Simplistic Basketball Probability Problem

[rk7697]

A basketball player has a 75% probability in making a free throw shot. If the basketball player attempts two free throws, what is the probability that they will make at least one of the two shots?

A question building onto the previous one: If the player has three free throws, what's the probability that they will score at least two of them?

I did 1-.25^2 because there's two shots and apparently 1 subtracted by the complementary ratio of one is helpful in this case, but I don't understand why. The reason being is because I understand that this is the probability that the player will miss and that you square it because there's two shots(I think), but I don't understand why you cannot take the .75^2 anymore than you can approach the problem in the way I'm supposed to. I'm sure that my lack of comprehension is apparent. I'm currently in an online class and this topic is barely explained. I'm not requesting for someone to calculate the answer for me, but rather to help explain the conceptual aspect. This unit has led to an abundance of confusion. I understand if you wanted to know the probability of the player making n shots in a row, .75^n, but I don't get the "at least" aspect of this problem. For the following question, I don't even know how to approach this. I would really appreciate some help, thanks.

 

[Prof]

Your idea to compute 1 - .25^2 is right. Here is why. Any time you want the probability of "at least one" that means you want "1 or more". Now making 1 or more could be making one shot or making two shots in this situation. This would give you two different problems to solve: making one and missing one, and making two and missing 0. In this case, the opposite of what you want is simpler. The only thing you do not want is to miss two shots. So what is the chance of missing two shots? .25^2 = .0625 (so about 6% of the time the player misses 2 shots). This means the rest of the time, the player makes at least one shot (about 100%-6% = 94%). Now 100% is the same as 1 so if you use decimals instead of percents, you would compute
1 - .25^2 = .9375 which is the answer.

 

[JeffM]

You used a short-cut that is valid, but the short cut itself may not make clear why it is valid or why it is useful..

Although you do not say so, the problem (if it is carefully written) should say that the probability of making a shot is independent of the results of previous shots. There is no valid reason to make that assumption unless it is given in the statement of the problem itself. It might be the case that the player's confidence goes down if he misses and goes up if he makes so that the probability of success after the first shot changes with the result of previous shots. If that plausible case were the true case, you would not have the information required to solve the problem. Of course, badly written problems abound, and they make students guess, UGH! So check for that word "independent" in your problem. Without it, we are reduced to guesswork.

But if the probability of success is independent of previous success or failures, we have enough information to solve the problem.

You throw the ball twice. There are four possibilities.

Success followed by success. Probability = (3/4)(3/4) = 9/16.
Success followed by failure. Probability = (3/4)(1/4) = 3/16.
Failure followed by success. Probability = (1/4)(3/4) = 3/16.
Failure followed by failure. Probability = (1/4)(1/4) = 1/16.

One of those has to happen. It is a certainty. And the probability of a certainty is 1. Let's check.

[MATH]\dfrac{9}{16} + \dfrac{3}{16} + \dfrac{3}{16} + \dfrac{1}{16} = \dfrac{9 + 6 + 6 + 1}{16} = \dfrac{16}{16} = 1.[/MATH]
You are interested in the probability that any one of the first three events occurs. Note that, as we have defined them, they are mutually exclusive. So we can just add their probabilities up.

[MATH]\dfrac{9}{16} + \dfrac{3}{16} + \dfrac{3}{16} = \dfrac{9 + 6 + 6}{16} = \dfrac{15}{16}.[/MATH]
But another (and computationally simpler) way to think about this "at least" problem is to frame it as the event that the fourth case does not occur. Either A occurs or does not occur. Mutually exclusive but exhaustive. Probability of A + probability of not A ALWAYS = 1.
So the probability of A ALWAYS = 1 - the probability of not A.

Thus the probability that one of the first three cases applies equals 1 minus the probability that none of the first three cases applies, which equals the probability of one minus the probability that the fourth cases applies.

[MATH]1 - \dfrac{1}{16} = \dfrac{15}{16} =\dfrac{9}{16} + \dfrac{3}{16} + \dfrac{3}{16}.[/MATH]
And now we see why 9/16 is not the correct answer. It covers only one of the three cases that we are interested in.

The words "independent," "at least," and "at most," are very important in probability questions. Look for them. Understand what they mean. The concepts of "mutually exclusive" and "exhaustive" are also very important, but they frequently must be identified without any explicit clues.

One last point. A lot of basic probability involves techniques for reducing the number of required computations. With 2 shots, we had to look at 4 possible outcomes. We did not really need a short cut. If we were talking about 10 shots, we would have to look 1,024 possibilities. Now a short cut looks eminently desirable.

This answer is a very round-about way to say what the prof already said. I suggest that to get all this firmly in your own mind that you yourself do your follow on problem BOTH WAYS, see that both methods necessarily give the same answer, and see why the prof's method is computationally more efficient.

And please do not hesitate to ask whatever follow-up questions you may have. I personally found probability theory quite unintuitive at first, and so I sympathize with beginning students who wonder if they have lost their minds. You have not, and basic probability theory becomes intuitive once you fully understand the meaning of words like " independent" and "mutually exclusive."