Question

Accountant Ian Somnia's infamous napping at work has given rise to a challenge from colleague I. M. Tarde. At the company retreat, the two will take turns trying to stay awake during a sequence of 5-minute company training films. The probability that Somnia will fall asleep during a given film is 0.8, while the probability that Tarde does is 0.7. The first one to stay awake (as determined by a panel of alert judges) wins a $1000 bonus. If Somnia and Tarde alternate watching films, with Somnia going first, what are the chances that Tarde wins the bonus? (Hint: A typical sequence of films for which Tarde wins the bonus is NNNY, where N = “not awake” and Y = “awake.”)

Answer 1

Solution

Back-up Theory

Sum of an infinite Geometric Progression when common ratio r < 1 is: S = a/(1 – r)..................... (1)

Now to work out the solution,

Let

S_Y represent the event that Somnia is awake; S_N represent the event that Somnia is not awake.

Similarly,

T_Y represents the event that Tarde is awake; T_N represents the event that Tarde is not awake.

Then, with Somnia going first, the sequences of events that will lead to Tarde’s winning are:

S_NT_Y, S_NS_NS_NT_Y, S_NS_NS_NS_NS_NT_Y, ..............................................................................

Thus, probability that Tarde wins

= P(S_NT_Y) + P(S_NS_NS_NT_Y) + P(S_NS_NS_NS_NS_NT_Y) + .................................................to infinity

= (0.8 x 0.3) + (0.8 x 0.8 x 0.8 x 0.3) + (0.8 x 0.8 x 0.8 x 0.8 x 0.8 x 0.3) + ................to infinity

= (0.8 x 0.3) + (0.8² x 0.8 x 0.3) + (0.8⁴ x 0.8 x 0.3) + ..................................................to infinity

The above is an infinite geometric series with first term, a = (0.8 x 0.3) and common ratio, r = 0.8².

So, vide (1),

the sum = 0.24/(1 – 0.64)

= 0.24/0.36

= 0.6667 Answer

DONE