Problems based on LANE (C5) #1. You have a standard deck of playing cards (52 ca

Question

Problems based on LANE (C5) #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 to do this: (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)!] * p k * q (n-k) (this formula is also in Lane around p-206 and Illowsky around p-62) In this equation “n” is the number of tosses or 5 in this problem. “p” and “q’ are your binomial chances of success and failure and MUST add up to 1.00 In this case p = 0.75 and q = 0.25, (For a normal coin the probabilities of heads/tails are equal at 0.5 so in that case p and q would both be 0.5, but not here.) “k” is the number of successes: 4 or 5 in this problem since we said “at least 4”. PLUG IN THE VALUES AND THEN ADD THE CALCULATED PROBABILITIES UP to get the total probability of getting heads 4 or 5 times. REMEMBER that 0! = 1 FOR EXAMPLE: You may notice that the first part of the equation with the factorials is the COMBINATIONS equation. These are the number of ways you can get “heads”. For example if you toss the weighted coin 4 times, how many ways can you get 4 “heads” (n = 4 and k = 4 here) combinations = 4!/[4! * (4 – 4) !] = 4!/(4!*0!) = 1 Only 1 way to get 4 “heads” meaning that you must get “heads” on each of the 4 tosses. How about getting 3 heads (not 5) in 4 tosses? (n = 4 and k = 3 here) 4!/[3!*(4 – 3)!] = 4!/(3!*1!) = 4 (just 4 ways to get 3 heads. This means that you get “tails” on either toss 1, 2, 3, or 4 and heads on the 3 tosses you don’t get tails. BUT, now we have to consider the PROBABILITIES (since this is a weighted coin) of heads/tails and how they affect the probabilities of these combinations. So, we multiply the number of combinations we calculated for 4 and 5 heads times those probabilities and then add those up to get the combined probability of tossing 4 or 5 heads out of 5 tosses with this particular coin.

#5 HOMEWORK PROBLEM: CALCULATE THIS TOTAL PROBABILITY OF GETTING 4 OR 5 HEADS IN 5 TOSSES WITH THIS WEIGHTED COIN USING THE ABOVE COMPLEX EQUATION. An additional insight is that once you have calculated and totaled your binomial probabilities of tossing 4 or 5 heads (out of 5 tosses), if you subtract that probability from 1.00 you will have the probability that you only toss 0, 1, 2 or 3 heads out of those 5 tosses. Depending on HOW a binomial problem is stated, it may be easier to calculate the (1 – probability) but not in this case as you would have had FOUR different k-values to use: 0, 1, 2 and 3 rather than just 4 and 5. BY THE WAY WHAT IS THAT PROBABILITY OF GETTING ONLY 0, 1, 2 or 3 HEADS OUT OF 5 TOSSES? THERE IS A SIMPLER WAY OF SOLVING THESE COMPLEX BINOMIAL PROBLEMS. LATER WE WILL IDENTIFY THE NORMAL APPROXIMATION OF THE BINOMIAL FORMULA, A MUCH SIMPLER APPROACH. IT IS NOT AS PRECISE, BUT CLOSE ENOUGH IN MOST CASES.

Explanation / Answer

1) Total number of cards = 52

Total number of aces = 4

I) probability of drawing an ace = 4/52

ii) probability of second ace = 3/51

iii) probability of drawing a third ace = 2/50

iv) probability of drawing a fourth ace = 1/49

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