What is the probability of flipping a coin 6 times and having the coin land heads every time?

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As the other answer has said you need to count the number of permutations of $\displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).

The total number of permutations of six dissimilar objects is $\displaystyle 6! = 720$.

When there are two groups comprising $3$ identical objects, the number of permutations becomes: $\displaystyle \frac{720}{3!3!} = 20$.

If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $\displaystyle \frac{20}{64} = \frac{5}{16}$.

That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $\displaystyle (\frac 12 + \frac 12)^6$, which is $\displaystyle \binom 63 (\frac 12)^3(\frac 12)^3 = \frac{5}{16}$ as before.

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Probability is a part of mathematics that deals with the possibility of happening of events. It is to forecast that what are the possible chances that the events will occur or the event will not occur. The probability as a number lies between 0 and 1 only and can also be written in the form of a percentage or fraction. The probability of likely event A is often written as P(A). Here P shows the possibility and A shows the happening of an event. Similarly, the probability of any event is often written as P(). When the end outcome of an event is not confirmed we use the probabilities of certain outcomes—how likely they occur or what are the chances of their occurring.

To understand probability more accurately we take an example as rolling a dice:

The possible outcomes are — 1, 2, 3, 4, 5, and 6.

The probability of getting any of the outcomes is 1/6. As the possibility of happening of an event is an equally likely event so there are same chances of getting any number in this case it is either 1/6 or 50/3%.

Formula of Probability

Probability of an event, P(A) = (Number of ways it can occur) ⁄ (Total number of outcomes)

Types of Events

Equally Likely Events: After rolling dice, the probability of getting any of the likely events is 1/6. As the event is an equally likely event so there is same possibility of getting any number in this case it is either 1/6 in fair dice rolling.
Complementary Events: There is a possibility of only two outcomes which is an event will occur or not. Like a person will play or not play, buying a laptop or not buying a laptop, etc. are examples of complementary events.

Solution:

Use the binomial distribution directly. Let us assume that the number of heads is represented by x (where a result of heads is regarded as success) and in this case X = 6
Assuming that the coin is unbiased, you have a probability of success ‘p’(where p is considered as success) is 1/2 and the probability of failure ‘q’ is 1/2(where q is considered as failure). The number of trials is represented by the letter ’n’ and for this question n = 7.
Now just use the probability function for a binomial distribution:
P(X = x) = nCxpxqn-x
Using the information in the problem we get
P(X = 6) = (7C6)(1/2)6(1/2)1
= 7 × 1/64 × 1/2
= 7/128
Hence, the probability of flipping a coin 7 times and getting heads 6 times is 7/128.

Similar Questions

Question 1: What is the probability of flipping a coin 20 times and getting 10 heads?

Answer:

Each coin can either land on heads or on tails, 2 choices.
(According to the binomial concept)
This gives us a total of 220 possibilities for flipping 20 coins.
Now how many ways can we get 10 heads? This is 20 choose 10, or (20C10)
This means our probability is (20C10)/220= 184756⁄1048576 ≈ .1762

Question 2: What is the probability of 6 heads in 6 coins tossed together.?

Solution:

6 coin tosses. This means,
Total observations = 36(According to binomial concept)
Required outcome → 6 Heads {H,H,H,H,H,H}
This can occur only ONCE!

Thus, required outcome = 1
Probability (6 Heads) = (1⁄2)6 = 1/64

The probability of some event happening is a mathematical (numerical) representation of how likely it is to happen, where a probability of 1 means that an event will always happen, while a probability of 0 means that it will never happen. Classical probability problems often need to you find how often one outcome occurs versus another, and how one event happening affects the probability of future events happening. When you look at all the things that may occur, the formula (just as our coin flip probability formula) states that

probability = (no. of successful results) / (no. of all possible results).

Take a die roll as an example. If you have a standard, 6-face die, then there are six possible outcomes, namely the numbers from 1 to 6. If it is a fair die, then the likelihood of each of these results is the same, i.e., 1 in 6 or 1 / 6. Therefore, the probability of obtaining 6 when you roll the die is 1 / 6. The probability is the same for 3. Or 2. You get the drill. If you don't believe me, take a dice and roll it a few times and note the results. Remember that the more times you repeat an experiment, the more trustworthy the results. So go on, roll it, say, a thousand times. We'll be waiting here until you get back to tell us we've been right all along. Go to the dice probability calculator if you want a shortcut.

But what if you repeat an experiment a hundred times and want to find the odds that you'll obtain a fixed result at least 20 times?

Let's look at another example. Say that you're a teenager straight out of middle school and decide that you want to meet the love of your life this year. More specifically, you want to ask ten girls out and go on a date with only four of them. One of those has got to be the one, right? The first thing you have to do in this situation is look in the mirror and rate how likely a girl is to agree to go out with you when you start talking to her. If you have problems with assessing your looks fairly, go downstairs and let your grandma tell you what a handsome, young gentleman you are. So a solid 9 / 10 then.

As you only want to go on four dates, that means you only want four of your romance attempts to succeed. This has an outcome of 9 / 10. This means that you want the other six girls to reject you, which, based on your good looks, has only a 1 / 10 change of happening (The sum of all events happening is always equal to 1, so we get this number by subtracting 9 / 10 from 1). If you multiply the probability of each event by itself the number of times you want it to occur, you get the chance that your scenario will come true. In this case, your odds are 210 * (9 / 10)4 * (1 / 10)6 = 0.000137781, where the 210 comes from the number of possible fours of girls among the ten that would agree. Not very likely to happen, is it? Maybe you should try being less beautiful!