When a certain coin is flipped, the probability of heads is 0.5. If the coin is flipped 6 times, what is the probability that there are exactly 3 heads?
Coin flipped 6 times
To calculate the probability, we divide the number of outcomes with three heads by the total number of possible outcomes.
To calculate the number of outcomes with three heads, we use combinations to find the ways to select 3 heads out of 6 tosses. That's 6C3 = 20, calculated from the combinations formula:
nCr = n!/(r!(n-r)!)
To calculate the total number of outcomes, we recognize that each flip has two possibilities, tails and heads. With two options for 6 flips, we get:
2 x 2 x 2 x 2 x 2 x 2 = 64 total number of outcomes.
The probabiliity is then: 20/64 = 5/16.
FAQ: What are the other ways to approach this? Can I do it without combinations?
Trying to do this without the combination formula is like trying to hammer a nail into a wall using a screwdriver instead of a hammer. Basically, you can do it, but it's much more difficult. Instead, it's important to understand how the combination in this question works and why we need it.
The combination formula is key no matter what we do. The only practical alternative to the answer given in the video is to avoid the combination formula by simply writing out every possible outcome with 3 heads.
Otherwise, we would do everything the same as is given in the video. Clearly, this is not the most elegant solution. It takes too long. And here's the kicker--to know that there are 20 ways to get 3 heads in 6 flips, the ONLY formula shortcut is the combination formula.
If we were to use the probability formulas "alone," we would still need to get the number 20 somehow. That is, we could find the probabilities for each of the cases listed above (By probability of A and B, that's 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2) and add them together (Probability of A or B). But how many of these cases are we adding? 20. And how did we get 20? Either through the combination formula or through simply listing.
Any other solutions would be even more circuitous, and they would still end up dealing with the simple truth that there is a combination built into this problem, one way or another.
FAQ: Can I solve this with the general binomial formula?
Yes, you can also use the method discussed in the Binomial Situation lesson video. That approach effectively has you go through the same process. The probability portion of that formula comes out to be (1/2)^6. And then you must multiply by the number of combinations: 6C3. Thus, you end up with 6C3 * (1/2)^6 = 20 * 1/64 = 20/64 = 5/16.
Watch the lessons below for more detailed explanations of the concepts tested in this question. And don't worry, you'll be able to return to this answer from the lesson page.