Probability and paper wads
Male and females were throwing paper in a trash can. The total shots made by both gender was 46, males made 23 and females made 23. Missed shots for both gender were 66, males missed 31 shotes while females missed 35 shots. The total amount of shots taken by males is 54 while the total amount of shots taken by females is 58. All in all 112 shots were taken by both genders.
Questions (Check if I got them right)
1.) What is the probability that a shot was made?
I got 46/112
2.) What is the probability that a shot was taken by a female?
I got 58/112
3.) What is the probability a shot was made and the shooter was a female?
I got 23/112
4.) What is the probability a shot was made or the shooter was a female?
I got 81/112
5.) Given that a shot was made, what is the probability the shooter was a male?
I got 23/46
6.) Are making the shot and being male independent?
I said not because P(B/A) does not equal P(B)
7.) Assuming the probabilities above remain true for future shots, what is the probability two shots in a row both land in the waste basket?
I got 0.168
The probability that a shot would be made is 46/112 or 0.41, then multiplied it by itself (0.41 x 0.41) because it asks us the probability two shots in a row both get in.
8.) What if the events are not independent? (THIS IS THE QUESTION THAT I DON'T GET, WHAT IS IT ASKING ME???????)
Re: Probability and paper wads
I agree with 1, 2, 3, 4, 5, 6, 7: well done.
For number 8 this just means that P(A and B) != P(A)P(B)
When an event is independent it means that conditioning on that event doesn't change the outcome. Typically we write this as P(A|B) = P(A) for any event B not including the null event which is not a proper event in a probability space even though it is accepted as a special kind of set member in set theory.
If you have a sequence of events in any probability space, then P(A|B) = P(A) for independence (i.e. its memory-less) or if not then P(A|B) != P(A) for dependent and non-memory less events.