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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.

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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?

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    $\begingroup$ I tend to disagree with the solution of the first problem. The answer should be $$P(X=3)+P(X=4)+P(X=5)=\binom530.8^30.2^2+\binom540.8^40.2^1+\binom550.8^50.2^0$$ Here $X$ denotes the number of successful shots. It has binomial distribution with parameters $n=5$ and $p=0.8$. Concerning the second problem: for calculating the probability that the first $2$ picks result in blue markers only the first $2$ picks are relevant. Observe that $0.8^3$ is the probability that e.g. the first $3$ shots are succesful (or e.g. the last $3$ shots). Not at least $3$ shots. $\endgroup$ – drhab Jul 27 '19 at 8:40
  • $\begingroup$ Yes, the solution to the 1st question is absolutely correct, I just extracted the part where my query lies which is having to multiply WITH the failure rate when the 2nd question doesn't require you to do so... So essentially, the difference lies in the position of the markers picked? $\endgroup$ – Hannah Tang Jul 27 '19 at 8:44
  • $\begingroup$ Why wouldn't the probability of making the first 3 shots be equal to the probability of making at least 3 shots though? The position matters? $\endgroup$ – Hannah Tang Jul 27 '19 at 8:53
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    $\begingroup$ Observe that there are at least $3$ succesful shots if the first $3$ shots are succesful but also if the last $3$ shots are succesful. So the probability on at least $3$ succesful shots will be larger than the probability that the first $3$ shots are succesful. $\endgroup$ – drhab Jul 27 '19 at 9:00
  • $\begingroup$ That makes a lot of sense!!! Thank you very much! $\endgroup$ – Hannah Tang Jul 27 '19 at 9:02
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I disagree with the solution of the first problem.

The answer should be: $$P(X=3)+P(X=4)+P(X=5)=\binom530.8^30.2^2+\binom540.8^40.2^1+\binom550.8^50.2^0\tag1$$ Here $X$ denotes the number of successful shots. It has binomial distribution with parameters $n=5$ and $p=0.8$.

Observe that $0.8^3$ is the probability that e.g. the first $3$ shots are succesful (or also e.g. the last $3$ shots). Not at least $3$ shots however.

Further observe that $0.8^30.2^2$ is the probability that only the first $3$ shots are succesful (or e.g. only the last $3$ shots).


Concerning the second problem: for calculating the probability that the first $2$ picks result in blue markers only the first $2$ picks are relevant.


P.S. The question has been edited in such a way that mentioning of $(1)$ is not relevant anymore.

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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.

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  • $\begingroup$ My apologies... I did not include the full question in this post because I just wanted to focus on my main query. I have edited the 2nd question. It is supposed to be 2 out of 3 draws, I failed to include that part previously. The part I don't understand is he just ignores the results of the 3rd drawing. In what situations can I just calculate success rates and in what situations must I multiply the success rate with the failure rate is what I am fuzzy about $\endgroup$ – Hannah Tang Jul 27 '19 at 9:11
  • $\begingroup$ I made my comment an answer now. At first hand I had doubts on my interpretation. $\endgroup$ – drhab Jul 27 '19 at 9:12
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Since you have changed the question a little bit, let me supplement my original answer by giving you some useful first principles. They are easy to remember, and I think you will be less confused in future if you think of any other methods you are taught as being laid on top of the first principles, which are universal.

Let's take the first question as a concrete example. $5$ is easier to write than $N$, $3$ is easier to write than $M$, $0.8$ is easier to write than $p$, $0.2$ is easier to write than $1-p$.

The perfectly reliable method

  1. Enumerate all the possibilities.
  2. Calculate the probability of each.
  3. Add up the probabilities of the ones you are interested in.

This always works.

In the case of your first example, denoting success by $S$ and failure by $F$, you have $2^5=32$ possibilities: $FFFFF$, $FFFFS$, $FFFSF$,… all the way up to $SSSSS$. Get a tall piece of paper and list them all, one to a line.

Now calculate the probability of each one. $FFFFF$ will be $0.2\times 0.2\times 0.2\times 0.2\times 0.2$, $FFFFS$ will be $0.2\times 0.2\times 0.2\times 0.2\times 0.8$, and so on.

If you have got this right, the sum of all the probabilities will be $1$.

Next, highlight all the lines that correspond to an outcome you actually want: $FFSSS$, $FSFSS$, and so on. (I can't remember whether you want "exactly 3 successes" or "at least 3 successes", but whichever it is that you want, highlight those lines).

Finally, add up the probabilities of all the lines you have highlighted, and you have your answer.

The less boring method

Boredom is a great educator. Since you are not a computer, you will have notice that many of your sums are the same. $0.2\times 0.2\times 0.2\times 0.2\times 0.8$ equals $0.2\times 0.2\times 0.2\times 0.8\times 0.2$, and so on. Being human, you will soon have stopped calculating and will have started copying the result of the last identical calculation. You will have established this principle:

The probability of a given sequence depends not on the arrangement of the $F$s and $S$s in it, but only on how many there are. The probability is $0.8^{N(S)}0.2^{N(F)}$.

So now you know that you only have six cases to consider: with 0, 1, 2, 3, 4 or 5 successes. Your new method is:

  1. Enumerate all the distinct numbers-of-symbols (6 of them in total).
  2. For each number-of-symbols, calculate the probability.
  3. For each number-of-symbols, calculate how many ways there are of it happening ("five $F$s" can only occur in one way, "four $F$s and one $S$" can occur in five ways, and so on).
  4. Multiply the probability (step 2) by the number of ways (step 3).

You can now manage with a much shorter piece of paper! There are only six probability calculations to be made.

Write down "0, 1, 2, 3, 4, 5" - the number of successes in each case.

In the next column, write down "1, 5, 10, 10, 5, 1" - the number of possible arrangements of $F$s and $S$s that match each case.

In the next column, write down "$0.2^5$", "$0.2^4 0.8^1$", "$0.2^3 0.8^2$", "$0.2^2 0.8^3$", "$0.2^1 0.8^4$", "$0.8^5$" - the probability of one of the arrangements in each case.

In the final column, write down the product of the previous two columns: "$1\times 0.2^5$", "$5\times 0.2^4 0.8^1$", "$10\times 0.2^3 0.8^2$", "$10\times 0.2^2 0.8^3$", "$5\times 0.2^1 0.8^4$", "$1\times 0.8^5$".

You can add them up to make sure that they all add up to $1$.

Now highlight the rows you actually want: either just row $3$, or rows $3$, $4$, and $5$, depending on whether the question is "exactly 3 successes" or "at least 3 successes".

Add up the probabilities of the highlighted rows.

Algebra and understanding

Having done all this, you can make it more symbolic. You can replace the numbers with variables. You can learn formulae. But none of this is anything new. The basic principles are still as I have outlined them, and if ever you get stuck or confused, remember that the first principles are still there, and still true, and you can always fall back onto them if the advanced formulae do not seem obvious.


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  • $\begingroup$ Thank you very much!!! $\endgroup$ – Hannah Tang Jul 27 '19 at 18:05
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I only want to add that the first question involves three separate WANTED outcomes with three DIFFERENT numbers of failure. The first outcome has 2 failures, the second has 1 failure, the third has 0 failures. So the failure rate is important in each wanted outcome.

The second question involves only one WANTED outcome, first two are blue, with only one number of failures, 0. Thus, the failure rate in each of the first two is 0.2^0 = 1 and it’s not worth writing it, so we simply write 0.8(0.8).

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