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**cnmath16** **42. From the numbers 3 to 12 inclusive, two numbers are chosen at random.**

a) What is the cardinality of the sample space? **Cardinality is the number of elements in the set. Your set is [3,12] = {3,4,5,6,7,8,9,10,11,12}. How many elements? **

b) List the event in which one number is a factor of the other number and find its cardinality. **An event is a set of outcomes, and is a subset of the sample space. From the set 12 has factors 3, 4, 6. 11 has no factors. 10 has 5. 9 has 3. 8 has 4. 7 has none. 6 has 3. 5 has none. 4 has none. 3 has none. So we have how many pairs there?:**

(12,3) (12,4) (12,6) (10,5) (9,3) (8,4) (6,3). There are 7 of them, but they can also happen in the other order!

c) List the event in which one number is a factor of the other number and find its cardinality. ** Same question twice?**

d) List the event in which the sum of the numbers is 15 and one number is a factor of the other number. **What combinations of these numbers sum to 15? 12+3... 10+5... 9+6...8+4+3... 8+7... which of these 5 has one number being a factor of the other? 12+3... 10+5...8+4+3... How many combinations of these can you get? Well you have 12+3...3+12...10+5...5+10...8+4+3...8+3+4...4+8+3. ..4+3+8...3+4+8...3+8+4. That is your event set.**

**43. From a group of 20 men and 16 women, 2 people are chosen at random. Find the probability of:**

a) 1 man and 1 woman being chosen **Total number of people... 36. Probability of picking a man is 20/36. Probability of a woman is 16/36. And the probability of both these occuring is the product of the two probabilities.**

b) 2 men being chosen ** When you first choose a man there is 20/36. Then there is one less man and one less person in the total set, so the probability of a 2nd man is 19/35. Again, the probability of both these occuring is the product.**

c) 2 women being chosen ** Same as above, but with new numbers!**

**44. A coin is tossed three times...**

a) What is the probability of getting three heads? ** Probability of one head is 1/2. So the probability of 3 heads is the product of the 3 probabilities of one head. $\displaystyle \frac{1}{2} \times \frac{1}{2} \times\frac{1}{2}$ **

b) What is the probability of getting exactly two heads? **Well to get exactly two heads you need to toss twice. There are 2 possible outcomes in each toss and hence 4 possible outcomes over 2 tosses. There are: (h,t) (t,h) (h,h) (t,t). What's the probability of one outcome occuring out of a possible 4? **