10: Binomial Distribution

Bernoulli Trials

To continue our exploration of discrete distributions, we will look at situations that have two disjoint possibilities.

Bernoulli Trials

For math symbols to represent a Bernoulli trial, the events $\{1,0\}$ have respective probabilities $p$ and $1-p$.

For example, for one flip of a coin $$P(\text{heads})=p,P(\text{tails})=1-p$$

one coin, but not necessarily fair

Arrangements

Permutations

Permutations (and the number of permutations) are the arrangements when order matters

Combinations

Combinations (and the number of combinations) are the arrangements when order does not matter

Flipping 3 fair coins, what is the probability that heads will be observed exactly twice?

Possibility Spaces

3 Coins

coin <- c("H", "T")
df <- data.frame(expand.grid(coin, coin, coin)) |>
  tidyr::unite("obs", c("Var1", "Var2", "Var3"),
               sep = "", remove = FALSE)
# print
dput(df$obs)

c("HHH", "THH", "HTH", "TTH", "HHT", "THT", "HTT", "TTT")

4 coins

coin <- c("H", "T")
df <- data.frame(expand.grid(coin, coin, coin, coin)) |>
  tidyr::unite("obs", c("Var1", "Var2", "Var3", "Var4"),
               sep = "", remove = FALSE)
# print
dput(df$obs)

c("HHHH", "THHH", "HTHH", "TTHH", "HHTH", "THTH", "HTTH", "TTTH", 
"HHHT", "THHT", "HTHT", "TTHT", "HHTT", "THTT", "HTTT", "TTTT"
)

5 coins

coin <- c("H", "T")
df <- data.frame(expand.grid(coin, coin, coin, coin, coin)) |>
  tidyr::unite("obs", c("Var1", "Var2", "Var3", "Var4", "Var5"),
               sep = "", remove = FALSE)
# print
dput(df$obs)

c("HHHHH", "THHHH", "HTHHH", "TTHHH", "HHTHH", "THTHH", "HTTHH", 
"TTTHH", "HHHTH", "THHTH", "HTHTH", "TTHTH", "HHTTH", "THTTH", 
"HTTTH", "TTTTH", "HHHHT", "THHHT", "HTHHT", "TTHHT", "HHTHT", 
"THTHT", "HTTHT", "TTTHT", "HHHTT", "THHTT", "HTHTT", "TTHTT", 
"HHTTT", "THTTT", "HTTTT", "TTTTT")

Choose

$$(\genfrac{}{}{0ex}{}{n}{k})={\displaystyle \frac{n!}{k!(n-k)!}}$$

• said ``n choose k’’
• This choose operator keeps track of the number of permutations in a certain combination
• note $0!=1$ (to avoid dividing by zero)

from The Simpsons

Binomial Distribution

Binomial Distribution

$$P(x=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$$

• $0\le k\le n$, where $n$ and $k$ are whole numbers
• $0\le p\le 1$

Example: Squirtle

Binomial Distribution

$$P(x=k)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$$

• $0\le k\le n$, where $n$ and $k$ are whole numbers
• $0\le p\le 1$

Historically, Squirtle defeats Charizard 32% of the time. If there are 5 battles, what is the probability that Squirtle wins exactly 2 times?

Battle Arena

battle <- c("S", "C")
df <- data.frame(expand.grid(battle, battle, battle, battle, battle)) |>
  tidyr::unite("obs", c("Var1", "Var2", "Var3", "Var4", "Var5"),
               sep = "", remove = FALSE)
# print
dput(df$obs)

c("SSSSS", "CSSSS", "SCSSS", "CCSSS", "SSCSS", "CSCSS", "SCCSS", 
"CCCSS", "SSSCS", "CSSCS", "SCSCS", "CCSCS", "SSCCS", "CSCCS", 
"SCCCS", "CCCCS", "SSSSC", "CSSSC", "SCSSC", "CCSSC", "SSCSC", 
"CSCSC", "SCCSC", "CCCSC", "SSSCC", "CSSCC", "SCSCC", "CCSCC", 
"SSCCC", "CSCCC", "SCCCC", "CCCCC")

But these observations have different weights!

Example: Charizard

Charizard

Historically, Charizard defeats Squirtle 68% of the time. If there are 5 battles, what is the probability that Charizard wins exactly 3 times?

Symmetry

Property

The previous two examples had the same answer, which is true due to a symmetry property in the choose operator:

$$(\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n}{n-k})$$

Squirtle

Charizard

R code

k <- 0:5
pk <- dbinom(k, 5, 0.32)
k_bool <- k == 2
df <- data.frame(k, pk, k_bool)

df |>
  ggplot(aes(x = k, y = pk, 
             color = k_bool, fill = k_bool)) +
  geom_bar(stat = "identity") +
  geom_label(aes(x = k, y = pk, 
                 label = round(pk, 4)),
             color = "black", fill = "white") +
  labs(title = "2 Squirtle Wins",
       subtitle = "n = 5, k = 2, p = 0.32, P(k = 2) = 0.3220",
       caption = "Math 32",
       x = "wins",
       y = "probability") +
  scale_color_manual(values = c("black", "#ca7721")) +
  scale_fill_manual(values = c("gray70", "#297383")) +
  theme(
    legend.position = "none",
    panel.background = element_blank()
  )

# plotly::ggplotly(ex2_plot)

k <- 0:5
pk <- dbinom(k, 5, 0.68)
k_bool <- k == 3
df <- data.frame(k, pk, k_bool)

df |>
  ggplot(aes(x = k, y = pk, 
             color = k_bool, fill = k_bool)) +
  geom_bar(stat = "identity") +
  geom_label(aes(x = k, y = pk, 
                 label = round(pk, 4)),
             color = "black", fill = "white") +
  labs(title = "3 Charizard Wins",
       subtitle = "n = 5, k = 3, p = 0.68, P(k = 3) = 0.3220",
       caption = "Math 32",
       x = "wins",
       y = "probability") +
  scale_color_manual(values = c("black", "#de5138")) +
  scale_fill_manual(values = c("gray70", "#e53800")) +
  theme(
    legend.position = "none",
    panel.background = element_blank()
  )

# plotly::ggplotly(ex2_plot)

How do we pick between $p$ and $1-p$?

At first, it does not matter how you define the binomial setting for what corresponds to $p$ and what corresponds to $1-p$, but you need to be consistent in the rest of the task for how you defined your variables and use the value(s) for $k$.

Parameters

The notation $X\sim Ber(p)$ is read as “random variable $X$ has a Bernoulli distribution with parameter $p$”. Compute the expected value and variance for a Bernoulli trial.

Parameters

The notation $X\sim Bin(n,p)$ is read as ``random variable $X$ has a binomial distribution with parameters $n$ and $p$’’. Compute the expected value and variance for a binomial distribution.

We are assuming that the $n$ trials are from each other, where independence in probability means that

$$P(\{X_i{\}}_{i=1}^n)={\displaystyle \prod _{i=1}^{n}P(X_i)}$$

In other words, we are sampling the Bernoulli trial $n$ times with replacement, so we can simply multiply the results from the previous example by $n$.

$$\begin{array}{|ccc|}\hline \mathbf{\text{mean}}& \mu & np\\ \mathbf{\text{variance}}& {\sigma }^2& np(1-p)\\ \mathbf{\text{standard deviation}}& \sigma & \sqrt{np(1-p)}\\ \hline\end{array}$$

Looking Ahead

• due Fri., Feb. 10:
  • WHW4
  • JHW2
  • Demographics Survey Part 1
• Be mindful of before-lecture quizzes

No lecture session for Math 32:

• Feb 20, Mar 10, Mar 24

Exam 1 will be on Wed., Mar. 1

• more information in weekly announcements

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