Question

Найти вероятность того, что в пяти независимых испытаниях событие появиться ровно три раза, если в каждом испытании вероятность появления события равна 0.9

Answer 1

P=0,9,
q = 0,1;
(p+q)^5 = (p^5) + 5*(p^4)*q +10*(p^3)*(q^2) + 10*(p^2)*(q^3) + 5*p*(q^4) + +q^5;
искомая вероятность = 10*(p^3)*(q^2) = 10*(0,9^3)*(0,1^2)=
= 10*0,9*0,9*0,9*0,1*0,1 = 9*0,81*0,01 = 9*81/10000 = 729/10000 = 0,0729.

Answer 2

Calculating the Probability of an Event Occurring Exactly Three Times in Five Independent Trials

To calculate the probability of an event occurring exactly three times in five independent trials, we can use the binomial probability formula. The formula for the probability of getting exactly k successes in n trials, when the probability of success in each trial is p, is given by:

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

Where: - n is the number of trials - k is the number of successes - p is the probability of success in each trial - (n choose k) is the binomial coefficient, which represents the number of ways to choose k successes out of n trials

Applying the Formula to the Given Scenario

In this case, we have: - n = 5 (five independent trials) - k = 3 (exactly three successes) - p = 0.9 (probability of the event occurring in each trial)

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

P(X=3) = (5 choose 3) * (0.9^3) * ((1-0.9)^(5-3))

Calculation

Using the formula: P(X=3) = (5 choose 3) * (0.9^3) * (0.1^2)

From the search results, we have the following relevant information: - Source provides the formula for calculating the probability of exactly k successes in n trials, given the probability of success in each trial. - Source also gives an example with n=5 and p=0.8, which demonstrates the calculation process for a similar scenario.

Plugging the values into the formula: P(X=3) = (5 choose 3) * (0.9^3) * (0.1^2) = 10 * 0.729 * 0.01

Therefore, the probability of the event occurring exactly three times in five independent trials, with a probability of 0.9 for each trial, is approximately 0.06561.

This calculation aligns with the information provided in Source which demonstrates a similar calculation process for a different scenario.