Question

Пять раз брошена коcть. Определить вероятноcть того, что три раза выпадала цифра меньшая двух.

Answer 1

Меньше двух может выпасть только 1, значит: $p=C^3_5 *(1/6)^3*(5/6)^2=\\ =\frac{5!}{3!*2!}*1/216*25/36= 10*25/7776=125/3888$

Answer 2

Problem Statement

You have thrown a die five times and want to determine the probability that the digit rolled is smaller than two on three occasions.

Solution

To solve this problem, we need to calculate the probability of rolling a digit smaller than two on three out of five throws of a die.

The probability of rolling a digit smaller than two on a single throw of a fair six-sided die is 1/6, as there is only one digit (1) that satisfies this condition out of the six possible outcomes (1, 2, 3, 4, 5, 6).

To calculate the probability of a specific number of successful outcomes in a series of independent events, we can use the binomial probability formula:

P(k) = C(n, k) * p^k * (1-p)^(n-k)

Where: - P(k) is the probability of getting exactly k successful outcomes, - C(n, k) is the number of combinations of n items taken k at a time, - p is the probability of a single successful outcome, - (1-p) is the probability of a single unsuccessful outcome, - n is the total number of trials.

In this case, we want to calculate the probability of getting three successful outcomes (rolling a digit smaller than two) out of five trials (throws of the die), with a probability of success (p) of 1/6.

Using the binomial probability formula, we can calculate the probability as follows:

P(3) = C(5, 3) * (1/6)^3 * (5/6)^(5-3)

Calculating the values:

C(5, 3) = 5! / (3! * (5-3)!) = 10

Substituting the values into the formula:

P(3) = 10 * (1/6)^3 * (5/6)^2

Calculating the probability:

P(3) = 10 * (1/216) * (25/36) = 250/7776 ≈ 0.0321

Therefore, the probability of rolling a digit smaller than two on three out of five throws of a die is approximately 0.0321.