TomJerry
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Question:
The probability that a driver stopping at petrol station will have only his car tyre checked is 0.012, the probability that he will have the oil checked is 0.29 and the probability that he will have both oil and tyre's checked is 0.07. What is the probability that a driver stopping at the station will have neither his car tyres nor oil checked
Solution:
Let A be he checks his tyres
P(A) = 0.012
P(A') = 1 - 0.012
Let B be he checks oil
P(B) = 0.29
P(B') = 1 - 0.29
P(A intersection B) = 0.07
We need to find P(A' union B')
P(A' union B') = P(A') + P(B') - P(A' intersection B')
I m stuck here don't know how to go forward from here
The probability that a driver stopping at petrol station will have only his car tyre checked is 0.012, the probability that he will have the oil checked is 0.29 and the probability that he will have both oil and tyre's checked is 0.07. What is the probability that a driver stopping at the station will have neither his car tyres nor oil checked
Solution:
Let A be he checks his tyres
P(A) = 0.012
P(A') = 1 - 0.012
Let B be he checks oil
P(B) = 0.29
P(B') = 1 - 0.29
P(A intersection B) = 0.07
We need to find P(A' union B')
P(A' union B') = P(A') + P(B') - P(A' intersection B')
I m stuck here don't know how to go forward from here
