# Thread: Probability, permuations & combinations help

1. ## Probability, permuations & combinations help

I don't have the exact answers for these questions, I only know whether the answers would be right or wrong, my answers below are wrong though

1.

solved

2.

[solved]

solved

5.
[solved]

6.[solved]!!
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well I don't know which categorary Permuations and combinations fall under, sorry if I posted in the wrong forum. Any help would be greatly appreciated

2. Originally Posted by kakab00
I don't have the exact answers for these questions, I only know whether the answers would be right or wrong, my answers below are wrong though

1.

I thought it would be 95/100 X 0.4%(in case the test was accurate) + 5/100X99.6%(in case the test was inaccurate) but it appears to be wrong..
The conditional probability $\displaystyle P(Has\ cancer|Test\ says\ so)$ is needed here.

From the definition of conditional probability

$\displaystyle P(Has\ cancer|Test\ says\ so) = \frac{P(Has\ cancer\ AND\ Test\ says\ so)}{P(Test\ says\ so)}.$

You have calculated the denominator. The numerator is just the case "the test was accurate."

3. Originally Posted by JakeD
The conditional probability $\displaystyle P(Has\ cancer|Test\ says\ so)$ is needed here.

From the definition of conditional probability

$\displaystyle P(Has\ cancer|Test\ says\ so) = \frac{P(Has\ cancer\ AND\ Test\ says\ so)}{P(Has\ cancer)}.$

You have calculated the denominator. The numerator is just the case "the test was accurate."
?? I don't quite get what you mean here, if I had calculated the denominator from my workings then wouldn't the probability be >1 if the numerator is just "the test was accurate"(0.95 in this case)

4. Originally Posted by JakeD
The conditional probability $\displaystyle P(Has\ cancer|Test\ says\ so)$ is needed here.

From the definition of conditional probability

$\displaystyle P(Has\ cancer|Test\ says\ so) = \frac{P(Has\ cancer\ AND\ Test\ says\ so)}{P(Test\ says\ so)}.$

You have calculated the denominator. The numerator is just the case "the test was accurate."
Originally Posted by kakab00
?? I don't quite get what you mean here, if I had calculated the denominator from my workings then wouldn't the probability be >1 if the numerator is just "the test was accurate"(0.95 in this case)
OK, ignore my description of what you did. Calculate $\displaystyle P(Has\ cancer\ AND\ Test\ says\ so),$ which is part of your calculation of the denominator.

5. Originally Posted by JakeD
OK, ignore my description of what you did. Calculate $\displaystyle P(Has\ cancer\ AND\ Test\ says\ so),$ which you've already done.
ok.. I got it, so it's 0.95x0.004 = 0.0038

and use this value divided by the value I found in my first post right?

6. Originally Posted by JakeD
OK, ignore my description of what you did. Calculate $\displaystyle P(Has\ cancer\ AND\ Test\ says\ so),$ which is part of your calculation of the denominator.
Originally Posted by kakab00
ok.. I got it, so it's 0.95x0.004 = 0.0038

and use this value divided by the value I found in my first post right?
Right. Notice I edited my posts. I said the denominator was P(Has cancer). It is actually P(Test says so) and this is what you calculated.

7. new question addedd, it's on probability too:

8. Hello, kakab00!

4. There are eight seats at a round table, equally spaced.
Find the number of ways in which two women, one married couple,
and one family of four people can be seated at the table

(a) if their are no restriction to which seats they must take.
There are: .$\displaystyle (8-1)! \:=\:7!\:=\:5040$ ways.

(b) if the married couple must be seated together
and the family of four must be seated together.
Duct-tape the married couple together: .$\displaystyle \{\text{Husband,Wife}\}$
Duct-tape the family-of-four together: .$\displaystyle \{F_1,F_2,F_3,F_4\}$

Then there are four "people" to arrange: .$\displaystyle \boxed{\text{Husband-Wife}},\;\boxed{F_1\text{-}F_2\text{-}F_3\text{-}F_4},\;\boxed{\text{woman}},\;\boxed{\text{woman} }$
. . and they can be seated in: .$\displaystyle (4-1)! \:=\:3! \:=\:6$ ways.

