Answer: 0.1
Step-by-step explanation:
Let F = must stop at first signal .
F’ = do not have to stop at first signal .
S = must stop at second signal .
S’ = do not have to stop at second signal.
As given , we have
P(F)=0.40
P(S)=0.50
P(F∪S)=0.60
Use formula , [tex]P(F\cap S)=P(F)+P(S)-P(F\cup S)[/tex]
i.e. [tex]P(F\cap S)=0.40+0.50-0.60=0.30[/tex]
The probability that she must stop at the first signal but not the second signal :
[tex]P(F\cap S')=P(F)-P(F\cap S)=0.40-0.30=0.10[/tex]
Hence, the probability that she must stop at the first signal but not the second signal= 0.1
