Permutations
1) You have two
dice. You roll both dice. What is they probability that you will
roll snake eyes (that is both die land as 1). The sample space is
all the possible outcomes rolling the dice. To determine this take
each possibility for one die along with all the possibilities for the
second die like so:
die 1
die 2
1
1,2,3,4,5,6 6 outcomes
2
1,2,3,4,5,6 6 outcomes
3
1,2,3,4,5,6 6 outcomes
4
1,2,3,4,5,6 6 outcomes
5
1,2,3,4,5,6 6 outcomes
6
1,2,3,4,5,6 6 outcomes
For die 1 we get 36
outcomes.
We are not done, we
must consider die 2 compared to die 1.
For each landing of
die2 there are 6 possible landings for die 1, so for die 2 we get
another 36 possibilities. (think of it this way, let die one be green
and die two be red.)
So our possible outcome
(sample) space contains 72 outcomes. Snake eyes occurs twice in
this space so we have 2/72 = 1/36 = 0.028 = 2.8%.
Once again you
have two dice. What is the probability that you get a sum of 5?
Well, the sample space
here is the same as the previous one; 72 pairing of the die.
The sum 5 can occur as
1 + 4, 4 + 1, 2 + 3, and 3 + 2 which is 4 ways. So our probability
of getting the sum 5 is 4 / 72 = 0.056 = 5.6%.
Consider example 2
above. What sum of the thrown dice has the highest probability of
occurring?
To answer this question it’s best to use a table like so:
Die 1 outcomes
along leftmost column, die 2 outcomes along top row, sums in the table.
|
|
1
|
2
|
3
|
4
|
5
|
6
|
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
Notice
there 7 occurs most in this table 6 times, so it has the highest
probability of occurring 6/36 = 1/6 = 0.167 = 16.7%. On the other
extreme, there is only one way to get a sum of 2 or 12, with each
having a probability of 1/36 = 0.028 = 2.8%. (Note: this table
shows 36 outcomes, not 72 as previously, realizing that each sum has
2 pairings that are indistinguishable (die of the same color) we can
divide this sample space by two.)
There are 26
marbles in a bowl. 4 are red, 16 are green, and 6 are black. You
draw one marble from the bowl, what is the probability the marble is
red or black?
Sample
space is 26. There are 4 ways to choose a red and 6 ways to choose
a black, so we have 4/26 + 6/26 = 10/26 = 38.5%
5) You have 4
pair of green socks, 6 pair of black socks, 10 pair of white socks
and 16 pair of red socks in your drawer, none folded and all mixed
together. Suppose in the dark you want to make sure you choose 2 of
the same color. How many socks do you need to choose to make sure?
Well, once you've
chosen your first sock, it is one of the 4 colors, say green. Now
the second choice, if it is green, you're done. Any other color and
you must move onto the next choice since you have two socks of
different colors. The third choice could be one of these two, then
you're done, but it could be a third color. So you move on. The
fourth choice gives you a match or does not; if a match you're done,
else you have one of each color. The fifth choice must give you two
matching socks since it must be one of the colors you already have.
What's the probability
that fifth choice will be black? There are 72 socks initially and
we've chosen one of each color to this point, so we only have 72 - 4
= 68 socks to choose from. Since we already have one black sock,
that gives us 6 pair -1 = 11 black socks out of 68 socks to choose,
so we have 11/68 = 0.162 = 16.2%.
This is another FREE ALGEBRA PRINTABLE presented to you from the
Algebra section of
K12math.com