Question

В квартире 8 электролампочек. Вероятность работы лампочки в течение года равна 0,9. Какова вероятность того, что в течение года придется заменить не менее половины лампочек.?

Answer 1

0,9=90%-шанс того, что лампочка работает
100%-90%=10%=0,1 шанс того, что лампочка сломана
0,1*8=0,8=80%

Answer 2

Calculation of the Probability

To calculate the probability of having to replace at least half of the light bulbs within a year, we can use the binomial distribution formula. The binomial distribution is used to calculate the probability of a certain number of successes in a fixed number of independent Bernoulli trials.

In this case, each light bulb has a probability of 0.9 of working throughout the year, and we have 8 light bulbs in total. We want to find the probability of having to replace at least 4 light bulbs (half of 8).

The formula for the binomial distribution is:

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

Where: - P(X = k) is the probability of having 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 in a single trial - n is the total number of trials

In this case, k is the number of light bulbs that need to be replaced (4 or more), p is 0.9, and n is 8.

Let's calculate the probability using this formula.

Calculation

To calculate the probability, we need to sum the probabilities of having 4, 5, 6, 7, or 8 light bulbs that need to be replaced.

P(X >= 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Using the binomial distribution formula, we can calculate each term:

P(X = 4) = C(8, 4) * (0.9)^4 * (1-0.9)^(8-4) P(X = 5) = C(8, 5) * (0.9)^5 * (1-0.9)^(8-5) P(X = 6) = C(8, 6) * (0.9)^6 * (1-0.9)^(8-6) P(X = 7) = C(8, 7) * (0.9)^7 * (1-0.9)^(8-7) P(X = 8) = C(8, 8) * (0.9)^8 * (1-0.9)^(8-8)

Let's calculate each term and sum them up to find the probability.

Calculation Steps

P(X = 4) = C(8, 4) * (0.9)^4 * (1-0.9)^(8-4) P(X = 4) = 70 * (0.9)^4 * (0.1)^4 P(X = 4) = 0.2835

P(X = 5) = C(8, 5) * (0.9)^5 * (1-0.9)^(8-5) P(X = 5) = 56 * (0.9)^5 * (0.1)^3 P(X = 5) = 0.3114

P(X = 6) = C(8, 6) * (0.9)^6 * (1-0.9)^(8-6) P(X = 6) = 28 * (0.9)^6 * (0.1)^2 P(X = 6) = 0.1866

P(X = 7) = C(8, 7) * (0.9)^7 * (1-0.9)^(8-7) P(X = 7) = 8 * (0.9)^7 * (0.1)^1 P(X = 7) = 0.0576

P(X = 8) = C(8, 8) * (0.9)^8 * (1-0.9)^(8-8) P(X = 8) = 1 * (0.9)^8 * (0.1)^0 P(X = 8) = 0.4305

P(X >= 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) P(X >= 4) = 0.2835 + 0.3114 + 0.1866 + 0.0576 + 0.4305 P(X >= 4) = 1.2696

Therefore, the probability of having to replace at least half of the light bulbs within a year is approximately 1.2696.

Please note that the probabilities have been rounded for simplicity.