Recent content by sol59

Thank you so much for your help! I've managed to solve all other problems but this one was too difficult for me:) Thanks again:)!

0,736*0.716=0.527

1-0.264=0.736

The probability of no arrivals in (9:10, 9:50) is P(Y=0)=e-4/3=0.264 Thank you for your patience.

0?

That's interesting...so there is an unlimited number of students that can arrive in that interval. But how to express it mathematically?

I think it meets that requirement but it is also said that 2 students per hour come into the class.

I don't know:( P(Y=2)=0.234 but I don't understand what to do with it

9:00-9:10 P(Y=0)=((2*1/6)0*(e-1/3))/0!=0.716 9:10-9:50 1 student P(Y=1)=((2*2/3)1*(e-4/3)/1!=0.351

0 (and between 9:50-10 too)...but I still don't know how to solve it.

9:10-9:50?

The fiftieth minute minus the time when the first student came?

The second student has to come in the fiftieth minute at the latest?

The second student has to come 40 minutes later at most after the first student.

Well I need to find out if the time of waiting for arrival of the second student is in the interval <= 50 minutes. And the arrival of the first student has to be in the interval >= 10 minutes.