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Hence, the probability is:
0.14 or 14%
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: [tex]P(A^c\bigcap B^c)[/tex]
Also, the events A complement and B complement will be independent.
Hence, we get:
[tex]P(A^c\bigcap B^c)=P(A^c)\times P(B^c)[/tex]
Based on the data we have:
[tex]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[/tex]
Hence, we get:
[tex]P(A^c\bigcap B^c)=0.70\times 0.20[/tex]
i.e.
[tex]P(A^c\bigcap B^c)=0.14[/tex]
which in percent is:
[tex]P(A^c\bigcap B^c)=14\%[/tex]
