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Need help solving these problems 1-10 for stats. Thank you!! #1. You have a standard deck of playing cards (52 cards). What is the probability of
Need help solving these problems 1-10 for stats. Thank you!!
#1. You have a standard deck of playing cards (52 cards). What is the probability of pulling an ACE from this deck?
What is the probability of pulling a second ace? Then, a third ace? And, finally the fourth ace? What is the combined probability of pulling all four aces? (All of this is WITHOUT replacement, of course).
#2. You toss a pair of dice, once. (a) What are the odds (probability) that BOTH are EVEN numbers and their SUM equals "6"? (b) That both are ODD numbers and total "6" ? MAKE SURE TO SHOW YOUR SETUP AND WORK DETAILS
#3. Here are 5 office staff members: Jim, Joan, Jeff, John, and Jane. You must assign them to five different clients. (This is the same kind of problem as re-arranging 5 different letters of the alphabet)
(a) How many different ways can you do these assignments (or re-arrange 5 letters) ?
(b) How many ways can you assign any 3 of the five (to different clients) ?
(c) What if it did NOT matter who went to which client and you wanted to assign only 3 of the five?
#4. 5,000 vehicles a day go through or around Baltimore continuing North on I-95. There are 4 major routes (H) I-295 the old Harbor Tunnel, (K) I-95 the new F.S. Key Tunnel, (B) I-695-s over the Sparrow's Point Bridge, and (T) I-695-n past Towson: 35% use K, 20% use T, 25% use H and 20% use B. If you randomly select a vehicle approaching Baltimore from the South, what is the probability it will take route:
(a) H or K or B or T ?
(b) B and T
(c) K or H or B
#5. A BINOMIAL problem has just TWO alternatives: heads/tails, right/wrong, black/white, yes/no, etc. An easy problem would be:
(a) You toss a normal (balanced) coin 6 times. What is the probability that you get 6 HEADS ?
(b) BUT, what if the coin is weighted so that heads comes up 75% (0.75) of the time and tails 25% ?
What is the probability that if you toss this coin 5 times you get at least 4 heads ? This problem requires a different formula since 4 wins and 5 wins are discrete numbers ( you can't have 4.39 wins for example). You must calculate the probability of EACH options, 4 wins and then 5 wins in this case:
YOU NEED TO USE THE BINOMIAL FORMULA WHICH IS MORE COMPLEX.
P (k) = n! / [k! *(n-k)!]
