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.

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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lidaralbany lidaralbany · Answer 2 · 2019-07-23T07:34:28-04:00

Answer:

Hence, the probability is:

0.14 or 14%

Step-by-step 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:

\(P(A^c\bigcap B^c)=P(A^c)\times P(B^c)\)

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:

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

i.e.

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

which in percent is:

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