Question

Из 20 филиалов банка 10 являются региональными. Для проверки случайным образом выбраны 5 филиалов. Каков вероятность того, что среди них окажется: а) не менее 3-х региональных филиалов; б) хотя бы один региональный

Answer 1

$a)\; \; P= \frac{C_{10}^3+C_{10}^4+C_{10}^5}{C_{20}^5} = \frac{\frac{10\cdot 9\cdot 8}{3!}+\frac{10\cdot 9\cdot 8\cdot 7}{4!}+\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5!}}{\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{5!}} =\\\\= \frac{120+210+252}{15504} =0,0375387\approx 0,0375\\\\b)\; \; P=1-\frac{C_{10}^5}{C_{20}^5}=1- \frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{20\cdot 19\cdot 18\cdot 17\cdot 16} \approx 0,9837$

Answer 2

Probability of having at least 3 regional branches out of 5 randomly selected branches

To calculate the probability of having at least 3 regional branches out of 5 randomly selected branches, we need to consider the total number of ways to select 5 branches out of 20 and the number of ways to select at least 3 regional branches out of 10.

The total number of ways to select 5 branches out of 20 can be calculated using the combination formula:

Total number of ways to select 5 branches out of 20 = C(20, 5) = 20! / (5! * (20-5)!) = 15,504

The number of ways to select at least 3 regional branches out of 10 can be calculated by considering the combinations of selecting 3, 4, or 5 regional branches out of 10:

Number of ways to select 3 regional branches out of 10 = C(10, 3) = 10! / (3! * (10-3)!) = 120

Number of ways to select 4 regional branches out of 10 = C(10, 4) = 10! / (4! * (10-4)!) = 210

Number of ways to select 5 regional branches out of 10 = C(10, 5) = 10! / (5! * (10-5)!) = 252

Therefore, the total number of ways to select at least 3 regional branches out of 10 is:

Total number of ways to select at least 3 regional branches out of 10 = 120 + 210 + 252 = 582

Now, we can calculate the probability by dividing the number of favorable outcomes (selecting at least 3 regional branches) by the total number of possible outcomes (selecting any 5 branches):

Probability of having at least 3 regional branches out of 5 randomly selected branches = 582 / 15,504 ≈ 0.0376

So, the probability of randomly selecting 5 branches and having at least 3 of them be regional branches is approximately 0.0376.

Probability of having at least one regional branch out of 5 randomly selected branches

To calculate the probability of having at least one regional branch out of 5 randomly selected branches, we can use the complement rule. The complement of having at least one regional branch is having no regional branches at all.

The number of ways to select 5 non-regional branches out of 10 can be calculated using the combination formula:

Number of ways to select 5 non-regional branches out of 10 = C(10, 5) = 10! / (5! * (10-5)!) = 252

Therefore, the number of ways to have at least one regional branch out of 5 randomly selected branches is:

Number of ways to have at least one regional branch out of 5 randomly selected branches = Total number of possible outcomes - Number of ways to select 5 non-regional branches out of 10 = 15,504 - 252 = 15,252

Now, we can calculate the probability by dividing the number of favorable outcomes (having at least one regional branch) by the total number of possible outcomes (selecting any 5 branches):

Probability of having at least one regional branch out of 5 randomly selected branches = 15,252 / 15,504 ≈ 0.9838

So, the probability of randomly selecting 5 branches and having at least one of them be a regional branch is approximately 0.9838.