Probabilities?! will you please help me help my little sister?

Question

Suppose a contest offers $1,000 to the person who can guess the winning four digit number. How many possibilities are there?

There are 10 women and 8 men in a club. How many different commities of 6 people can be selected from the group if equal numbers of men and women are to be on the committee?

Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles. Find the probability of both marbles being white.

I don't even know how to start these. It's been forever since I've done these. I know there has to be a simple way! Thanks!

Answer 1

10000 if 0000 is included , basically 1 to 9999 plus zeros

not sure of second one

1/16 for first white marble and 1/15 for second, so 1/240 for 2 white ones

Answer 2

there are 10,000 four digit numbers.

0000, 0001, 0002, .... , 9999

mathematically you find this are there are 10 different values for each of the four digits

10 * 10 * 10 * 10 = 10,000

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If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr

there are six people who will be selected, 3 males and 3 females.

10 C 3 = 120 combinations of 3 females
8 C 3 = 56 combinations of 3 males

120 * 56 = 6720 different committees

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Two ways to solve the marble problem.

one: to get draw both marbles at the same time and have two white marbles you have:

6 / 16 * 5 / 15 = 0.125 probability of drawing a both with marbles.

The other way to do this is to use the hypergeometric distribution.

you have
K = number of items to be drawn = 2 marbles
N = total objects = 16 marbles
M = number of objects of a given type = 6 white marbles

P(X = x | N, M, R) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K )

If you have n objects and chose r of them, the number of combinations is:
n! / ( r! (n-r)! )
this can be written as nCr

the N C K is the total number of possible combinations of K objects drawn from N objects.
the M C x is the number of combinations of getting x objects of the given type
the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn.

P( X = 2 ) = (6 C 2) * (10 C 0) / (16 C 2) = 0.125

Answer 3

C(n,m) is the number of ways to choose m things out of a population of n. C(n,m) = n!/m!(n-m)!

1. there are 10 choices for each digit. these are independent. 10^4 = 10000
2. C(10,3) * C(8,3) =120 * 56 = 6720
3. C(14,2) ways to draw 2 marbles from the bag. 91
C(6,2) ways to draw 2 marbles. 15
15/91

Answer 4

6 white + 4 red + 6 green = 16 marbles
probability 1st is white = 6/16 = 3/8
if 1st white then probability 2nd is white = (6-1)/(16-1) = 5/15 = 1/3
combined probability (3/8)x(1/3) = 3/24 = 1/8

Answer 5

The first answer is simple. It is 10^4 =10000.

However, the next two are not so simple. I suggest you draw out all the possibilities and the probabilities of each (tree diagram). That is the easiest way to calculate conditional probabilities.

Answer 6

a million. risk that he gets below 5 on the 1st roll = 4/6=2/3 (4 numbers below 5 on a die) risk that he gets even type on the 2d = 3/6=a million/2 (3 of the 6 numbers are even) the two activities are autonomous, so which you would be able to multiply them and get a million/3. 2. same concept. The risk of having heads on the 1st toss is a million/2. risk of having tails, on the 2d or third toss, is a million/2. Multiply a million/2 circumstances a million/2 circumstances a million/2 and you get a million/8.