# Math Help - Probability question...

1. ## Probability question...

Takeisha and Antoine are part of a group of 8 women and 8 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Takeisha is a female and Antoine is a male.)

a) What is the probability that Takeisha and Antoine will be paired together as partners?
b) What is the probability that Takeisha will be paired with a woman?
c) What is the probability that each of them will be paired with a person of the same sex, i.e. Takeisha with a woman and Antoine with a man?
d) What is the probability that each of them will be paired with a woman?

Can anyone help me with this, I'm stuck!! =/....thank you!!!!

2. Originally Posted by nikkik
Takeisha and Antoine are part of a group of 8 women and 8 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Takeisha is a female and Antoine is a male.)

a) What is the probability that Takeisha and Antoine will be paired together as partners?
b) What is the probability that Takeisha will be paired with a woman?
c) What is the probability that each of them will be paired with a person of the same sex, i.e. Takeisha with a woman and Antoine with a man?
d) What is the probability that each of them will be paired with a woman?

Can anyone help me with this, I'm stuck!! =/....thank you!!!!
In these sorts of problems the probability is:

(number of favourable cases)/(total number of cases)

This is because we set things up so each case is equally likely.

a) T can be pired with any of the 17 others, this is the total number of cases. In only one of these is T paired with A, this is the number of
favourable cases. So the required probability is:

p = 1 /17 ~= 0.0588.

b) T can be pired with any of the 17 others, this is the total number of cases. As there are six other women, in 6 of these cases T is paired with
another woman, this is the number of favourable cases. So the required probability is:

p = 6 /17 ~= 0.3529.

RonL

3. Thank you very much!!! =)

I also had another question I was having problems solving

Suppose there is a ski club with 4 members. Each member normally hangs his (or her) ski poles on a hook, with each member using a separate hook. If the cleaning person takes down all the poles to wash the wall, and puts them back up randomly with two poles to each of the 4 hooks, what is the probability that the poles will be paired correctly, i.e. that each hook will hold two poles belonging to the same member?

4. Hello, nikkik!

Here's one approach to this problem.
. . (There are many others.)

Suppose there is a ski club with 4 members. Each member normally hangs
his (or her) ski poles on a hook, with each member using a separate hook.
If the cleaning person takes down all the poles to wash the wall,
and puts them back up randomly with two poles to each of the 4 hooks,
what is the probability that the poles will be paired correctly,
i.e. that each hook will hold two poles belonging to the same member?

Consider the four hooks: 1, 2, 3, 4
. . and the eight ski poles: A, A, B, B, C, C, D, D.

Two poles will be placed on Hook-1.
The first can be any pole: . $\frac{8}{8}\:=\:1$
The second must be its mate: . $\frac{1}{7}$

Two poles will be placed on Hook 2.
The first can be any of the remaining 6 poles: . $\frac{6}{6}\:=\:1$
The second must be its mate: . $\frac{1}{5}$

Two poles will be placed on Hook 3.
The first can be any of the remaining 4 poles: . $\frac{4}{4}\:=\:1$
The second must be its mate: . $\frac{1}{3}$

Two poles will be placed on Hook 4.
The remaining two poles are a matching pair: $\frac{2}{2}\:=\:1$

The probability is: . $1\cdot\frac{1}{7}\cdot1\cdot\frac{1}{5}\cdot1\cdot \frac{1}{3}\cdot1\;=\;\boxed{\frac{1}{105}}$

5. Thanks so much, that was very helpful!! I'm still having trouble with the last two parts of the first question I posted though, I got the first two but can't seem to figure these out->

Takeisha and Antoine are part of a group of 8 women and 8 men who will participate in a chess tournament. They will be paired up randomly into teams of two. (Takeisha is a female and Antoine is a male.)

What is the probability that each of them will be paired with a person of the same sex, i.e. Takeisha with a woman and Antoine with a man?

What is the probability that each of them will be paired with a woman?

6. There are $D = \frac{{16!}}{{\left( {8!} \right)\left( {2^8 } \right)}}$ ways to form unordered pairs of contestants. There are 7 ways that T can be paired with another female; likewise 7 ways that A can be paired with another male. This leaves 12 unassigned people. There are $\frac{{12!}}{{\left( {6!} \right)\left( {2^6 } \right)}}$ ways to form unordered pairs of these.

Note that there are $\frac{{8!}}{{\left( {4!} \right)\left( {2^4 } \right)}} = {105}$ ways to form unordered pairs of the ski poles.