Ok so you have a 1/40 chance of getting the first number correct. You also have a 1/40 chance of getting the second number correct and so on.
So you need to beat the odds of
1/40 x 1/40 x 1/40 x 1/40 x 1/40 x 1/40 =
To understand why we multiply the odds together, we can use a simplified model. Say we want 6 heads in a row in a coin toss. For the first flip, you got the possibility for getting a head or a tail. So the odds are 1/2 for a head.
Now for 2 flips, you have four possibilities.
- Head - Head
- Head - Tail
- Tail - Head
-Tail - Tail
Now there is only one possibility out of four for a head-head combination. So the odds are 1/4.
For every additional flip, each of the combination branches off into two more with an added heads or tails, doubling the # of possible combinations whilst there will only be one correct all heads combination.
Thank you so much, you did a wonderful job of explaining that. I understand perfectly now.
What if the ball is not returned after it is drawn?
I suppose it would
1/40 1/39 1/38 e.t.c and your chances of getting all six would be eqaul to timesing them all together.
I have another question though, in this case, what would the chances of getting say, two balls correct? How would you figure that out?
Thanks again for the great explanation.
This number needs to be multiplied by 6! because the order of the numbers is not important.
That is not correct since the numbers you choose could be drawn, but not in the order that you chose them in.
The correct answer is , where is the number of ways of choosing 6 balls from 40 balls.
I'm really confused now. In the first instant (when the balls are returned), you say the chances are 49!, what does this mean and why is it the case (can you show the working).
Same with when the ball is not returned.
Sorry for all the questions, my math is really bad and i'm determinded to understand this.
Thanks for all the help so far.