Question

1. Производится пять независимых выстрелов по цели. Вероятность попадания при каждом выстреле равна 1/3 . Чему равна вероятность того, что число попаданий будет заключено в пределах от 1 до 3?

Answer 1

Ответ:

Пошаговое объяснение:

Вероятность попадания

p = 1/3

Вероятность промаха:

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

1 попадание (один раз попал, четыре раза промахнулся):

P(1) = (1/3)¹·(2/3)⁴ = 2⁴ / 3⁵

2 попадания (два  раза попал, три раза промахнулся):

P(2) = (1/3)²·(2/3)³ = 2³ / 3⁵

3 попадания (три  раза попал, два раза промахнулся):

P(3) = (1/3)³·(2/3)² = 2² / 3⁵

Искомая вероятность:

P = P(1)  + P(2) + P(3)

P = 2⁴ / 3⁵ + 2³ / 3⁵ + 2² / 3⁵ =

= (2⁴ + 2³ + 2²) / 3⁵ =  28 / 243

Answer 2

Probability of hitting the target

To calculate the probability that the number of hits will be between 1 and 3, we need to consider the different combinations of hits that fall within this range.

Let's denote a hit as "H" and a miss as "M". The possible combinations of hits and misses that fall within the range of 1 to 3 hits are as follows:

- 1 hit and 4 misses: HMMM - 2 hits and 3 misses: HHMM, HMHM, MHHM - 3 hits and 2 misses: HHHM, HHMH, HMHH, MHHH

To calculate the probability of each combination, we need to multiply the probability of hitting (1/3) by the probability of missing (2/3) for each hit or miss in the combination.

Let's calculate the probabilities for each combination:

- 1 hit and 4 misses: (1/3) * (2/3) * (2/3) * (2/3) * (2/3) = 32/243 - 2 hits and 3 misses: (1/3) * (1/3) * (2/3) * (2/3) * (2/3) = 16/243 (for each combination) - 3 hits and 2 misses: (1/3) * (1/3) * (1/3) * (2/3) * (2/3) = 8/243 (for each combination)

To find the total probability, we sum up the probabilities of each combination:

Total probability = (32/243) + 3 * (16/243) + 4 * (8/243) = 32/243 + 48/243 + 32/243 = 112/243

Therefore, the probability that the number of hits will be between 1 and 3 is 112/243.