Probability----NEED HELP ASAP WITH HW

[guest82]

1. Suppose 4 dogs and 3 cats line up at random for a free hamburger. Find the probability that the first 4 positions are NOT all filled by dogs.

2 A jury of 9 people is selected at random from an available pool of 12 women and 13 men.
a. Find the probability that the jury selected has exactly 6 men?
b. Find the probability that the jury selected has more men then women?

Thanks

 

[tkhunny]

P(the first four ARE dogs) = ??

 

[soroban]

Hello, guest82!

(1) Suppose 4 dogs and 3 cats line up at random for a free hamburger.
Find the probability that the first 4 positions are NOT all filled by dogs.

I will assume that the seven animals are distinguishable.

There are: .\(\displaystyle 7!\) possible orderings.


Suppose the first 4 positions are filled by dogs.
. . They have the form: .\(\displaystyle DDDDCCC\)

The four dogs can be arranged in \(\displaystyle 4!\) ways.
The three cats can be arranged in \(\displaystyle 31\) ways.
. . They can be arranged in \(\displaystyle 4!\cdot 3!\) ways.

\(\displaystyle \text{Hence: }\(\text{first 4 }are\text{ dogs}) \;=\;\frac{4!\,3!}{7!} \:=\:\frac{1}{35}\)

\(\displaystyle \text{Therefore: }\(\text{first 4 are }not\text{ all dogs}) \;=\;1-\frac{1}{35} \;=\;\frac{34}{35}\)

(2) A jury of 9 people is selected at random from a pool of 12 women and 13 men.

\(\displaystyle \text{There are: }\:{25\choose9} \:=\:\frac{25!}{9!\,16!} \:=\:2,042,975\text{ possible juries.}\)

a. Find the probability that the jury selected has exactly 6 men.

\(\displaystyle \text{We want 6 men: }\:{13\choose6}\:=\:1716\text{ ways.}\)
\(\displaystyle \text{We want 3 women: }\:{12\choose3} \:=\:220\text{ ways.}\)

. . \(\displaystyle \text{Hence, there are: }\:1716 \times 220 \:=\:337,520\text{ juries with 6 men.}\)


\(\displaystyle \text{Therefore: }\(\text{6 men}) \;=\;\frac{377.520}{2,042,975} \;=\;\frac{6,864}{36,145}\)

b. Find the probability that the jury selected has more men then women?

\(\displaystyle \text{We want: }\:\text{(5M, 4W) or (6M, 3W) or (7M, 2W) or (8M, 1W) or (9M, 0W)}\)

. . \(\displaystyle \begin{array}{cccc} P(\text{5M, 4W)} &=& \dfrac{{13\choose5}{12\choose4}}{{25\choose9}} \\ \\[-3mm] P(\text{6M, 3W)} &=& \dfrac{{13\choose6}{12\choose3}}{{25\choose9}} \\ \\[-3mm] P(\text{7M, 2W)} &=& \dfrac{{13\choose7}{12\choose2}}{{25\choose9}} \\ \\[-3mm] P(\text{8M, 1W)} &=& \dfrac{{13\choose8}{12\choose1}}{{25\choose9}} \\ \\[-3mm] P(\text{9M, 0W)} &=& \dfrac{{13\choose9}{12\choose0}}{{25\choose9}} \\ \end{array}\)

\(\displaystyle \text{ . . . and add them up.}\)


I'll wait in the car . . .
.