probability question

Question

Probability of stopping at first traffic light is 0.7 and the probability of stopping at 2nd traffic light is 0.4. what is the probability of stopping at

a)both light

b) none of the traffic light

 

Answer 1

Hi Parveen.  There is an assumption that must be made when answering this.  We assume that both events are independent of one another (which I am not sure is true).  Assuming this, when we talk about the probability of event1 occurring AND event2 occurring we can multiply the individual probabilities.  So in this case:

 

P(stopping at 1st) = 0.7

P(stopping at 2nd) = 0.4

 

P(stopping at 1st) AND P(stopping at 2nd) = 0.7 * 0.4 = 0.28

 

As for part b let's reword it.  This is the same as saying "What is the probability of not stopping at light one AND not stopping at light two".  We don't have these probabilities specified, but with a little thought we can get them.  Since there are only two options for each event -- stopping or not stopping -- we can say that the probability of not stopping is simply 1 - P(stopping).  Thus:

 

P(not stopping at 1) = 1-P(stopping at 1) = 1 - 0.7 = 0.3

P(not stopping at 2) = 1 - 0.4 = 0.6 

 

Therefore, P(stopping at neither light) = P(not stopping at 1) AND P(not stopping at 2) = 0.3 * 0.6 = 0.18