Probability Question!

• Apr 18th 2007, 06:42 AM
mattsh
Probability Question!
Hi everyone,

I need help on a probability question please. Thanks in advance. Here it is:

1/3 of the people in a club are men.
The number of men in the club is n.

a) Write down an expression, in terms of n, for th enumber of people in the club.

b) Two of the people in the club are chosen at random. The probability that both these people are men is one tenth. Calculate the number of people in the club.

Thanks for any help!
Matt
• Apr 18th 2007, 07:10 AM
Jhevon
Quote:

Originally Posted by mattsh
Hi everyone,

I need help on a probability question please. Thanks in advance. Here it is:

1/3 of the people in a club are men.
The number of men in the club is n.

a) Write down an expression, in terms of n, for th enumber of people in the club.

let P be the number of people in the club. 1/3 of the people in the club are men, so if n is the number of men:

n = (1/3)P
=> P = 3n

Quote:

b) Two of the people in the club are chosen at random. The probability that both these people are men is one tenth. Calculate the number of people in the club.
The probability of choosing one man is n/3n = 1/3, the probability of choosing the second man will be (n – 1)/(3n – 1) since after choosing one man, we have one less man in the group of men to choose from as well as one less person in the club in general to choose from. We want both these probabilities to happen at the same time, so we multiply them. So we have:

1/3 * (n – 1)/(3n – 1) = 1/10 .........since 1/10 is the probability of these two events happening at the same time.

=> (n – 1)/(3n – 1) = 3/10
=> 10(n – 1) = 3(3n – 1) ...............cross multiplied
=> 10n – 10 = 9n – 3
=> n = 7 .....................................the number of men in the club

=> P = 3n = 3(7) = 21 ...................the number of people in the club.