# The probability [closed]

The probability that I have to wait at the traffic lights on my way to school is 0.25. Find the probability that ,on two consecutive mornings,I have to wait on at least one morning.

-

## closed as off-topic by Normal Human, Claude Leibovici, hardmath, M. Vinay, drhabJul 2 '14 at 8:46

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." – Normal Human, Claude Leibovici, hardmath, M. Vinay, drhab
If this question can be reworded to fit the rules in the help center, please edit the question.

Is "The probability" really the most useful question title you can come up with? – Henning Makholm Jan 27 '13 at 15:11

The probability that on a certain day, you don't have to wait is $$1-\frac14=\frac34$$

So, the probability that on two consecutive mornings, you don't have to wait at all is $$\left(\frac34\right)^2=\frac9{16}$$

So, the probability that on two consecutive mornings, you have to wait on at least one morning is $$1-\frac9{16}=\frac7{16}$$

Alternatively, if $W$ represents waiting and $N$ represents no wait,

the probability of waiting is $P(W)=\frac14$ and the probability of not waiting $P(N)=1-\frac14=\frac34$

Hence, our required probability is $$P(W)P(N)+P(W)P(W)+P(N)P(W)=\frac14\frac34+\frac14\frac14+\frac34\frac14=\frac{3+1+3}{16}=\frac7{16}$$

-
Why is the probability of not having to wait two mornings in a row the square of the probability of not having to wait on any given morning? – Dilip Sarwate Jan 27 '13 at 14:53
What is probability of throwing two heads in two throw of one a single unbiased coin? – lab bhattacharjee Jan 27 '13 at 14:55
"What is probability of throwing two heads in two throw of one a single unbiased coin?" You tell me; I don't know. There is the assumption of independent trials which needs to be made, and this is much more easy to justify in my mind when talking of coin tosses than the corresponding assumption about having to wait at traffic lights on two successive mornings. To put it another way, please edit your answer to at least mention that you are assuming independence of the events on two successive mornings. – Dilip Sarwate Jan 27 '13 at 15:03
How can the waiting on two successive mornings be dependent on each other? – lab bhattacharjee Jan 27 '13 at 15:05
I remember reading about a (Danish?) mathematician who, as a prisoner-of-war in WWII, actually tossed a coin a large number of times and recorded the results. After thousands of tosses, the coin was so beat up that it was hard to tell the sides apart. – Dilip Sarwate Jan 27 '13 at 15:18