Example: A game of dice is played in which a $6$ is considered a "win" and all other outcomes a "lose." This game is played $20$ times. Find the probability that, of these $20$ games, exactly $5$ are a "win."

$P(\text{Win}) = \frac{1}{6} \quad P(\text{Lose}) = \frac{5}{6}$

$\binom{20}{5} \times \left(\frac{1}{6}\right)^5 \times \left(\frac{5}{6}\right)^{15} = :highlight[0.129] \quad (3 \text{sf})$

(Combinations: $5$ $wins$, $15$ $losses$)

A biased coin is tossed with $P(H) = 0.7$ considered a win. If this is done $25$ times, find the probability of exactly $12$ wins.

$\binom{25}{12} \times 0.7^{12} \times 0.3^{13} = :highlight[0.0115] \quad (3 \text{sf})$

The steps for calculating the probability on a calculator (images of calculator screen are provided) show:

1. Navigate to Distribution.
2. Select Binomial PD.
3. Input:

• $x = 12$
• $n = 25$
• $p = 0.7$

4. Result: $P = :highlight[0.01147575294]$

Example The probability that a letter gets lost in the post is $0.01$. On a particular day, a postman has a batch of $2000$ letters. Model this as a binomial distribution, stating the distribution.

$X \sim B(2000, 0.01)$

where $X$ is the number of lost letters (a lost letter is considered a "win").

Find $P(X = 26)$, the probability of exactly $26$ letters being lost:

$P(X = 26) = \binom{2000}{26} \times 0.01^{26} \times 0.99^{1974} = :highlight[0.0342]$

Example: Let $X \sim B(10, 0.2)$. Find $P(X \leq 3)$.

$P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X \leq 3) = 0.1074 + 0.2688 + 0.3020 + 0.2013 = :highlight[0.879] \quad (3 \text{sf})$

A quick way is to use the Binomial CD option on the calculator.

More steps are shown in the calculator screenshots with a final result: $P(X \leq 3) = :highlight[0.8791261184]$

Example: Let $X \sim B(15, 0.3)$. Find $P(X \leq 7) = :highlight[0.950]$.

Find $P(X \leq 12) = P(X \leq 11) = :highlight[0.9999]$ ($4 sf$)

(Note: $\leq$ so we must modify the question)

Find $P(X > 4)$ :

Note: For $>$ or $\geq$ , it helps to list out the outcomes we do want and therefore those we don't want.

Do Want: $4, 5, 6,$ $\dots$, $15$

Don't Want: $0, 1, 2, 3 ,$ i.e., $\leq$ $3$

So,

$P(X > 4) = 1 - P(X \leq 3)$

$1 - 0.2969 = :highlight[0.703] \quad (3 \text{sf})$

Find $P(X > 10)$ :

Do Want: $11, 12,$ $\dots$, $15$

Don't Want: $10, 9,$ $\dots$, $0$

$1 - P(X \leq 10) = 1 - 0.9993 = :highlight[6.72 \times 10^{-4}]$

Example: For $X \sim B (12, 0.1)$, find $P(8 \leq X \lt 11)$. Do Want: $8, 9, 10$

Diagram illustrating values $0$ to $12$, highlighting $8, 9,$ and $10$ in red.

$P(X \leq 10) - P(X \leq 7) = P(8 \leq X \lt 11)$