![]() ![]() Hence, the number of ways to get exactly 2 heads in 3 flips is 3C2 and not 3P2. So we divide by the number of ways the heads can be arranged among themselves (2! in this example) to get the actual number of events that will satisfy the condition exactly 2 heads in 3 flips. If heads A was obtained on flip 1 and heads B was obtained on flip 3, this is equivalent to heads B being obtained on flip 1 and heads A being obtained on flip 3 since they are both heads, so we are overcounting if we use permutations. I'll list them out:īut as you can see, this means we are differentiating between heads A and heads B, which is unnecessary. Hence there are 6 permutations for flipHeadsA and flipHeadsB. We have a total of 3 flips to choose from initially, so for flipHeadsA we have 3 options, and now we have only 2 options left for flipHeadsB. Now if we had to choose 2 flips out of these 3 to have heads, in how many ways could we choose them? Let's call these two flips flipHeadsA and flipHeadsB. Here the total number of outcomes is 2^3 = 8. Let me try and illustrate this through a smaller example, suppose we were dealing with the probability of getting exactly 2 heads in 3 flips. Since one "heads" is exactly the same as another, the ordering of these 3 heads does not matter, hence the number of ways is 8C3. To find the number of outcomes out of these 2^8 possible outcomes that have exactly 3 heads we need to figure out how many ways we can choose exactly 3 heads in 8 flips. He doesn't divide by the total number of permutations, he's dividing by the total number of possible outcomes which is 2^8. ![]()
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