Here is the problem:
To get to work, a commuter must cross train tracks. The time the train arrives varies slightly from day to day , but the commuter estimates he'll get stoped on about 15% of work days. During a certain 5-day work week , what is the probability that he

a) gets stopped on Monday and again on Tuesday. What I did here is I multiplied the prob of Tuesday and Monday (1/5*1/5= 1/25) and then found 15 % out of it. So the answer is .006. Is it right?

B) gets stopped for the first time on Tuesday?
First I found the probability that it will not come on Monday (1-1/5=4/5) and I have no idea what to do next... Can you help me?
Thank you!:wave:

a) this is .15 * .15 (15% chance on any day of the week, so this is the probability of getting stopped on Monday and Tuesday)

b) this is .85 * .15 (in order for the first stop to be on Tuesday, he must not get stopped on Monday, so this is the probability of not getting stopped on Monday and getting stopped on Tuesday)

Thank you very much for your help! Now I got it...

Also there is a question for the same problem:

D) Gets stopped at least once during the week?

Is not it P(Ac)= 1- .85= .15 And then P(A) = .15*.15=0.0225

Is it right?

The probability of getting stopped at least once is the same as 1 minus the probability of never getting stopped at all.

P(Never getting stopped) = not stopped on Monday AND not stopped on Tuesday AND ... AND not stopped on Friday

= .85 *.85 * .85 * .85 *.85 = .85^5 = .4437

1 - P(never getting stopped) = 1 - .4437 = .5563

which is also the probability of not getting stopped 4 times or fewer, so you could use the cumulative probability distribution function of a binomial experiment where p=.84, trials = 5 and successes =4 or less to get .5563!

::yup: Sorry, could not resist, and needed a break from grading exams...

cheers
jerry