Question

1) Saturn's moon, Titan, is known to have clear evidence of stable bodies of surface liquid. New data received indicates that there is a 27.28% chance of there being life within of one of the bodies of liquid. An exploratory probe is launched to gather more data. During the trip, an electrical failure compromises some of the craft's instrumentation. Therefore, it is believed that every 4th piece of data (on average) collected by the probe should be viewed as false and/or inconsequential. The probe sends back a data package to suggest that there are indeed signs of life in one of the bodies of liquid on Titan. (25pts.) a. What is the probability that there is not life on Titan? b. A new probe with a more robust software system is launched a year later and sends back a new data package with a reliability rate of 84.2%. What is the new expectation of there being no life within one of the bodies of surface liquid?

Answer 1

The total probablity that probe sends a data to suggest signs of life is the sum of

(i) Life exists, and probe sends correct data

(ii) Life does not exist, and probe sends false i.e. inconsequential data

The required probability is the Bayesian probability that life does not exist, given that probe sent a positive sign

$\mathbb{P}(\textup{Life-sign})=\mathbb{P}(\textup{Life})\times \mathbb{P}(\textup{Correct})+\mathbb{P}(\textup{No-life})\times \mathbb{P}(\textup{False})$

$=0.2728\times \frac{3}{4}+(1-0.2728)\times \frac{1}{4}=0.3864$

$\mathbb{P}(\textup{No-life}|\textup{Life-sign})=\frac{\mathbb{P}(\textup{No-life }\cap \textup{ Life-sign})}{\mathbb{P}(\textup{Life-sign})}$

P(No-life) P(False) P(Life-sign) (1 -0.2728) 1/4 + 0.3864 0.4705

In second part just solve above problem with 0.842 instead of 3/4

P(Life-sign) = 0.2728 x 0.842 + (1 - 0.2728) X (1 - 0.842) = 0.3445952

$\mathbb{P}(\textup{No-life}|\textup{Life-sign})=\frac{\mathbb{P}(\textup{No-life }\cap \textup{ Life-sign})}{\mathbb{P}(\textup{Life-sign})}$

$=\frac{(1-0.2728)\times(1-0.842)}{0.3445952}\approx0.3334$