Mathematics
tracmac:

the probability that a young person plays soccer in the summer is 0.57. The probability that a young person plays hockey in the winter is 0.83. Assuming that they are independent events, if a young person is chosen at random, what is the probability that they play hockey or soccer?

AZ:

since they're independent events probability of playing hockey or soccer = probability of playing soccer + probability of playing hockey so just add the two numbers

SmokeyBrown:

Hold on, I get what AZ is getting at, but I don't think that result makes sense. I mean, adding together the probability of 0.57 with 0.83 gives us 1.4, which is greater than 1. But we know that probabilities always have to be between 0 and 1

SmokeyBrown:

Actually, I think there are 4 possibilities hidden in this question: A student could play soccer but not hockey; a student could play hockey but not soccer; a student could play both soccer and hockey; or a student could play neither soccer nor hockey

SmokeyBrown:

I think I should make a quick clarification here. When we want to compute the probability of multiple *independent* events happening, we can multiply the probabilities of the individual events happening alone, in order to get the probability of all events happening together. We can consider the classic example of flipping coins. The outcome of one coin flip does not affect the other, so we can consider them to be independent. Now, we know that flipping heads is a 1/2 chance because there are only 2 *equally likely* outcomes: H or T. Flipping 2 consecutive heads is a 1/4 chance because there are 4 *equally likely outcomes: HH, HT, TH, TT. It's no coincidence that 1/4 happens to be 1/2*1/2

SmokeyBrown:

These are pretty important probability concepts, so we should make sure to understand them. Anyway, now that that review is out of the way, I'll get back to the problem

SmokeyBrown:

We know are given the probabilities that a student *does* play either sport, and from that we can also consider the probability that they *do not* play either sport. Since the probability of playing soccer is 0.57, the probability of *not* playing soccer is (1.0-0.57) = 0.47. Likewise, the probability of playing hockey is 0.83, so the probability of *not* playing hockey is (1.0-0.83) = 0.17

SmokeyBrown:

Now, the probability that a student doesn't play soccer *and* doesn't play hockey would be: (probability *not* play soccer) * (probability *not* play hockey) or 0.47 * 0.17 = 0.0799

SmokeyBrown:

@smokeybrown wrote:
We know are given the probabilities that a student *does* play either sport, and from that we can also consider the probability that they *do not* play either sport. Since the probability of playing soccer is 0.57, the probability of *not* playing soccer is (1.0-0.57) = 0.47. Likewise, the probability of playing hockey is 0.83, so the probability of *not* playing hockey is (1.0-0.83) = 0.17
Apologies, the probability that a student does *not* play soccer should be 0.43

SmokeyBrown:

@smokeybrown wrote:
Now, the probability that a student doesn't play soccer *and* doesn't play hockey would be: (probability *not* play soccer) * (probability *not* play hockey) or 0.47 * 0.17 = 0.0799
And the correct calculation would be 0.43*0.17 = 0.0731

SmokeyBrown:

We can actually calculate the probabilities for all possible scenarios: Student plays soccer but doesn't play hockey: 0.57 * 0.17 = 0.0969 Student doesn't play soccer but does play hockey: 0.43*0.83 = 0.3569 Student plays both soccer and hockey: 0.57 * 0.83 = 0.4731 Student does not play soccer and does not play hockey: 0.43 * 0.17 = 0.0731

SmokeyBrown:

Do you notice anything about the probabilities of the 4 scenarios? They all add up to exactly 1.0! This is as we would expect. It just means that something out of all possibilities has to occur. It's also a good sign that we did our calculations correctly

SmokeyBrown:

So, to (finally) actually answer your question, it depends what the question means by "or". If the question means *either* soccer or hockey (but not both), we can take the sum of the probabilities for scenario 1 and 2: 0.0969 + 0.3569 = 0.4538 If the question means soccer or hockey or both, we can take the sum of the probabilities for scenarios 1, 2, and 3: 0.0969 + 0.3569 + 0.4731 = 0.9269

SmokeyBrown:

I think that's probably what AZ was thinking of, actually. It is correct that you can get the total probability of independent outcomes by summing their individual probabilities; the issue is that we didn't want to consider "soccer" and "hockey" as separate outcomes. Rather, we want to consider the outcomes that occur when "soccer" and "hockey" interact with each other

SmokeyBrown:

To be more clear about that, you can imagine a bag full of 3 red balls, 2 black balls, and 5 white balls. If you want to get the probability of picking a red ball or picking a black ball, you would indeed add the probability of picking a red ball (3/10) and the probability of picking a black ball (2/10) to get 5/10, which would be right. Only, this situation isn't like that situation

SmokeyBrown:

Putting the two examples together, I guess you could say that "soccer" and "hockey" aren't picked from the same "bag"? That's why the same strategy doesn't work

AZ:

Ah, thank you for the very thorough explanation!! I, perhaps in my haste, didn't read it completely or pay attention to the numbers. I agree with your your answer. They play soccer in the summer and hockey in the winter so these events would not be mutually exclusive. P(A or B) = P(A) + P(B) - P(A and B)

AZ:

That would also result in the same answer that you calculated using two different methods

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