So how do I think about that? Well, let's go back to one So I say 3/8 heads, but 3 of myįlips are going to be heads and the rest are going to be tails. The probability of getting exactly 3 out of 8 heads. Out the probability- I'm going to flip a coin eight timesĪnd it's a fair coin. Playlist, and I think you'll learn in this video. Went off into permutations and combinations in the probability This is also the number of combinations of 3 flips that result in 2 heads, calculated above (if it's still unintuitive, it can help to think about the case where we get HHH or TTT, which would correspond respectively to 6 permutations or 1 combination of flips).įinally, to get the probability of getting exactly 2 heads in 3 flips, we divide the number of 'success' events, in this case, of getting exactly 2 heads, by the number of total outcomes, so 3/8. We can see that 3 of those outcomes contain exactly 2 heads (HHT, HTH, THH). Now moving to the realm of outcomes in terms of heads and tails, we've 8 (2^3) possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. So we've 3 possible combinations of 3 flips that result in 2 heads. We get the number of combinations by dividing the permutations by 2 (= k! = 2! = 2x1), so 3 combinations: AB/BA, AC/CA and BC/CB. The flips permutations that would result in 2 heads are AB, AC, BC, BA, CB, CA.

And we're interested in 2 'heads' events (analogous to 'sitting' state of people). Flips are the 'unique entity', and we can name them like people: flip A, flip B, flip C (instead of using numbers, which might be confusing). Reducing the problem space to getting 2 heads in 3 flips helped. Especially the analogy to people and seats from the previous videos. This means that the probability of a certain sequence depends on the number of Bells that are in the sequence.Ī sequence containing 3 Bells has the probabilityīut now that we have that probability, all we need to do is multiply it by the number of sequences containing 3 Bells, which again is the same as the number of ways we can choose which 3 wheels are Bells = 8!∕(3! ∙ 5!) However, now the two possibilities have different probabilities: In the case with the fruit machine, each wheel can be seen as having two possible outcomes, Bell or No Bell. So, we want to know in how many ways we can choose which 3 coins are Heads, which of course must be the exact same as the number of ways we can choose which 3 people are sitting. However, since each coin is equally likely to land Heads as to land Tails, then each sequence of 8 coins is also equally likely to occur, and thereby the probability that 3 coins are Heads is equal to the number of sequences that consist of 3 Heads and 5 Tails, divided by the total number of possible sequences. The difference is that for the 8 people we are asked the number of ways that we can choose which 3 are sitting, while for the 8 coins we are asked the probability that 3 are Heads and the others are Tails. The coins and the people aren't really that different.Įach coin has two possible outcomes, Heads or Tails, and each person also has two possible outcomes, sitting or standing.

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.

Permutations and combinations examples