# Normal Approximation to find probability of stopping at a red light at least 15 times [closed]

Samantha and Julian are driving to a party out of town. They must pass through $40$ stoplights before they reach their destination. Each hour there is $35$ minutes of green light, $10$ minutes of yellow light and $15$ minutes of red light at each stoplight. Assume all lights change randomly. During Samantha and Julian’s drive, what is the probability that they will have to stop more than $15$ times? (Samantha is driving and never brakes for yellow.)

-

## closed as off-topic by 900 sit-ups a day, martini, Mathmo123, user1729, Daniel Fischer♦Jul 30 '14 at 10:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, martini, Mathmo123, user1729, Daniel Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.

Welcome here.

As all lights change randomly, this gives us $\frac 14$ probability of red light (lets call it success)and $\frac 34$ otherwise (lets call it failure). Now you want to calculate the probaility of $15$ successes in $40$ trials . This is what binomial distribution is. http://en.wikipedia.org/wiki/Binomial_distribution

For more than $15$ success. You'll have to add up the probabilities of 16 success,... to 40 success or calculate the complementary probability. One way is to really add these up.

Alternative way is to use the normal approximation to binomial , i.e, $\text {Binomial} (n,p) \equiv \text {Normal} (np,np(1-p)$. Then the sum becomes an integral.

Hope this helps. Also you can consider continuity correction. http://en.wikipedia.org/wiki/Continuity_correction

-
Yes, but I'm not looking for 15 successes. I'm looking for at least 16 successes (more than 15 times). – user48754 Nov 9 '12 at 5:43
ok, i'll edit my answer. – TheJoker Nov 9 '12 at 5:44
thanks so much!! – user48754 Nov 9 '12 at 6:30
if you are happy with the answer, you can accept it... – TheJoker Nov 9 '12 at 7:27