$$

The answers are about $22.35\%$ and $4.73\%$. The second event excludes all outcomes with exactly two failures, which explains the decrease. Replacing $p=0.15$ by $0.85$ while still counting failures would answer a different question.

Let $X$ be the number of runs producing the target among the chosen $n$ runs. For each fixed choice of $n$, $X\sim\operatorname{Bin}(n,0.30)$. No successes has probability $(0.70)^n$, so we need

$$\begin{aligned}\Pr(X\ge1)&=1-(0.70)^n\ge0.95,\\(0.70)^n&\le0.05.\end{aligned}$$

Testing consecutive whole-number values near the target gives

$$\begin{aligned}n=8:&\quad1-(0.70)^8=0.94235199,\\n=9:&\quad1-(0.70)^9=0.959646393.\end{aligned}$$

Eight runs fall below $0.95$; nine exceed it. Each additional run multiplies the no-success probability by $0.70<1$, so the chance of at least one success increases with $n$. Every smaller positive number also fails the target. The minimum is therefore nine runs, giving probability $0.9596$ to four decimal places, or about $95.96\%$.

This is a probability target, not a guarantee of success. The design chooses nine runs in advance; stopping at the first success would describe a different experiment.