Binomial variables
Last updated
Last updated
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.
X = # of heads from flipping a coin 5 times
possible outcomes from 5 flips:
prob (score) = 70% or 0.7
prob (miss) = 30% or 0.3
P(Exactly 2 scores in 6 attempts) =
f = prob of making attempts
each trial can be classified as a "success" or "failure"
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:
each free-throw is a "make" (success) or a "miss" (failure)
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}
each shot is either a make or miss
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}
Do they all have the same probability?
Let's return to our original strategy to answer the question:
Let's return to our original strategy to answer the question:
P(Exactly scores in attempts) =
a set number of trials ()
the probability of success () is the same for each trial
free-throws
probability she makes a free-throw is
Steph promises to buy Luke ice cream if he makes or more of his free-throws.
What is the probability that he makes or more of the free throws?
Luke gets ice cream if he makes or 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:
trials (shots)
probability that he makes a shot is
Layla has a coin that has a chance of showing heads each time it is flipped. She is going to flip the coin times. Let represent the number of heads she gets.
What is the probability that she gets more than heads?
For each flip, we know . To find the probability that all flips are heads, we can multiply probabilities since flips are independent:
We'll come back and use this result later. Next, we need to find (the probability that she gets heads).
Getting heads in attempts means Layla needs to get heads and tail. For each flip, we know and .
Let's start by finding the probability of getting heads followed by tail:
This isn't the entire probability though, because there are other ways to get heads from flips (for example, THHHH). How many different ways are there? We can use the combination formula:
Each of the ways has the same probability that we already found:
So we can multiply this probability by since that is how many ways there are to get heads in flips.
Ira ran out of time while taking a multiple-choice test and plans to guess on the last questions. Each question has possible choices, one of which is correct. Let the number of answers Ira correctly guesses in the last questions.
What is the probability that he answers fewer than questions correctly in the last questions?
The probability that Ira gets fewer than questions correct in the questions is equivalent to the probability that he gets or question correct. So we can find those probabilities and add them them together to get our answer:
There are possible choices for each question, so we know and . Answering questions correctly is equivalent to answering all questions incorrectly. We can multiply probabilities since we are assuming independence:
βWe'll come back and use this result later. Next, we need to find (the probability that he answers question correctly).
Answering question correctly in the last questions means Ira needs to get question correct and questions not correct. There are possible choices for each question, so we know and .
Since we are assuming independence, let's multiply probabilities to find the probability of getting question correct followed by questions not correct:
This isn't the entire probability though, because there are other ways to get question correct from questions (for example, NNNNNC). How many different ways are there? We can use the combination formula:
There are ways to get question correct in questions. Do they all have the same probability?
Each of the ways has the same probability that we already found:
βSo we can multiply this probability by since that is how many ways there are to get question correct in questions:
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