# Homework A2 – Compound Probability

HomeworkA2 – Compound Probability

CompoundProbabilityQuestion1:

1. For a .250 hitter:

1. (1-0.200)=0.832

Question2:

Aperson can get a blackjack by either drawing a 10 card followed by anace or draw an ace followed by a 10card.

Probabilityof drawing 10 cards at first is

ii.Probability of first obtaining an ace then a 10 value

Therefore,the probability of obtaining a blackjack is

Thismeans that the probability of getting a blackjack in a single deck ofcards is 0.0483 = 4.83%

Question3: Probability (destroying a single silo) =

Probabilityof destroying all silos

Ifwe have 10 such targets, then the probability that all 10 will bedestroyed if we send 2 bombs against each of them (requiring 20bombs) is approximately 1. This implies that there is a high chanceof hitting the silo.

Question4

Forthe transmission work, all the five systems will be required. Whenany of the systems fail to work, then the transmission cannot bemade.

Parta:

Partb: The ‘A’ portion of the system has two components.The system can remain at A when at least one of these two componentsremains operable. The two components will be in parallel. Therefore

Thisequation has taken two parts of ‘A’. The first part involves eachof the parts working 98% of the time and failing 2% of the time. Thiscreates a unit that works 99.96% of the time and fails to work 0.04%of the time. Such an improvement in reliability is commendable buthappens at the expense of the equipment cost.

Thus

Partc: Adding a third component to each of the systems. Similarsteps are followed as in part ‘b’ above. The first step involvesworking out the probability that ‘A’ works.

Thisimplies that ‘A’ is extremely reliable working at 99.9992% oftime with a failure of 0.0008% of time. When these reliable units arecombined in the system, then the system gives the below data.

Thisimplies an extremely high success probability for the data to besent.

Question5: a)

b)The dice should be rolled continuously until the sum is 6 in whichcase it’s a win or roll a 7 to lose. The game never ends until anyof the two scenarios occur. Therefore the probability that the sum is6 given that the sum of 6 or 7 has occurred is represented as:

Therefore,for conditional probability

Inthis case, A is the event that the dice is 6 and B is the event wherethe sum is either 6 or 7. In this case

Sincethere are 5 to roll 6 with two dice and 6 ways to roll a seven,therefore

c)

d)

Therefore,the probability of winning when the first roll is 6 is

Similarly,when the initial roll is 4, probability is 1/36, initial roll is fivehas a probability of 2/45, when initial roll is 8, and theprobability is 25/396, when the initial role is 9, probability is2/45, and finally, when initial roll is 10, probability is 1/36.

Therefore,the probability of winning the game is the sum total of the aboveprobabilities

2/9+1/36+2/45+25/396+25/396+2/45+1/36=244/495= 0.4929

Reference

Weiers, R., Brian, J., &amp Lawrence, H. (2005). Introduction to Business Statistics. Belmont, CA:: Thomson Brooks/Cole.