But for each of these seatings,
. . the married couple can be rearranged in $\displaystyle 2!=2$ ways
. . and the family-of-four can be rearranged in $\displaystyle 4!= 24$ ways.

Therefore, there are: .$\displaystyle 6 \times 2 \times 24 \:=\:288$ possible seatings.

(c) if the married couple must sit together
and the two women must not be seated directly opposite each other.
This is much trickier; I had to baby-talk my way through it.

Seat the married couple.. .Call their seats #1 and #2.
The diagram looks like this:
Code:
             Husband
[1]   Wife
[8]    |   [2]
\   |   /
\ | /
[7]- - - * - - - [3]
/ | \
/   |   \
[6]    |     [4]
[5]

Now we will seat the two women: $\displaystyle \{W_1,W_2\}$

Select one of the women; there are 2 choices.

Suppose she sits in chair #5 or #6; 2 choices.
Since the other woman can not be seated opposite.
. . then the remaining five people can be seated in $\displaystyle 5! =$ 120 ways.

Hence, there are: .$\displaystyle 2 \times 2 \times 120 \:=$ 480 seatings.

Select one of the women; there are 2 choices.

Suppose she sits in chair #3, 4, 7,or 8 . . . 4 choices.
. . Then the other woman must not sit in the opposite chair; she has 4 choices.
Then the family-of-four can be seated in $\displaystyle 4! =$ 24 ways.

Hence, there are: .$\displaystyle 2 \times 4 \times 4 \times 24 \:=$ 768 seatings.

So there are: .$\displaystyle 480 + 768 \:=$ 1248 seatings.

But originally, we placed Husband in chair #1 and Wife in chair #2.
. . Of course, they could be reversed.
This doubles the number of seatings.

Therefore, the number of seatings is: .$\displaystyle 2 \times 1248 \:=$ 2496

9. Therefore, the number of seatings is: . 2496

The answer I was given was just 1248, maybe it's because it's a round table?

10. Hello, kakab00!

The events A and B are independent.
If it given that: .$\displaystyle P(A) = 0.25,\;P(A' \cap B')= 0.15$

(a) What is the value of $\displaystyle P(B)$ ?
By DeMorgan's Law: .$\displaystyle A'\cap B' \:=\:(A \cup B)'$
So we have: .$\displaystyle P(A' \cap B') \:=\:P(A \cup B)' \:=\:0.15$
. . Hence: .$\displaystyle P(A \cup B) \:=\:0.85$

Events A and B are independent.
.. Hence: .$\displaystyle P(A \cap B) \:=\:P(A)\!\cdot\!P(B) \:=\:(0.25)P(B)$

Formula: .$\displaystyle P(A \cup B) \:=\:P(A) + P(B) -P(A \cap B)$

So we have: . .$\displaystyle 0.85 \:=\:0.25 + P(B) - (0.25)P(B)$

Then: .$\displaystyle 0.60 \:=\:(0.75)P(B)\quad\Rightarrow\quad P(B) \:=\:\frac{0.60}{0.75}$

Therefore: .$\displaystyle P(B) \:=\:0.8$

(b) What is the value of $\displaystyle P(A' \cup B')$ ?
We know that: .$\displaystyle P(A \cap B) \:=\:P(A)\!\cdot\!P(B)\:=\:(0.25)(0.8) \:=\:0.20$
. . Hence: .$\displaystyle P(A \cap B)' \:=\:0.8$

By DeMorgan's Law: .$\displaystyle A'\cup B' \:=\:(A \cap B)'$

Therefore: .$\displaystyle P(A' \cup B') \:=\:P(A \cap B)' \:=\:0.80$

(c) It is given that events B and C are mutually exclusive,
. . .$\displaystyle P(A \cup C) = P(A)$ and $\displaystyle P(A \cap C) = P(C)$
What is the largest possible chance that event C will occur?
Since $\displaystyle P(A \cup C) = P(A)$ and $\displaystyle P(A \cap C) = P(C)$, then: .$\displaystyle C \subset A$
. . (Circle $\displaystyle C$ is completely contained in circle $\displaystyle A$.)

