Question

Монета брошена 10 раз. Найти вероятность того, что «герб» выпадет ровно 4 раза.

Answer 1

Вероятность вычисляем  двумя способами:

1).  по формуле Бернулли,  для нашего случая

P=C*p^n

C= n!/k!(n-k)! = 10*9*8*7*6*5*4!/4!*6*5*4!= 5040/24=210

k- выпадение герба

P- вероятность выпадения  герба 4 раза при 10 бросках

p- вероятность выпадения герба при одном броске -   1/2
n- общее количество бросков  n=10
P=210*(1/2)^10 = 210/1024=0,205

 

2). Классическим способом:

P = m/N,

N  - число всех равновозможных исходов  = 2^n,  (где 2 исходы бросания герб или решка, n – число бросков),

 N= 2^10=1024  

Благоприятных событий  m = C k/m = n!/k!(n-k)!= 210, где k- выпадение герба

P = 210/1024=0,205

Answer 2

To find the probability of getting exactly 4 "heads" (герб) when flipping a coin 10 times, we can use the binomial probability formula. The binomial probability formula is:

P(X = k) = C(n, k) * p^k * q^(n-k)

where:

• P(X = k) is the probability of getting exactly k successes (in this case, 4 "heads").
• C(n, k) is the number of combinations of n items taken k at a time, which is also known as the binomial coefficient. It can be calculated as C(n, k) = n! / (k! * (n-k)!).
• p is the probability of getting a "head" in a single coin flip.
• q is the probability of getting a "tail" in a single coin flip (which is equal to 1 - p).
• n is the total number of trials (in this case, 10 coin flips).
• k is the number of successful outcomes we want (in this case, 4 "heads").

For a fair coin, p = q = 0.5.

Let's calculate the probability:

p = 0.5 (probability of getting a "head") q = 1 - p = 0.5 (probability of getting a "tail") n = 10 (total number of coin flips) k = 4 (number of "heads" we want)

C(10, 4) = 10! / (4! * 6!) = 210

P(X = 4) = C(10, 4) * p^4 * q^(10-4) P(X = 4) = 210 * (0.5)^4 * (0.5)^6 P(X = 4) = 210 * 0.0625 * 0.015625 P(X = 4) = 0.328125

So, the probability of getting exactly 4 "heads" in 10 coin flips is approximately 0.3281 or 32.81%.