A cooker company announced that the life time T of the cooker defined as the…
A cooker company announced that the life time T of the cooker defined as the amount of time (in years) the car works properly until it breaks down satisfies
P(Tt)=e-t/5 , for all t0
For example, the probability that the cooker lasts more than (or equal to) 2 years is
P(T2)=e-2/5 =0.6703
Devdas purchased that cooker and use it for two years without any problems. What is the probability that it breaks in third year?
Answer: B. 0.1813 — We want the probability that the cooker breaks during the third year given it survived the first two years, i.e. P(2 ≤ T < 3 | T ≥ 2). Given survival…
- A.
0.2813
- B.
0.1813
- C.
0.6703
- D.
0.2412
Attempted by 6 students.
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Correct answer: B
We want the probability that the cooker breaks during the third year given it survived the first two years, i.e. P(2 ≤ T < 3 | T ≥ 2).
Given survival function: P(T ≥ t) = e^{-t/5}.
Compute P(T ≥ 2) = e^{-2/5} ≈ 0.670320046.
Compute P(T ≥ 3) = e^{-3/5} ≈ 0.548811636.
Unconditional probability of failing during year 3: P(2 ≤ T < 3) = P(T ≥ 2) − P(T ≥ 3) ≈ 0.670320046 − 0.548811636 = 0.12150841.
Conditional probability given survival to year 2: (0.12150841) ÷ (0.670320046) ≈ 0.181269247 ≈ 0.1813.
Answer: approximately 0.1813.