Question

Монету подбрасывают 5 раз. найдите вероятность, что монета упадет лицом на герб 3 раза.

Answer 1

Problem Statement

We are given that a coin is tossed 5 times, and we need to find the probability that the coin lands on heads (H) 3 times.

Solution

To solve this problem, we can use the concept of binomial probability. The binomial probability formula is given by:

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

Where: - P(X = k) is the probability of getting exactly k successes - C(n, k) is the number of combinations of n items taken k at a time - p is the probability of success on a single trial - q is the probability of failure on a single trial - n is the total number of trials

In this case, the probability of getting heads (H) on a single toss is 1/2, and the probability of getting tails (T) is also 1/2. So, p = 1/2 and q = 1/2.

We want to find the probability of getting 3 heads (H) out of 5 tosses. So, k = 3 and n = 5.

Plugging these values into the binomial probability formula, we get:

P(X = 3) = C(5, 3) * (1/2)^3 * (1/2)^(5-3)

Simplifying the expression:

P(X = 3) = C(5, 3) * (1/2)^3 * (1/2)^2

Now, let's calculate the values:

- C(5, 3) = 5! / (3! * (5-3)!) = 10 - (1/2)^3 = 1/8 - (1/2)^2 = 1/4

Substituting these values back into the equation:

P(X = 3) = 10 * (1/8) * (1/4)

Simplifying further:

P(X = 3) = 10/32 = 5/16

Therefore, the probability that the coin lands on heads (H) 3 times out of 5 tosses is 5/16.