# [SOLVED] Probablility

• Nov 27th 2007, 01:17 AM
Melinda
[SOLVED] Probablility
A Carton of eggs contains three 55g eggs, five 59g eggs and two 52 g eggs. Work out the probability of choosing two 55 eggs at random?

I would really appriciate it if someone could show me how to do this as i need to know it for an upcoming exam. Thankyou
• Nov 27th 2007, 02:44 AM
PaulRS
Welcome, you are taking pairs I see, thus the possible outcomes are the 2-combinations (It doesn't matter the order) of those elements. The positive case is getting ( 55egg, 55egg ) :)

The number of possible cases is $\binom{10}{2}$
And the positive ones, just 1

Therefore : $p(two 55 eggs) = \frac{1}{\binom{10}{2}}=\frac{1}{\frac{10!}{2!\cdo t{8!}}}=\frac{1}{45}$
• Nov 27th 2007, 10:17 AM
Soroban
Hello, Melinda!

There are two approaches to this problem (at least) . . .

Quote:

A carton of eggs contains three 55g eggs, five 59g eggs and two 52g eggs.
Find the probability of choosing two 55g eggs at random

I'll assume that the two eggs are drawn without replacement.

[1] There are 10 eggs and 3 of them are 55g eggs.
The probability that the first egg drawn is a "55" is: . $\frac{3}{10}$
The probability that the second egg is a "55" is: . $\frac{2}{9}$
. . Therefore: . $P(\text{two 55s}) \:=\:\frac{3}{10}\cdot\frac{2}{9} \:=\:{\color{blue}\frac{1}{15}}$

[2] There are: . ${10\choose2} \:=\:45$ possible pairs of eggs that can be drawn.
There are: . ${3\choose2} \:=\:3$ ways to draw two "55s".
. . Therefore: . $P(\text{two 55s}) \:=\:\frac{3}{45} \:=\:{\color{blue}\frac{1}{15}}$

• Nov 27th 2007, 01:35 PM
PaulRS
Oh, I confused the number of 55g eggs with the 52 g eggs (Doh)