Binomial variables

coin flip -> P(H) = 0.6; P(T) = 0.4

X = # of heads after 10 flips of my coin

• made up of independent trails (flip)

• each trail can be classified either as success or failure

• fixed # of trails

• probability of success on each trail is constant

Inference -> The act or process of deriving logical conclusions from premises known or assumed to be true.

Distribution

X = # of heads from flipping a coin 5 times

possible outcomes from 5 flips: $2 * 2 * 2 * 2 * 2 = 2^5 = 32$

$P(X=0) = \frac{1}{32} =\frac{_5 C_0}{32} \Rightarrow _5 C_0 = \frac{5!}{0! * (5-0)!} = \frac{5!}{5!} = 1$

$P(X=1) = \frac{5}{32} = \frac{_5 C_1}{32} \Rightarrow _5C_1 = \frac{5!}{1!*(5-1)!}=\frac{5!}{1!}=5$

$P(X=2) = \frac{10}{32} = \frac{_5 C_2}{32} \Rightarrow _5C_2 = \frac{5!}{2!*(5-2)!}=\frac{5!}{2!*3!}=\frac{5*4*3*2}{2*3*2}=10$

$P(X=3) = \frac{10}{32} = \frac{_5 C_3}{32} \Rightarrow _5C_3 = \frac{5!}{3!*(5-3)!}=\frac{5!}{3!*2!}=10$

$P(X=4) = \frac{5}{32} = \frac{_5 C_4}{32} \Rightarrow _5C_4 = \frac{5!}{4!*(5-4)!}=\frac{5!}{1!}=5$

$P(X=5) = \frac{1}{32} =\frac{_5 C_5}{32} \Rightarrow _5 C_5 = \frac{5!}{5! * (5-5)!} = \frac{5!}{5!} = 1$

Example:

prob (score) = 70% or 0.7

prob (miss) = 30% or 0.3

P(Exactly 2 scores in 6 attempts) =

$\binom {6}{2} * 0.7^2 * 0.3^4 = 15 * 0.49 * 0.0081 = 0.05935$

$\Longrightarrow \text{SSMMMM} = 0.7 * 0.7 * 0.3 * 0.3 * 0.3 * 0.3 = (0.7)^2 * (0.3)^4$

$\Longrightarrow _6C_2 = \binom{6}{2}=\frac{6!}{2!*(6-2)!} = \frac{6*5*4*3*2*1}{(2*1(4*3*2*1)}=15$

Generalizing k scores in n attempts:

P(Exactly $k$ scores in $n$ attempts) = $\large\binom{n}{k}* f^k * (1-f)^{n-k}$

f = prob of making attempts

A binomial probability problem has these features:

• a set number of trials ($\blueD{n}$)

• each trial can be classified as a "success" or "failure"

• the probability of success ($\greenD{p}$) is the same for each trial

• results from each trial are independent from each other

Here's a summary of our general strategy for binomial probability:

Using the example from Problem 1:

• $n=3$ free-throws

• each free-throw is a "make" (success) or a "miss" (failure)

• probability she makes a free-throw is $\greenD{p}=\greenD{0.90}$

• assume free-throws are independent

\begin{aligned}P(\text{makes 2 of 3 free throws}) &= \, _3\text{C}_2 \cdot(\greenD{0.90})^{2} \cdot (\maroonD{0.10})^1 \\ \\ &=3\cdot0.81\cdot0.10 \\ \\ &=3\cdot0.081 \\ \\ &=0.243\end{aligned}

In general...

P(\text{exactly }k \text{ successes})=\,_n\text{C}_k \cdot p^k \cdot (1-p)^{n-k}

Challenge Problem

Steph promises to buy Luke ice cream if he makes $3$ or more of his $4$ free-throws.

What is the probability that he makes $3$ or more of the $4$ free throws?

