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## Integrated math 3

### Course: Integrated math 3>Unit 11

Lesson 1: Binomial probability

# Binomial probability (basic)

AP.STATS:
UNC‑3 (EU)
,
UNC‑3.B (LO)
,
UNC‑3.B.1 (EK)

## Problem 1: Building intuition with free-throws

Steph makes 90, percent of the free-throws she attempts. She is going to shoot 3 free-throws. Assume that the results of free-throws are independent from each other.
She wants to find the probability that she makes exactly 2 of the 3 free-throws.
problem A
If she makes 2 of the free-throws, how many free-throws does that mean she needs to miss?

problem b
Find the probability that she makes her first 2 free-throws and misses her third free-throw.
P, left parenthesis, start text, m, a, k, e, comma, space, m, a, k, e, comma, space, m, i, s, s, end text, right parenthesis, equals

problem c
"Make, make, miss" isn't the only way Steph can make make 2 free-throws in 3 attempts.
Find the probability that she makes her first free-throw, then misses the second, and then makes her third free-throw.
P, left parenthesis, start text, m, a, k, e, comma, space, m, i, s, s, comma, space, m, a, k, e, end text, right parenthesis, equals

problem d
Steph could also make 2 free-throws if her results are "miss, make, make".
Find the probability that she misses her first free-throw and makes her next 2 free-throws.
P, left parenthesis, start text, m, i, s, s, comma, space, m, a, k, e, comma, space, m, a, k, e, end text, right parenthesis, equals

problem E
Use the combination formula to verify that these 3 ways represent every way we can arrange 2 makes in 3 attempts.
start subscript, n, end subscript, start text, C, end text, start subscript, k, end subscript, equals, start fraction, n, !, divided by, left parenthesis, n, minus, k, right parenthesis, !, dot, k, !, end fraction
start subscript, 3, end subscript, start text, C, end text, start subscript, 2, end subscript, equals
ways

problem f
Now put it all together to find the probability that she makes exactly 2 of the 3 free-throws.
P, left parenthesis, start text, m, a, k, e, s, space, 2, space, o, f, space, 3, space, f, r, e, e, space, t, h, r, o, w, s, end text, right parenthesis, equals, P, left parenthesis, start text, S, space, end text, start text, S, space, end text, start text, F, end text, right parenthesis, plus, P, left parenthesis, start text, S, space, end text, start text, F, space, end text, start text, S, end text, right parenthesis, plus, P, left parenthesis, start text, F, space, end text, start text, S, space, end text, start text, S, end text, right parenthesis
P, left parenthesis, start text, m, a, k, e, s, space, 2, space, o, f, space, 3, space, f, r, e, e, space, t, h, r, o, w, s, end text, right parenthesis, equals

## Generalizing from Problem 1: Building a formula for future use

We saw in Problem 1 that different orders of the same outcome each had the same probability.
We can build a formula for this type of problem, which is called a binomial setting. A binomial probability problem has these features:
• a set number of trials left parenthesis, start color #11accd, n, end color #11accd, right parenthesis
• each trial can be classified as a "success" or "failure"
• the probability of success left parenthesis, start color #1fab54, p, end color #1fab54, right parenthesis 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:
\begin{aligned} &P\left( \overset{\text{getting exactly some}}{\text{# of successes}}\right) \\\\ &=\left( \overset{\text{# of}}{\text{arrangements}}\right) \cdot \left( \overset{\text{probability}}{\text{of success}}\right)^{\left( \overset{\text{# of}}{\text{successes}}\right)} \cdot \left( \overset{\text{probability}}{\text{of failure}}\right)^{\left( \overset{\text{# of}}{\text{failures}}\right)} \end{aligned}
Using the example from Problem 1:
• n, equals, 3 free-throws
• each free-throw is a "make" (success) or a "miss" (failure)
• probability she makes a free-throw is start color #1fab54, p, end color #1fab54, equals, start color #1fab54, 0, point, 90, end color #1fab54
• 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, left parenthesis, start text, e, x, a, c, t, l, y, space, end text, k, start text, space, s, u, c, c, e, s, s, e, s, end text, right parenthesis, equals, start subscript, n, end subscript, start text, C, end text, start subscript, k, end subscript, dot, p, start superscript, k, end superscript, dot, left parenthesis, 1, minus, p, right parenthesis, start superscript, n, minus, k, end superscript
Try using these strategies to solve another problem.

