Question #1) Consider four events A, B, C and D which are completely exhaustive

Question

Question #1) Consider four events A, B, C and D which are completely exhaustive and mutually exclusive. If P(A) = .2, P(B) = 0.3 and P(C) = 0.4 then calculate the following showing full details or by explaining how we know the answer

Question #2) Assume in question #1 we know that there exists another event X with a 30% probability of occurrence. IF we assume X is independent from each of A, B, C and D, then calculate the following showing full details or by explaining how we know the answer

Question #3) A new trend in fertility assisted births is to include gender selection as part of the criteria. A new company Girls-R-Us claims that by using their procedures they can give a woman a 75% chance that if conceived her child will be female. If we have 8 mothers-to-be each expecting one child, determine the sample space and sample space probabilities using a tree for all the possible gender outcomes (for example one is F-F-F-F-F-F-F-F with 10.01% probability). Present the results in a clear organized table.

Hint: there are 256 cases…. Use excel to list them saves a lot of time

Question #4) Consider rolling a pair of 6 sided dice. For each of the following events, determine and list all the possible sample points that can arise and calculate the probability of the event. For example X ?{Roll snake-eyes or Box Cars} = { [1,1], [6,6]} and P(X) = 5.56%.

Question #5) Assume in Question #3 we are interested in the discrete Random Variable
X? {The number of male children in the 8 births}.

Question #6) On the final examination I am thinking of having 30 Multiple Choice questions where each question has exactly 4 options of which only one will be considered correct. We are interested in the random variable X which is the number of questions you get correct if you just randomly guess each question without reading it.

Question #7) Showing all details and diagrams, determine the following probabilities for the standard normal distribution

Question #8) Showing all details and diagrams, determine the following probabilities for the given normal distribution

Explanation / Answer

Q1: Given P(A)=0.2, P(B)=0.3, P(C)=0.4 P(D)=1-P(A)-P(B)-P(C)=1-0.2-0.3-0.4 = 0.1 P(A U B U C U D) = P(A)+P(B)+P(C)+P(D) = 1 P(A and B ) =P(A) +P(B) = 0.2 +0.3 = 0.5

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects