RheyDCD11 wrote:superstar2011 wrote:If my answers were not helpful (incorrect, misleading or confusing), I apologize in advance. I read your post this very, very late at night.
Thank you for your answers. Unfortunately, the problem is tht the answers are not the same as the back of the textbook...even though there have been times that they are wrong...my teacher "assured" me that the answers for these particular questions are correct but again thank you. If you are at all curious to what the answers were they were: 35/324 for the first one so approximately 11% and then 5/84 approximately being 6% for the second one.
I guess that's the problem with doing this late at night. I'll repost my answers in case they are of any help.
1) What I forgot here is that there are TWO red aces in the deck; an ace of hearts and an ace of diamonds. So there are not two, but
three cards in the deck in which you would like to draw. My methods, however, were correct.
At least one of your draws has a red ace or a red joker. It's possible that you draw a red joker and a red ace both times.
In any given draw, your chance of drawing one of the cards is
3/54 = 1/18.
Using the same approaches as I did before,
Approach 1: Union (Summing the two draws, subtracting the intersection)
P(Event) = 1/18 (first draw) + 1/18 (second draw) - (1/18)^2 (drawing both times)
= 1/9 - 1/324
= 36/324 - 1/324
= 35/324 = about 10.8%
Approach 2: Exponential distribution (waiting time, until you draw one of the desired cards)
P(Event) = 1/18 (success after one draw) + (1/18)(17/18) (success after one failed draw)
= 1/18 + 17/324
= 18/324 + 17/324
= 35/324 = about 10.8%
2) I think the problem with this question is that it wasn't specific, but rather ambiguous; apparently the order in which the committee members were picked
mattered. (I don't know why, were they assigned different roles?)
Okay, so I will use the counting principle that I used before, except with permutations instead of combinations (now that order matters).
In this example, you choose 3 people (out of 5) from the union and 1 person (out of 5) not from the union.
So, N(Event)= (5P3)*(5P1) = 5!/2! * 5!/4! = 60*5 = 300
There are 300 possible ways to pick 3 union members and 1 non-union member in a particular order. Let's find the sample set now, where you are simply picking 4 members from a list of 10.
N(S)= 10P4 = 10!/6! = 5040
Computing the probability of event E, P(Event)= N(Event)/N(S) =
300/5040 = 5/84 (divide both numbers by 60) or about 5.95%.
The world is not over and I'm still kicking butt!