Questions

Raipur’s Swami Vivekananda Airport was ranked first in Customer Satisfaction Index (CSI) Survey among 49 airports in the country. It has achieved 4.84 score on a five point scale index. It was followed by Udaipur (2nd), Amritsar (3rd) and Dehradun (4th) airports which have scored 4.75, 4.74 and 4.73 respectively. This is third consecutive time in the past two years, Raipur airport has received this recognition.

While a test plan is being created, risks involved in testing the product are to be taken into consideration along with possibility of their occurrence and the damage they may cause along with solutions; if any. Detailed study of this is called as Risk Analysis.

Some of the risks could be:

New Hardware 

New Technology 

New Automation Tool 

Sequence of code delivery 

Availability of application test resources 

Identify and describe the risk magnitude indicators: High, Medium and Low

High magnitude means the effect of the risk would be very high and non-tolerable. Company may face severe loss and its reputation is at risk. Must be tested.

Medium: tolerable but not desirable. Company may suffer financially but there is limited liability or loss of reputation. Should be tested.

Low: tolerable. Little or no external exposure. Little or no financial loss. Company’s reputation unaffected. Might be tested.

Three perspectives of Risk Assessment: Effect, Cause and Likelihood.

To assess risk by Effect, identify a condition, event or action and try to determine its impact.

To asses risk by Cause is opposite of by Effect. Begin by stating an undesirable event or condition and identify the set of events that could have permitted the condition to exist.

To asses risk by Likelihood is to determine the probability that a requirement will not be satisfied.

As you were asleep, none of the things could have happened without doing "open your eyes" first.

At least one black can be chosen in two ways from each box:
Now, probability of choosing at least one black ball from first box 

= 1/2 × [ (2C1× 2C1)/4C2 + 2C2/ 4C2 ]= 5/12

Probability of choosing at least one black ball from 2nd box
= 1/2 × [(4C1×12C1)/16C2 + 4C2/ 16C2] = 9/40

Final probability

= 5/12 + 9/40 = (50 + 27)/120 = 77/120