Answer to Question #126009 in Statistics and Probability for paula cortes

Question

The following line was given in a solution to a question.
P(A)= (6C3 x 43C3 ) + ( 6C4 x 43C2 ) divided by 49 6

What might the question have been? Be sure to identify and explain everything that you know is true about the question.

Answer 1

Solution:

Take for example Canadian lottery 6/49.

The question could be as follow:

"Let A be event that there are 3 winning numbers and 3 losing numbers or

4 winning numbers and 2 losing numbers. Determine the probability of event A"

Then P(A)=m/n.

₆C₃ - number of ways to draw 3 numbers from 6 winning numbers,

₄₃C₃ - number of ways to draw 3 numbers from 43 losing numbers,

₆C₄ - number of ways to draw 4 numbers from 6 winning numbers,

₄₃C₂ - number of ways to draw 2 numbers from 43 losing numbers,

m=₆C₃ x ₄₃C₃ +₆C₄ x ₄₃C₂.

₄₉C₆ - number of ways to draw 6 numbers from 49 numbers,

n=₄₉C₆

P(A)=(₆C₃ x ₄₃C₃ +₆C₄ x ₄₃C₂)/₄₉C₆=

=(6!/3!/3!*43!/3!/40!+6!/4!/2!*43!/2!/41!)/(49!/6!/43!)=37195/1997688.

Here _nC_k=n!/(k!(n-k)!) is binomial coefficient.