Question

Ming's travel plans include three independent flights. The chance of a fight being on time is (3)/(4). What is the probability that all of Ming's flights will be on time?

Answer 1

The probability that all three of Ming's flights will be on time is 27/64, which is approximately 0.421875 or 42.19%.

Step-by-step explanation:

The question asks for the probability that all of Ming's three flights will be on time, given that the chance of a single flight being on time is (3/4). Since the flights are independent, we can multiply the probabilities for each flight to find the combined probability. Therefore, the probability that all three flights will be on time is calculated as follows:

P(All three flights on time) = P(Flight 1 on time) × P(Flight 2 on time) × P(Flight 3 on time)

P(All three flights on time) = (3/4) × (3/4) × (3/4)

P(All three flights on time) = (27/64)

The probability that all of Ming's flights will be on time is 27/64, which can also be written as approximately 0.421875 or 42.19% when expressed as a percentage.

Answer 2

27/64

Step-by-step explanation:

If each flight's punctuality is independent of the others, the probability that all of Ming's flights will be on time can be calculated by multiplying the individual probabilities.

Given that the probability of a single flight being on time is
`(3)/(4)` the probability of all three flights being on time is:

`\text{Probability of all flights on time} = \left((3)/(4)\right) * \left((3)/(4)\right) * \left((3)/(4)\right)`

`\text{Probability of all flights on time} = \left((3)/(4)\right)^3`

Calculating this:

`\text{Probability of all flights on time} = \left((3)/(4)\right) * \left((3)/(4)\right) * \left((3)/(4)\right) = (27)/(64)`

Therefore, the probability that all three of Ming's flights will be on time is
`(27)/(64)`