There are 8 opposite-sex married couples at a party. Two people are chosen at random to win a door prize?

Question

a) what is the probability that the 2 people will be married to each other?
b) What is the probability that the 2 people will be of the same sex?
c) If 6 people are chosen, what is the probability that they are 3 married couples?

Answer 1

a) 1 in 15. Pick one person. Any will do. In order for (a) to be satisfied you have to pick the one person out of fifteen remaining that's the first person's spouse.

b) 7 in 15. Following similar logic from (a), once one person is picked 7 of the 15 remaining are going to have the same gender.

c) Wow, that's not an easy one! It winds up being 8C3/16C6, which is about 0.7%

Answer 2

Well ok you two do the easy ones I'll give c a crack.

6 people ok, 1/15 of choosing 1 married couple. 1/13 of getting that second one since 14 people remain. 1/11 of getting that third. Multiply all that crap out. Problem here is they would come in a substantial order.
1/15*1/13*1/11 would be odds they would get picked in order guy a girl a guy b girl b guy c girl c

try again
6 people, 3 couples, 18 total people. chance of a and his or her spouse being picked 5/15 (1/15 x 5 more picks). chance of b getting his spouse picked 3/13. 13 people remain and 3 chances eh? chance of c is 1/11.
5/15 * 3/13 * 1/11 is chance of them getting picked in any random order.
which is about 7.7%

actually that is too rounded, edit number 3!
chance (5/15+4/14+3/13+2/12+1/11)*(3/14+2/12+1/11)*(1/11). Logically that is odds of each person getting their spouse picked depending on when they are chosen. Problem now is that this assumes the spouseless people will be picked in the first, third, and fifth picks. Wow this problem is complicated.

lets just say... the odds are not very good eh?

Answer 3

a) 1 in 15
b) 7 in 15**
c) 1 in 143

Explanation:

a) You have 16_C_2 ways to pick 2 people from 16, and 8 of your choices are the 8 married couples. 16_C_2 is 16! / 2!(16-2)! = 120. 120/8 = 15.

b) You still have 16_C_2 combinations of 2 people, but the number of same-sex pairs is twice 8_C_2 (choosing a pair from 8, for each gender), so 120 / (2*28) = 120/56 = 15/7.

c) This one is tricky. There are 8_C_3 ways to draw 3 couples from 8 couples, which is 56 possible ways to choose 3 married couples from 8. There are 16_C_6 ways to draw 6 people from 16, or 8008 ways. 8008/56 = 143.

**I initially calculated this one wrong and corrected it when I saw my stupid error.

Answer 4

bottom of fractions, ways to choose 2 from 16, 16•15/2 = 120

a) 16 ways to choose 1st person, 1 way to choose spouse, divided by 2 arrangements, so (16/120)/2 = 1/15. alternatively, 8 couples to choose from, so 8/120 = 1/15.

b) (16•7/2)/120 = 56/120 = 7/15

c) 8C3 / 16C6 = [(8•7•6)/(1•2•3)] / [(16•15•14•13•12•11)/6!] =
56/(4•14•13•11) = 1/143

Answer 5

a) Once the first person is chosen, there are 15 choices for the second person, only one of whom is married to the first person. 1/15

b) Once the first person is chosen, there are 15 choices for the second person, only seven of whom are the same sex as the first person. 7/15

c) There are 16C6 ways to choose 6 people. There is only 1 way to choose the 6 to have 3 married couples. 1/(16C6)

Answer 6

a) 1/15
b) 1/2
c) Two answers at a time is my limit.

Answer 7

a)1/8

b)1/2

c) not sure havent done probabilities ina long time