Problem
Let’s consider data collected over multiple days, revealing the prevalence of defects in cars. The findings indicate that 20% of the cars exhibit defect 1, 25% have defect 2, and 30% possess defect 3. Additionally, 5% of the cars have both defects 1 and 2, 7.5% exhibit both defects 2 and 3, and 6% showcase both defects 1 and 3. Notably, a minority of 1.5% are affected by all three defects simultaneously.
Let’s compute the probabilities for the following events concerning a randomly selected car:
- The probability that defect 1 did not occur.
- The probability that at least one defect occurs.
- The probability that no defect occurs.
- The probability that defects 1 and 3 occur, but defect 2 does not.
Solution
n(D1) = 0.2
n(D2) = 0.25
n(D3) = 0.3
n(D1∩D2) = 0.05
n(D2∩D3) = 0.075
n(D1∩D3) = 0.06
n(D1∩D2∩D3) = 0.015
1) The probability that defect 1 did not occur.
Here we have to find D‘.
The probability that defect 1 did not occur = 1 – The
probability that defect 1 did occur
= 1 – 0.2
= 0.8
2)The probability that at least one defect occurs = P(D1∪D2∪D3)
= P(D1) + P(D2) + P(D3) – P(D1∩D2)
– P(D2∩D3) – P(D1∩D3)
+ P(D1∩D2∩D3)
=0.2 + 0.25 + 0.3 – 0.05 – 0.075 – 0.06 + 0.015
=0.58
3) The probability that no defect occurs
= 1 – The probability that at least one defect occurs
=1 – 0.58
=0.42
4) The probability that defects 1 and 3 occur, but defect 2 does not.
= P(D2∩D3) – P(D1∩D2∩D3)
= 0.075 – 0.015
=0.06
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