Question

Игральную кость бросают 7 раз, какова вероятность что число очков кратко 3,выпадет 2 раза? ​

Answer 1

На игральной кости расположены числа от 1 до 6

Кратные трем: 3 и 6

Их два из шести

Значит вероятность выпадения числа очков кратного трем равна

p=(2/6)=1/3

q=1-p=1-(1/3)=(2/3)

Повторные испытания с  двумя исходами.

По формуле Бернулли:

$P_{7}(2)=C^{2}_{7}p^{2}q^{7-2}=\frac{7!}{2!(7-2)!}\cdot (\frac{1}{3} )^2\cdot (\frac{2}{3} )^5=\frac{5!\cdot 6\cdot 7}{2!5!}\cdot \frac{1}{9} \cdot \frac{32}{243} =\frac{224}{729}$

Answer 2

Probability of Rolling a Number Less Than 3 Twice in 7 Rolls of a Die

To calculate the probability of rolling a number less than 3 (i.e., 1 or 2) twice in 7 rolls of a die, we can use the concept of binomial probability. The formula for binomial probability is:

P(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k))

Where: - n = number of trials (in this case, 7 rolls) - k = number of successful outcomes (in this case, 2 times rolling a number less than 3) - p = probability of success on an individual trial (in this case, the probability of rolling a number less than 3)

Calculation

The probability of rolling a number less than 3 on a fair six-sided die is 2/6 or 1/3. Using this information, we can calculate the probability of rolling a number less than 3 twice in 7 rolls.

P(X = 2) = (7 choose 2) * ((1/3)^2) * ((2/3)^(7-2))

P(X = 2) = (7 choose 2) * (1/9) * (32/243)

Using the combination formula (n choose k) = n! / (k! * (n-k)!) where n! denotes the factorial of n, we can calculate the combination.

P(X = 2) = (7! / (2! * 5!)) * (1/9) * (32/243)

P(X = 2) = (7*6 / 2*1) * (1/9) * (32/243)

P(X = 2) = 21 * (1/9) * (32/243)

P(X = 2) ≈ 0.266

Therefore, the probability of rolling a number less than 3 twice in 7 rolls of a die is approximately 0.266.

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