## Problem 2

Steph's little brother Luke only has a 20, percent chance of making a free-throw. He is going to shoot 4 free-throws.
What is the probability that he makes exactly 2 of the 4 free throws?
P, left parenthesis, start text, e, x, a, c, t, l, y, space, 2, space, m, a, k, e, s, end text, right parenthesis, equals

## 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?
P, left parenthesis, start text, 3, space, o, r, space, m, o, r, e, space, m, a, k, e, s, end text, right parenthesis, equals

## Want to join the conversation?

• probability of each shot is given as .2. My doubt is if they are independent events, then in each shot probability of success or failure should be .5 ( as he is equally likely to fail or succeed). I am not clear on how .2 probability is arrived at?
(1 vote)
• my friend, if I have understood you correctly, you are assuming that each shot has a FAIR chance of going in. That is, each shot you take has a 50% chance of going in. However, this is not the case for this given scenario. There exist a bias; the shooter has a 20% chance of making each shot. As for a 50% (or .5) chance scenario, a good example to consider is that of a fair coin. In this case, each coin flip has an equal opportunity to show heads or tails. Hence, there is no bias--the exact opposite scenario of that of the shooter in the above example.

Welp, hope that helps.
• How do you solve if you do not know the probability at all (i.e. 90% in first problem)? Or is there something I am not understanding about a question that asks probability of rolling a pair of dice x times and getting the sum of 10?
Example: Find the probability of throwing a sum of 10 at least 4 times in 9 throws.
• Assuming that the dice is fair and six-sided, the probability is 1 in 6 to get a certain value.
For example: The probability of getting a 4 on a dice is 1/6.
• I seriously don't get the combination formula. All I did was multiply the probabilities, like 3(9/10*9/10*1/10), which works just as well, but I would like to understand the combination formula a bit better.
• In Combinations, we count all the unordered ways in which we can write a particular sequence. In permutations, we count all the ordered ways to write a particular sequence. Here we have to choose 2 places from 3 places where each chosen will corresponds to making a basket and each non-chosen place will correspond to missing a basket. e.g. From 123 position number we can choose 12, 13 or 23 position number where steph's can make a basket in 3 consecutive trials. The same can be chosen with the help of 3C2. Hope it helps. Thanks
• Hi, in Problem 2 I don't understand why there are more chances of scoring 3 or more free throwns out of 4, than scoring actually 3 out of 4. All in all I don't understand why it is ok to add the independent probabilites, getting an even better one. Thanks!!
• Hi, when scoring 3 or more free throws out of 4, the chances are: getting 3 out of 4; and getting 4 out of 4, however, when actually scoring exactly 3 out of the four there is only a chance of getting 3 out of 4 and so is less of a chance. I hope this helps answer your question.
(1 vote)
• For the challenge problem, instead of getting P(3 successes) + P(4 successes), I got the P(1 success) (0.410), added it to the P(2 successes) (0.154) which we already got in the previous question, then took it away from 1. But instead of the given answer, I got 0.436. Where did I go wrong?

If I get the probability of 1 success by adding the non-1 successes given in the explanations, I get P(2)=.154+P(3)=.026+P(4)=.002, giving a prob of 2 or more successes of 0.182, therefore a prob of 1 success of 1-0.182=0.818 This is very different from the prob of 1 successs I get from following the formula which gives .410 What is going on here?

In fact, I checked my answers in a spreadsheet and found the total probability doesn't add up to 1. How can this be?
• I’m glad that you are in the habit of analyzing your answers, to check whether or not they make sense. What went wrong is that you did not include P(0 successes) in the list of probabilities that must add to 1.

Have a blessed, wonderful new year!
• OK I'm not getting this. In problem #1, why couldn't we use the basic probability formula: P(x)= number of possible ways X can occur/number of total possible outcomes = 3/8
i.e., there are 3 ways Steph can make 2 of 3 throws and there are 8 total possible outcomes (2x2x2)
?
(1 vote)
• The 8 possible sequences of makes and/or misses in 3 free throws are not all equally likely, because the probability of making any given free throw (that is, 90%) is a percentage other than 50%. The basic probability formula you wrote in your question only applies for equally likely outcomes!

Have a blessed, wonderful day!
• HOW to use it when P(x<any number)
(1 vote)
• Are you asking how to solve the challenge question if instead of "3 or more", it read "3 or less"? Then P(Makes 3 or less free throws)= P(makes 3)+P(makes 2)+P(makes 1)+P(makes 0)