Luke gets ice cream if he makes $3$ or $4$ free throws. We can find the probability of each of those outcomes and add the results together.Here's how we can think of this problem:

• $n=4$ trials (shots)

• each shot is either a make or miss

• probability that he makes a shot is $p=0.20$

• shots are independent

\begin{aligned}P(\text{makes 3 or more}) &= P(\text{makes 3 free-throws}) + P(\text{makes 4 free-throws}) \\ \\ & =\,_4\text{C}_3 \cdot (0.20)^3 \cdot (0.80)^1+\,_4\text{C}_4 \cdot (0.20)^4 \cdot (0.80)^0 \\ \\ &= \dfrac{4!}{(4-3)! \cdot3!} \cdot0.008 \cdot 0.80+\dfrac{4!}{(4-4)! \cdot4!} \cdot 0.0016 \\ \\ &=\dfrac{4 \cdot 3\cdot 2\cdot 1}{3\cdot 2 \cdot 1}\cdot0.008 \cdot 0.80 +\dfrac{4 \cdot 3\cdot 2\cdot 1}{4 \cdot 3 \cdot 2 \cdot 1} \cdot0.0016 \\ \\ &=4 \cdot0.008 \cdot 0.80+0.0016\\ \\ &=0.0256+0.0016 \\ \\ &=0.0272\end{aligned}

Quiz #1

Layla has a coin that has a $60\%$ chance of showing heads each time it is flipped. She is going to flip the coin $5$ times. Let $X$ represent the number of heads she gets.

What is the probability that she gets more than $3$ heads?

Finding $P(X=5)$

For each flip, we know $P(\blueD{\text{heads}})=\blueD{60\%}$. To find the probability that all $5$ flips are heads, we can multiply probabilities since flips are independent:

$\begin{aligned} P(X=5)&=(\blueD{0.60})(\blueD{0.60})(\blueD{0.60})(\blueD{0.60})(\blueD{0.60}) \\\\ &=(\blueD{0.60})^5 \\\\ &=0.07776 \end{aligned}$

We'll come back and use this result later. Next, we need to find $P(X=4)$ (the probability that she gets $4$ heads).

Finding $P(X=4)$

Getting $4$ heads in $5$ attempts means Layla needs to get $4$ heads and $1$ tail. For each flip, we know $P(\blueD{\text{heads}})=\blueD{60\%}$ and $P(\maroonD{\text{tails}})=\maroonD{40\%}$.

Let's start by finding the probability of getting $4$ heads followed by $1$ tail:

$P(\blueD{\text{HHHH}}\maroonD{\text{T}})=(\blueD{0.6})^4(\maroonD{0.4})=0.05184$

This isn't the entire probability though, because there are other ways to get $4$ heads from $5$ flips (for example, THHHH). How many different ways are there? We can use the combination formula:

$\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _5\text{C}_4&=\dfrac{5!}{(5-4)!\cdot4!} \\\\ &=\dfrac{5 \cdot \cancel{4 \cdot 3 \cdot 2 \cdot 1}}{(1) \cdot \cancel{4 \cdot 3 \cdot 2 \cdot 1}} \\\\ &=5 \end{aligned}$

Do they all have the same probability?

Each of the $5$ ways has the same probability that we already found:

$\begin{aligned} P(\blueD{\text{HHHH}}\maroonD{\text{T}})&=(\blueD{0.6})^4(\maroonD{0.4})=0.05184 \\\\ P(\blueD{\text{HHH}}\maroonD{\text{T}}\blueD{\text{H}})&=(\blueD{0.6})^4(\maroonD{0.4})=0.05184 \\\\ P(\blueD{\text{HH}}\maroonD{\text{T}}\blueD{\text{HH}})&=(\blueD{0.6})^4(\maroonD{0.4})=0.05184 \\\\ P(\blueD{\text{H}}\maroonD{\text{T}}\blueD{\text{HHH}})&=(\blueD{0.6})^4(\maroonD{0.4})=0.05184 \\\\ P(\maroonD{\text{T}}\blueD{\text{HHHH}})&=(\blueD{0.6})^4(\maroonD{0.4})=0.05184 \end{aligned}$

So we can multiply this probability by $5$ since that is how many ways there are to get $4$ heads in $5$ flips.

$\begin{aligned} P(X=4)&=5(0.6)^4(0.4) \\\\ &=5(0.05184) \\\\ &=0.2592 \end{aligned}$

Putting it all together

Let's return to our original strategy to answer the question:

$\begin{aligned} P(X>3)&=P(4\text{ heads})+P(5\text{ heads}) \\\\ &=P(X=4)+P(X=5) \\\\ &=5(0.6)^4(0.4)+(0.6)^5 \\\\ &=0.2592+0.07776 \\\\ &=0.33696 \\\\ &\approx0.34 \end{aligned}$

Quiz #2

Ira ran out of time while taking a multiple-choice test and plans to guess on the last $6$ questions. Each question has $4$ possible choices, one of which is correct. Let $X=$ the number of answers Ira correctly guesses in the last $6$ questions.

