Binomial probability practice problems

Check the setup: each of $n = 6$ seeds either germinates (success) or does not, the seeds are independent, and each germinates with probability $p = 0.8$, so this is binomial.

Write the binomial probability formula for exactly $x$ successes: $P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$, where $\binom{n}{x}$ is read as $n$ choose $x$. Here $x = 5$.

Count the arrangements: $\binom{6}{5} = 6$.

Raise the success probability to the power: $0.8^5 = 0.32768$.

Raise the failure probability to the power: $(1 - 0.8)^{6-5} = 0.2^1 = 0.2$.

Multiply the three pieces: $6 \times 0.32768 \times 0.2 = 1.96608 \times 0.2 = 0.393216$.

Identify the binomial parameters: $n = 40$ trial listeners and $p = 0.3$, the probability each one upgrades. Let $X$ be the number who upgrade.

Use the binomial mean formula $\mu = np$, where $\mu$ (mu) is the mean number of successes.

Compute the mean: $\mu = 40 \times 0.3 = 12$ upgrades.

Use the binomial standard deviation formula $\sigma = \sqrt{np(1-p)}$, where $\sigma$ (sigma) is the standard deviation.

Fill in the numbers under the root: $np(1-p) = 40 \times 0.3 \times 0.7 = 12 \times 0.7 = 8.4$.

Take the square root: $\sigma = \sqrt{8.4} \approx 2.90$ upgrades.

Recall the four conditions for a binomial random variable: each trial has two outcomes (success or failure), the number of trials $n$ is fixed in advance, the trials are independent, and the probability of success $p$ is the same on every trial.

(a) Drawing names without replacement shrinks and changes the pool after each draw, so the chance a drawn name is left-handed shifts from draw to draw and the draws are not independent. This fails the independent-trials and constant-$p$ conditions, so it is not binomial.

(b) Each call gives a yes or no answer (two outcomes), there is a fixed number of calls $n = 25$, random selection from a large city keeps the calls independent, and each adult has about the same probability of saying yes. All four conditions hold, so this is binomial.

(c) Opening boxes until the first rare item means the number of trials is not set in advance; it depends on when the first success happens. This fails the fixed-$n$ condition, so it is not binomial.

Set up the binomial model with $n = 4$ students and $p = 0.7$, and let $X$ be the number who show up. At most 2 means $X = 0$, $1$, or $2$, so add those three probabilities.

Compute $P(X = 0) = \binom{4}{0}(0.7)^0(0.3)^4 = 1 \times 1 \times 0.0081 = 0.0081$.

Compute $P(X = 1) = \binom{4}{1}(0.7)^1(0.3)^3 = 4 \times 0.7 \times 0.027 = 4 \times 0.0189 = 0.0756$.

Compute $P(X = 2) = \binom{4}{2}(0.7)^2(0.3)^2 = 6 \times 0.49 \times 0.09 = 6 \times 0.0441 = 0.2646$.

Add the three pieces: $P(X \le 2) = 0.0081 + 0.0756 + 0.2646 = 0.3483$.

Set up the binomial model with $n = 12$ and $p = 0.15$, and let $X$ be the number who click. Adding the probabilities of 1 through 12 clicks would take many terms, so use the complement trick: $P(X \ge 1) = 1 - P(X = 0)$.

Write the probability that no one clicks: $P(X = 0) = \binom{12}{0}(0.15)^0(0.85)^{12} = 1 \times 1 \times 0.85^{12}$.

Raise the failure probability to the 12th power by repeated squaring: $0.85^2 = 0.7225$, $0.85^4 = 0.7225^2 = 0.5220$, $0.85^8 = 0.5220^2 = 0.2725$, and $0.85^{12} = 0.85^8 \times 0.85^4 = 0.2725 \times 0.5220 = 0.1422$.

Subtract from 1: $P(X \ge 1) = 1 - 0.1422 = 0.8578$.

Set up the binomial model with $n = 8$ and $p = 0.05$, and let $X$ be the number that burn out. At least 2 is the complement of 0 or 1 burnouts: $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$.

Compute the needed powers of the failure probability $0.95$ by repeated multiplication: $0.95^7 = 0.6983373$ and $0.95^8 = 0.6634204$.

Find $P(X = 0) = \binom{8}{0}(0.05)^0(0.95)^8 = 1 \times 1 \times 0.6634204 = 0.6634204$.

Find $P(X = 1) = \binom{8}{1}(0.05)^1(0.95)^7 = 8 \times 0.05 \times 0.6983373 = 0.4 \times 0.6983373 = 0.2793349$.

Add them: $P(X = 0) + P(X = 1) = 0.6634204 + 0.2793349 = 0.9427553$.

Subtract from 1: $P(X \ge 2) = 1 - 0.9427553 = 0.0572447 \approx 0.0572$.

Set up the binomial model with $n = 50$ orders and $p = 0.08$, and let $X$ be the number of returns.

(a) Mean: $\mu = np = 50 \times 0.08 = 4$ returns.

(a) Standard deviation: $\sigma = \sqrt{np(1-p)} = \sqrt{50 \times 0.08 \times 0.92}$. Inside the root, $50 \times 0.08 = 4$ and $4 \times 0.92 = 3.68$, so $\sigma = \sqrt{3.68} \approx 1.92$ returns.

(b) Measure how far 9 returns sits from the mean in standard deviations: $\frac{9 - 4}{1.92} = \frac{5}{1.92} \approx 2.60$.

(b) Because 9 returns is about 2.60 standard deviations above the mean, and values more than 2 standard deviations out are uncommon, 9 returns is an unusually high number for a day.

(c) The mean of 4 says that over many busy days like this one, the shop would average about 4 returns for every 50 orders it ships.

Set up the binomial model with $n = 10$ tags and $p = 0.6$.

(a) Use the formula for exactly 7 successes: $P(X = 7) = \binom{10}{7}(0.6)^7(0.4)^3$, with $\binom{10}{7} = 120$.

(a) Evaluate the powers: $0.6^7 = 0.0279936$ and $0.4^3 = 0.064$.

(a) Multiply the pieces: $120 \times 0.0279936 = 3.359232$, then $3.359232 \times 0.064 = 0.2149908 \approx 0.2150$.

(b) Use the complement for at least 1 transmit: $P(X \ge 1) = 1 - P(X = 0)$, where $P(X = 0) = \binom{10}{0}(0.6)^0(0.4)^{10} = 0.4^{10} = 0.0001048576$.

(b) Subtract from 1: $P(X \ge 1) = 1 - 0.0001048576 = 0.9998951 \approx 0.9999$.

(c) Mean: $\mu = np = 10 \times 0.6 = 6$ tags. Standard deviation: $\sigma = \sqrt{np(1-p)} = \sqrt{10 \times 0.6 \times 0.4} = \sqrt{2.4} \approx 1.55$ tags.