# Probability Questions Help

1. May 5, 2012

### xChee

Need someone to check if my answers to these probability questions are right :\$

Q1: Two jokers are added to a standard 52 card deck. The deck is then shuffled, what is the probability that someone draws 5 cards and two of those 5 cards happen to be both jokers.

is this what you do? 5C3 / 54C5 ?

Q2: 7 members of the swimming club line up for a picture. What is the probability that Marry and Jory will sit together?

Q3: There is a 15% of winning a game, what are the odds against winning the game?

Q4: From a group of 7 juniors and 10 seniors, determine how many communities of 6 students can be chosen if 4 are juniors?

2. May 6, 2012

### Eutrophicati

Q1: Two jokers are added to a standard 52 card deck. The deck is then shuffled, what is the probability that someone draws 5 cards and two of those 5 cards happen to be both jokers.

is this what you do? 5C3 / 54C5 ?

This one looks like a hypergeometric distribution where you have dependent events (i.e. drawing one joker affects the probability of drawing a second). Don't let the name scare you if you haven't heard of it, it's nothing that fancy :)

I think it should be
(2C2 *49C3) / (54C5)

Population (N) = 54
Total population successes (M) = 2
Sample (n) = 5
Total sample successes (m) = 2

The formula for getting this is basically the (combination of successes * combination of failures) / (total combinations).

Or (MCm)*([N-M]C[n-m])/ (NCn)
It's not as complicated as it looks, you'll notice some patterns in the terms, the first value is M, N-M, and N; the second values are m, n-m, n.
Eminem 'n' Eminem 'n'
Maybe the 'min' can remind you of the minus sign. Dunno =p

(2C2 *49C3) / (54C5)

I'd like to help you with the other questions too, but I have a stats exam tomorrow, this question just caught my eye because I was just going over the different distributions XD. Oh, and you've got number 3 right.

3. May 13, 2012

### haruspex

Q1. Think of the 54 cards as the positions that the 2 jokers may occupy. 5 of them are the drawn 5. There are 54C2 ways of positioning both jokers, of which 5C2 put them in the drawn 5. Answer: 5C2/54C2.

Q2. M equally likely occupies any of 7 positions. In 2 cases there is a 1/6 chance J will be adjacent. In the other 5 cases the odds are 2/6. Answer: (2/7)*(1/6)+(5/7)*(2/6) = 12/42 = 2/7.
Alternatively, there are 7C2 = 21 pairs of positions M and J can occupy, of which 6 consist of adjacent positions. 6/21 = 2/7.

Q3. & Q4 you have correct.