If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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.
To think about this problem, let's break it up into smaller parts.
problem A
If she makes 2 of the free-throws, how many free-throws does that mean she needs to miss?
Choose 1 answer:
Choose 1 answer:

problem b
Find the probability that she makes her first 2 free-throws and misses her third free-throw.
Round your answer to the nearest hundredth if necessary.
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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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.
Round your answer to the nearest hundredth if necessary.
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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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.
Round your answer to the nearest hundredth if necessary.
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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
ways

problem f
Now put it all together to find the probability that she makes exactly 2 of the 3 free-throws.
Round your answer to the nearest hundredth if necessary.
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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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:
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
P(makes 2 of 3 free throws)=3C2(0.90)2(0.10)1=30.810.10=30.081=0.243\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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Want to join the conversation?

  • blobby green style avatar for user ameetranjan
    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?
    (0 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user D.Q.
      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.
      (42 votes)
  • blobby green style avatar for user Rachel Heldberg
    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.
    (4 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user mr.swedishfish
    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.
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user ajay06081993
      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
      (4 votes)
  • blobby green style avatar for user Gonzalo Ferreiro Volpi
    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!!
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Mohid Khan
      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)
  • blobby green style avatar for user First Last
    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?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user hnashawi
    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)
    Default Khan Academy avatar avatar for user
    • primosaur seed style avatar for user Ian Pulizzotto
      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!
      (3 votes)
  • mr pants teal style avatar for user YIMING JIA
    HOW to use it when P(x<any number)
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user shawn.pitner
    I am trying to find out what the odds are of getting at least 16 out of 24 trials correct where the probability of getting any one trial correct is .25 or 25%. My calculations give me Cumulative Probability: P(X > or = 16) 2.02210569919536E-05 or a probability of .0000202210569919536. Does that mean that the odds of getting at least 16 out of 24 correct with a probability of 25% on each trial would be equal to about 1 in 50,000?
    (0 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user robbyex1401
    Binomial probability distribution
    A disease is transmitted with a probability of 0.4, each time two indivuals meet. If a sick individual meets 10 healthy individuals, what is the probability that
    (a) exactly 2 of these individuals become ill.
    (b) less than 2 of these individuals become ill.
    (c) more than 3 of these individuals become ill.

    Explanation to this?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Daniel Lemmer
    For the "Steph promises to buy Luke ice cream if he makes 3 or more of his 4 free-throws" problem, the probability of 4 throws has, in the denominator, (4-4)! times 4!
    Since 4-4=0, should the demoinator not be 0 as well (in the solution, the this does not happen. Instead it says 4! divided by 4! in the next step).
    (1 vote)
    Default Khan Academy avatar avatar for user