Question

A person's home to work commute on a wintery morning involves traveling on Highway A, Highway B, and then Highway C. The person cannot make it to work if one or more of the roadway segments is icy. The probability of Highway A being icy is 0.1, Highway B being icy is 0.15, and Highway C being icy is 0.15. Assume the highways being icy is independent. Calculate the probability that the person will not make it to work.

Answer 1

Probability = 0.35

Explanation:

Given :-

• Probability (Highway A Icy) = 0.1
• Probability (Highway B Icy) = 0.15
• Probability (Highway C icy) = 0.15

So :-

• Probability (Highway A not icy) = 1 - 0.1 = 0.9
• Probability (Highway B not icy) = 1 - 0.15 = 0.85
• Probability (Highway C not icy) = 1 - 0.15 = 0.85

Probability of person not getting to work timely

= Probability [Even one of the highway A, B or C is icy]

= 1 - Probability [None of Highway A, B, C is icy]

Since these are independent events, So :-

= Prob. [Highway A not icy & Highway B not icy & Highway C not icy]

= 1 - ( 0.9 x 0.85 x 0.85)

= 1 - 0.65

= 0.35