How to find the number of distinct patterns possible?

bmsmith6

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Feb 15, 2017
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A fair coin is flipped 30 times.
  1. How many possible distinct patterns of heads and tails are possible?
  2. What is the probability of getting 3 heads?
  3. What is the probability of getting 15 heads?

I am not looking for someone to give me the answer. I am just unsure of how to begin this problem. Could someone refer me to some concepts to read about or some equations to look at?

Thanks.

Mason Smith
 
Well, I think two of the most beneficial math skills to develop are pattern recognition and critical thinking. They play a large role in all of math, but especially in probability and statistics. To that end, it's often beneficial to take a step back, and really think deeply about what the problem is asking you to do and what it means. Then, reduce it to its core elements and come up with the simplest version you can think of, and build up from there to find the underlying pattern. In this case we know that the coin is completely fair, meaning there are 2 outcomes and each one occurs 50% of the time. So, let's go through each sub-part:

1) How many possible distinct patterns of heads and tails are possible?

We've already established that flipping one coin has only two outcomes, but what about two coins? Well, if you flip heads on the first coin, then you can flip either heads or tails; similarly, if you flip tails on the first coin, you can flip either heads or tails. That means there are 2 * 2 = 4 outcomes. Let's add a third coin - we know the first two coins have four possibilities, and for each one of those four possibilities, you can flip either heads or tails on the third coin. So, there's 4 * 2 = (2 * 2) * 2 = 8 outcomes. Are you seeing a pattern yet? If not, consider what happens if you add a fourth coin then reevaluate.

What is the probability of getting 3 heads?

So, getting 3 heads means we also must get 27 tails. In this specific case, because the chance of heads is exactly the same as the chance of tails, that's a meaningless distinction. However, it does matter in cases where the probabilities aren't equal. Like, say, if a different coin were biased and came up heads 60% of the time, that would change the formula a bit. Each flip of the coin is an independent event, because it in no way affects the outcome of any future flip (be careful of the "Gambler's Fallacy" where people wrongly assume that, for instance, a coin that's come up heads 27 times in a row is "due" for a tails so betting on tails would be the smart move).

When considering what the events being independent means in the context of the problem, it may prove useful to consider other examples. If we let A stand for the event that we get heads on one coin flip and B stand for the event that we get heads on the second coin flip, we know that P(A and B) [the probability of both A and B happening) is 1/4. But how could we calculate that if we didn't already know it? Well, consider that P(A) = 1/2 and P(B) is also 1/2. How do those values relate to 1/4? Or another example might be drawing cards from a standard deck. If we let A stand for the event that we draw a Jack, and B stand for the event that we draw a Spade, then P(A and B) = 1/52. Here P(A) = 1/13 and P(B) = 1/4. How do those values relate to 1/52? This pagehttp://www.mathgoodies.com/lessons/vol6/independent_events.html may also prove helpful to you.

What we've done so far is all well and good, but it's not quite there yet. We've only found the probability of the first three coins coming up heads then the next 27 all coming up tails. But, maybe we got 5 tails in a row, then 3 heads, then 22 more tails; or maybe we got 18 tails, 1 head, 2 tails, 1 head, 7 tails, and 1 head. In other words, we have to account for the fact that the the three heads we get could be any of the 30 coins. So how many ways can be arrange 30 coins and pick three of them? We can see that we can pick any of the 30 coins, then we can pick any of the 29 remaining coins, and finally any of the 28 remaining coins. That makes a total of 30 * 29 * 28 = 24360 ways, each of which has the chance we found above of occurring. This pagehttps://www.mathsisfun.com/combinatorics/combinations-permutations.html has some more intuition about combinations and some formulas, to help you generalize the results to any picking any number of objects.

What is the probability of getting 15 heads?

This is the exact same as above, but different numbers. Recall that in order to get 15 heads, you must also get 15 tails. And there are "30 choose 15" (to use the language given on the linked page above) ways to arrange those 15 heads.
 
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