Question

Предполагая, что для шахматиста А в каждой партии равновероятны три ис-хода: выигрыш, ничья и проигрыш, найдите вероятность того, что из четырех партий шахматист А: а) не проиграет ни одной партии; б) проиграет хотя бы две партии.

Answer 1

А)Выйгрыш в каждой из партий имеет вероятность 1/3,а раз у него четыре партии,то 1/3^4=1/81
б)проигрыш в каждой партии так же имеет вероятность 1/3,то вероятность того,что он проиграет две будет равна 1/3*1/3=1/9

Answer 2

Calculating the Probability for Chess Player A

To calculate the probability for chess player A in each scenario, we can use the given information that in each game, there are three equally likely outcomes: win, draw, or loss.

a) Probability of Not Losing Any Game

The probability of not losing any game out of four can be calculated using the binomial probability formula, which is: \[ P(X=k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \] where: - n = total number of games - k = number of successful outcomes - p = probability of success in a single game

In this case, the probability of not losing any game is the same as winning or drawing all four games.

The probability of not losing any game is: \[ P(X=0) = \binom{4}{0} \times \left(\frac{2}{3}\right)^0 \times \left(\frac{1}{3}\right)^4 \]

Using the binomial coefficient formula, \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), we can calculate the probability.

b) Probability of Losing at Least Two Games

The probability of losing at least two games can be calculated by finding the complement of the event "not losing at least two games." This is the same as finding the probability of losing one or zero games.

The probability of losing at least two games is: \[ P(X \geq 2) = 1 - P(X=0) - P(X=1) \]

We can calculate the probability of losing at least two games using the binomial probability formula for losing one game and then subtracting the result from 1.

Let's calculate these probabilities.