Question

Carson drives to school the same way each day and there are two independent traffic lights on his trip to school. He knows that there is a 30% chance that he will have to stop at the first light and an 80% chance that he will have to stop at the second light. Determine the probability that he will NOT have to stop at either light.

Answer 1

Hence, the probability is:

0.14 or 14%

Explanation:

Let A denote the event that he stop at the first light.

and B denote the event that he stops at the second light.

Let P denote the probability of an event.

Also, we are given that:

There is a 30% chance that he will have to stop at the first light.

i.e.

P(A)=30%=0.30

and

80% chance that he will have to stop at the second light.

i.e.

P(B)=80%=0.80

Now, we are asked to find:

The probability that he will NOT have to stop at either light.

i.e. we are asked to find:
`P(A^c\bigcap B^c)`

Also, the events A complement and B complement will be independent.

Hence, we get:

`P(A^c\bigcap B^c)=P(A^c)* P(B^c)`

Based on the data we have:

`P(A^c)=1-P(A)\ and\ P(B^c)=1-P(B)\\\\i.e.\\\\P(A^c)=1-0.30\ and\ P(B^c)=1-0.80\\\\i.e.\\\\P(A^c)=0.70\ and\ P(B^c)=0.20`

Hence, we get:

`P(A^c\bigcap B^c)=0.70* 0.20`

i.e.

`P(A^c\bigcap B^c)=0.14`

which in percent is:

`P(A^c\bigcap B^c)=14\%`

Answer 2

The probability he will not stop is 14%.

The probability that he will not stop in the first light is 70% while the second light holds the probability of 20%.
0.7*0.2 = 0.14 or 14%

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