Answer:
a) 0.15
b) 0.2
c) 0.6
Step-by-step explanation:
We are given the following in the question:
A: Stopping at first signal
B: Stopping at second signal
P(A) = 0.35
P(B) = 0.55
Probability that he must stop at at least one of the two signals is 0.75
[tex]P(A\cup B) = 0.75[/tex]
a) P(at both signals)
[tex]P(A\cup B) = P(A) + P(B) - P(A\cap B)\\0.75 = 0.35 + 0.55 - P(A\cap B)\\P(A\cap B) = 0.35 + 0.55 - 0.75 = 0.15[/tex]
0.15 is the probability that motorist stops at both signals.
b) P(at the first signal but not at the second one)
[tex]P(A\cap B') = P(A) - P(A\cap B)\\P(A\cap B') = 0.35 - 0.15 = 0.2[/tex]
0.2 is the probability that motorist stops at the first signal but not at the second one.
c) P(at exactly one signal)
[tex]P(A\cap B') + P(A\cap 'B) = P(A\cup B) - P(A\cap B) \\P(A\cap B') + P(A\cap 'B) = 0.75 - 0.15 = 0.6[/tex]
0.6 is the probability that the motorist stops at exactly one signal.
