A coin is tossed 7 times. What is the probability of getting four heads. My teacher said that there is a formula involved but i don't know which one. How do i set it up?
With a fair coin the probability of a given 7-flip series is equal to ANY 7-flip series. That is, P(HHHHTTT) = P(TTTHHHH) = P(THTHTHT)....
There are 2^7 = 128 possible 7-filp series. This is the denominator.
There are 4-choose-7 ways to have a 4 heads in 7 filps. 4-choose-7 is 7!/(4!*(7-4)!) = 35.
So, the probability is 35/128
To determine the probability of getting four heads when a coin is tossed seven times, you can use the binomial probability formula. This formula calculates the probability of a specific number of successes (heads) in a fixed number of Bernoulli trials (coin tosses).
The binomial probability formula is:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes (heads),
C(n, k) is the number of combinations of n items taken k at a time,
p is the probability of getting a success (heads) on a single trial,
n is the total number of trials (coin tosses), and
k is the number of successes (heads) you are interested in.
In this case, you want to find P(X=4) for n=7 tosses, so you will use k=4. The probability of flipping a head is 1/2, so p=1/2.
Let's plug in the values:
P(X=4) = C(7, 4) * (1/2)^4 * (1-(1/2))^(7-4)
C(7, 4) can be calculated as follows:
C(7, 4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
Now we have all the values we need:
P(X=4) = 35 * (1/2)^4 * (1/2)^3
Simplifying further:
P(X=4) = 35 * (1/16) * (1/8)
P(X=4) = 35/128
Therefore, the probability of getting four heads in seven coin tosses is 35/128.