# Math Help - Basic Probability Questions, Part 1

1. ## Basic Probability Questions, Part 1

Note to this first part: I don't need help on these, I figured them out. Thanks, though

************

I was hoping someone could help me with these, as I do not understand them at all. I work graveyard and so I end up missing class. My professor doesn't have much patience to work with us individually, and we have an exam on Wednesday.

1.) A bag has 12 marbles of equal size - 5 red, 4 green, 3 blue. Two marbles are removed at random. Compute the following:

a. the event that the first marble is blue and the second is red.
b. the event that the 2 marbles are NOT both blue.
c. if three marbles are selected in order what is the probability that all three marbles selected are blue?

2.) Use combinations in answering each of the following questions and change your answers to the form 1/x where x is rounded to an integer: A seven card hand is dealt from an ordinary deck. What is the probability that...

a. the hand contains two clubs and three diamonds?
b. the hand contains exactly three aces?

2. Hello, amysteriousman!

2) Use combinations in answering each of the following questions
and change your answers to the form $\tfrac{1}{x}$ where $x$ is rounded to an integer.

A seven-card hand is dealt from an ordinary deck.
There are: . ${52\choose7} \;=\;133,\!784,\!560$ possible hands.

What is the probability that:

(a) the hand contains exactly two Clubs and three Diamonds?
There are: .13 Clubs, 13 Diamonds, and 26 Others.

We want: . .2 Clubs, 3 Diamonds, and 2 Others.

There are: . ${13\choose2}{13\choose 3}{26\choose2} \:=\:7,\!250,\!100$ ways.

The probability is: . $\frac{7,\!250,\!100}{133,\!784,\!560} \;=\;\frac{27,\!885}{515,\!556} \;=\;\frac{1}{18.45...} \;\approx\;\frac{1}{18}$

b. the hand contains exactly three Aces?
There are: .4 Aces, 48 Others.
We want: . 3 Aces, 4 Others.

There are: . ${4\choose3}{48\choose4} \:=\:778,\!320$ ways.

The probability is: . $\frac{778,\!320}{113,\!784,\!560} \;=\;\frac{9}{1547} \;=\;\frac{1}{171.88...} \;\approx\;\frac{1}{172}$