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
![P(X)=0.9\\\\P(Y)=0.8\\](https://tex.z-dn.net/?f=P%28X%29%3D0.9%5C%5C%5C%5CP%28Y%29%3D0.8%5C%5C)
a) We calculate the probability that the sprinkler head will be activated.
![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\\](https://tex.z-dn.net/?f=P%28X%5Ccup%20Y%29%3DP%28X%29%2BP%28Y%29-P%28X%5Ccap%20Y%29%5C%5C%5C%5CP%28X%5Ccup%20Y%29%3D0.9%2B0.8-P%28X%29P%28Y%29%5C%5C%5C%5CP%28X%5Ccup%20Y%29%3D1.7-0.9%5Ccdot%200.8%5C%5C%5C%5CP%28X%5Ccup%20Y%29%3D0.98%5C%5C)
Therefore, the probability is P=0.98.
b) We calculate the probability that the sprinkler head will not be activated.
![P=1-P(X\cup Y)\\\\P=1-0.98\\\\P=0.02\\](https://tex.z-dn.net/?f=P%3D1-P%28X%5Ccup%20Y%29%5C%5C%5C%5CP%3D1-0.98%5C%5C%5C%5CP%3D0.02%5C%5C)
Therefore, the probability is P=0.02.
c) We calculate the probability that both activation devices will work properly.
![P=P(X)\cdot P(Y)\\\\P=0.9\cdot 0.8\\\\P=0.72](https://tex.z-dn.net/?f=P%3DP%28X%29%5Ccdot%20P%28Y%29%5C%5C%5C%5CP%3D0.9%5Ccdot%200.8%5C%5C%5C%5CP%3D0.72)
Therefore, the probability is P=0.72.
d) We calculate the probability that only the device with reliability 0.9 will work properly.
![P=P(X)\cdot P(Y^c)\\\\P=0.9\cdot 0.2\\\\P=0.18\\](https://tex.z-dn.net/?f=P%3DP%28X%29%5Ccdot%20P%28Y%5Ec%29%5C%5C%5C%5CP%3D0.9%5Ccdot%200.2%5C%5C%5C%5CP%3D0.18%5C%5C)
Therefore, the probability is P=0.18.