What is the probability that he answers fewer than $2$ questions correctly in the last $6$ questions?

Strategy (without a fancy calculator)

The probability that Ira gets fewer than $2$ questions correct in the $6$ questions is equivalent to the probability that he gets $0$ or $1$ question correct. So we can find those probabilities and add them them together to get our answer:

$\begin{aligned} P(X<2)&=P(0\text{ correct})+P(1\text{ correct}) \\\\ &=P(X=0)+P(X=1) \end{aligned}$

Finding $P(X=0)$

There are $4$ possible choices for each question, so we know $P(\blueD{\text{correct}})=\blueD{25\%}$ and $P(\maroonD{\text{not}})=\maroonD{75\%}$. Answering $0$ questions correctly is equivalent to answering all $6$ questions incorrectly. We can multiply probabilities since we are assuming independence:

$\begin{aligned} P(X=0)&=(\maroonD{0.75})(\maroonD{0.75})(\maroonD{0.75})(\maroonD{0.75})(\maroonD{0.75})(\maroonD{0.75}) \\\\ &=(\maroonD{0.75})^6 \\\\ &\approx0.17798 \end{aligned}$

​We'll come back and use this result later. Next, we need to find $P(X=1)$ (the probability that he answers $1$ question correctly).

Finding $P(X=1)$

Answering $1$ question correctly in the last $6$ questions means Ira needs to get $1$ question correct and $5$ questions not correct. There are $4$ possible choices for each question, so we know $P(\blueD{\text{correct}})=\blueD{25\%}$ and $P(\maroonD{\text{not}})=\maroonD{75\%}$.

Since we are assuming independence, let's multiply probabilities to find the probability of getting $1$ question correct followed by $5$ questions not correct:

$P(\blueD{\text{C}}\maroonD{\text{NNNNN}})=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933$

This isn't the entire probability though, because there are other ways to get $1$ question correct from $6$ questions (for example, NNNNNC). How many different ways are there? We can use the combination formula:

$\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _6\text{C}_1&=\dfrac{6!}{(6-1)!\cdot1!} \\\\ &=\dfrac{6 \cdot \cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{(\cancel{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}) \cdot 1} \\\\ &=6 \end{aligned}$

There are $6$ ways to get $1$ question correct in $6$ questions. Do they all have the same probability?

Each of the $6$ ways has the same probability that we already found:

$\begin{aligned} P(\blueD{\text{C}}\maroonD{\text{NNNNN}})&=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933 \\\\ P(\maroonD{\text{N}}\blueD{\text{C}}\maroonD{\text{NNNN}})&=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933 \\\\ P(\maroonD{\text{NN}}\blueD{\text{C}}\maroonD{\text{NNN}})&=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933 \\\\ P(\maroonD{\text{NNN}}\blueD{\text{C}}\maroonD{\text{NN}})&=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933 \\\\ P(\maroonD{\text{NNNN}}\blueD{\text{C}}\maroonD{\text{N}})&=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933 \\\\ P(\maroonD{\text{NNNNN}}\blueD{\text{C}})&=(\blueD{0.25})(\maroonD{0.75})^5\approx0.05933 \end{aligned}$

​So we can multiply this probability by $6$ since that is how many ways there are to get $1$ question correct in $6$ questions:

$\begin{aligned} P(X=1)&=6(0.25)(0.75)^5 \\\\ &\approx6(0.05933) \\\\ &\approx0.35596 \end{aligned}$​

Putting it all together

Let's return to our original strategy to answer the question:

$\begin{aligned} P(X<2)&=P(0\text{ correct})+P(1\text{ correct}) \\\\ &=P(X=0)+P(X=1) \\\\ &=(0.75)^6+6(0.25)(0.75)^5 \\\\ &\approx0.17798+0.35596 \\\\ &\approx0.53394 \\\\ &\approx0.53 \end{aligned}$