# Math Help - Probability of Events - Helppp Please

1. ## Probability of Events - Helppp Please

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

a) What is the cardinality of the sample space?

b) List the event in which one number is a factor of the other number and find its cardinality.

c) List the event in which one number is a factor of the other number and find its cardinality.

d) List the event in which the sum of the numbers is 15 and one number is a factor of the other number.

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
b) 2 men being chosen
c) 2 women being chosen

44. A coin is tossed three times...
a) What is the probability of getting three heads?
b) What is the probability of getting exactly two heads?

2. Originally Posted by 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. $\frac{1}{2} \times \frac{1}{2} \times\frac{1}{2}$
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?
Mush.

3. Originally Posted by cnmath16
[snip]
44. A coin is tossed three times...
a) What is the probability of getting three heads?
b) What is the probability of getting exactly two heads?
Mush has made a small error in (b) because it's asking for 2 heads in 3 tosses ....

The number of ways you can get 2 heads in 3 tosses is $^3C_2 = 3$. So the answer is $3 \cdot \left(\frac{1}{2}\right)^3$.

Alternatively, you could list all eight possible outcomes and count how many have two heads ....

4. Originally Posted by mr fantastic
Mush has made a small error in (b) because it's asking for 2 heads in 3 tosses ....

The number of ways you can get 2 heads in 3 tosses is $^3C_2 = 3$. So the answer is $3 \cdot \left(\frac{1}{2}\right)^3$.

Alternatively, you could list all eight possible outcomes and count how many have two heads ....
Ah, woops. I didn't read the "coin tossed three times" part. My mistake. His bold bits and my bold bits... it's all too confusing!