Answer :
Answer:
a) Therefore, the probability is P=0.98.
b) Therefore, the probability is P=0.02.
c) Therefore, the probability is P=0.72.
d) Therefore, the probability is P=0.18.
Step-by-step explanation:
We know that: One type has a reliability of 0.9; that is, the probability that it will activate the sprinkler when it should is 0.9.
The other type, which operates independently of the first type, has a reliability of 0.8.
We get
[tex]P(X)=0.9\\\\P(Y)=0.8\\[/tex]
a) We calculate the probability that the sprinkler head will be activated.
[tex]P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.9+0.8-P(X)P(Y)\\\\P(X\cup Y)=1.7-0.9\cdot 0.8\\\\P(X\cup Y)=0.98\\[/tex]
Therefore, the probability is P=0.98.
b) We calculate the probability that the sprinkler head will not be activated.
[tex]P=1-P(X\cup Y)\\\\P=1-0.98\\\\P=0.02\\[/tex]
Therefore, the probability is P=0.02.
c) We calculate the probability that both activation devices will work properly.
[tex]P=P(X)\cdot P(Y)\\\\P=0.9\cdot 0.8\\\\P=0.72[/tex]
Therefore, the probability is P=0.72.
d) We calculate the probability that only the device with reliability 0.9 will work properly.
[tex]P=P(X)\cdot P(Y^c)\\\\P=0.9\cdot 0.2\\\\P=0.18\\[/tex]
Therefore, the probability is P=0.18.