Since $\displaystyle B$ and $\displaystyle C$ are mutually exclusive, their circles do not intersect.

The Venn diagram looks like this:
Code:
  * - - - - - - - - - - - - - - - - - - - - - - - - - *
|                                                   |
|           *---------*       *---------*           |
|         /      A      \   /      B      \         |
|       /                 ×                 \       |
|     /     0.05        /   \        0.60     \     |
|   *                 *       *                 *   |
|   |                 |       |                 |   |
|   |       *---*     |       |                 |   |
|   |      /     \    |  0.20 |                 |   |
|   |     *       *   |       |                 |   |
|   |     |   C   |   |       |                 |   |
|   *     *       *   *       *                 *   |
|     \    \     /      \   /                 /     |
|       \   *---*         ×                 /       |
|         \             /   \             /         |
|           *---------*       *---------*           |
|                                             0.15  |
* - - - - - - - - - - - - - - - - - - - - - - - - - *

Since circle $\displaystyle C$ could be equal to circle $\displaystyle A:\;\;P(C) \:\leq\:0.05$

11. Hello, kakab00!

Others may benefit from the exchange we've had here.

I finally took the time to work out #3.

Code:
      *
*   *
o       o    C
*           o
A *               o
o                   o
*                       *
* * * * o * o * o * * * * * *
B
(i) How many line segments are there joining any two points on different sides?
On side $\displaystyle A$, there are 2 choices of points.
Each can be joined to any of the other 7 points.
. . There are: .$\displaystyle 2 \times 7 \,=\,14$ line segments.

On side $\displaystyle B$, there are 3 choices of points.
Each can be joined to any of the 4 points on side $\displaystyle C$.
. . There are: .$\displaystyle 3 \times 4 \,=\,12$ line segments.

Therefore, there are: .$\displaystyle 14 +12 \:=\:\boxed{26}$ possible line segments.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

There are 9 points and there are: $\displaystyle {9\choose2} \,=\,36$ pairs of points.
But some of these join two points on the same side of the triangle.

There is 1 such pair on side $\displaystyle A$.
There are $\displaystyle {3\choose2} = 3$ such pairs on side $\displaystyle B$.
There are $\displaystyle {4\choose2} = 6$ such pairs on side $\displaystyle C$.
. . Hence, there are: $\displaystyle 1 + 3 + 6\:=\:10$ disallowed segments.

Therefore, there are: $\displaystyle 36 - 10 \:=\:\boxed{26}$ allowed line segments.

(ii) How many triangles can be formed from these points?

There are 9 points; we will select 3 of them.
. . There are: .$\displaystyle {9\choose3} \,=\,84$ sets of 3 points.
But some are sets of 3 collinear points, which do not form a triangle.
. . How many sets of three collinear points are there?

There is one set on side $\displaystyle B$.
There are: $\displaystyle {4\choose3} = 4$ sets on side $\displaystyle C$.
. . Hence, there are: $\displaystyle 1 + 4 \,=\,5$ sets of collinear points.

Therefore, there are: .$\displaystyle 84 - 6 \:=\:\boxed{79}$ triangles.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

On side $\displaystyle A$, there is 1 pair of points.
It can be joined with any of the other 7 points.
. . Hence, there are: $\displaystyle 1 \times 7 \:=\:7$ triangles.

On side $\displaystyle B$, there are: $\displaystyle {3\choose2} = 3$ pairs of points.
They can be joined with any of the other 6 points.
. . Hence, there are: $\displaystyle 3 \times 6 \:=\:18$ triangles.

On side $\displaystyle C$, there are: $\displaystyle {4\choose2} = 6$ pairs of points.
They can be joined with any of the other 5 points.
. . Hence, there are: $\displaystyle 6 \times 5 \:=\:30$ triangles.

We can also pick one point from each side.
There are 2 choices for side $\displaystyle A$, 3 for side $\displaystyle B$, 4 for side $\displaystyle C$.
. . Hence, there are: $\displaystyle 2 \times 3 \times 4 \:=\:24$ triangles.

Therefore, there are: $\displaystyle 7 + 18 + 30 +24 \:=\:\boxed{79}$ triangles.