On texas avenue between university drive and george bush drive, accidents occur according to a poisson process at a rate of three accidents per week. (a) what is the probability that a given day has no accidents? (b) whenever there is an accident, the risk is 1 in 8 that personal injury will result. what is the probability that february (which has four weeks) has at least one accident with a personal injury?

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.

too hard the other guy did it

step-by-step explanation: