Ahmed has a free-throw percentage of 80%. What is the probability that
he will make at least 3 out of 5 shots?
These question were from Khan Academy. As he solves the 1st question, he calculates this probability by multiplying the free-throw percentage of 0.8 for successful shots, WITH 0.2 for the shot he misses.
From that point on, I learnt that you need to factor in the probability of missing the shot in your calculation.
Which makes me confused when he solved the 2nd question:
Laila is playing a game where there are 4 blue markers and 6 red
markers in a box. The markers will be placed back after every draw. If out of the 3 markers, the first 2 she picks are blue, she will
win a total of \$300. Under any other outcome, she will win \$0.
To calculate the probability of the first 2 markers picked being blue, he simply multiplied (0.4) * (0.4), without factoring in the probability of picking the red/blue marker the 3rd time.
I am curious as to why this works. If the "formula" works for the marker question, then why isn't the probability of making at least 3 out of 5 free throws simply (0.8)(0.8)(0.8)? Why must we multiply by the miss rate for that question and not for the marker question? What probability does (0.8)(0.8)(0.8) represent in that case?
Best Answer
The answer to the first question is wrong.
To see this, suppose that the probability of success at each shot is $0.5$. That means that you can work out the probability without any calculation: just count among the $2\times 2\times 2\times 2\times 2=32$ possible results.
The probability of getting at least 3 successes is $\frac {16}{32}$.
The probability of getting precisely 3 successes is $\frac{10}{32}$.
Neither of these figures matches what the method quoted for the first example would give, which is $0.5^3\times 0.5^2=\frac 1 {32}$.
I have changed the probability to $0.5$ to show that the method given produces wrong results, in a case where it easily to see without calculation what the right result should be. A method which fails for $p=0.5$ will also fail for $p=0.8$ - its failure will be less easy to see, that’s all.
Since the method given is nonsensical, no wonder you are confused!
The second example is simple to understand if you are not confused by the first one. There is no third drawing. The probability of success on two drawings is simply the square of the probability of success on one.
EDIT: As drhab points out, the answer that is being given in the first example is to the question “What is the probability of three successes followed by two failures?” His solution of the question as you stated it is correct and he should make it into a proper answer instead of a comment.