In the SuperLottery, three balls are drawn (at random) from ten white balls numbered from 1 to 10, and one SuperBall is drawn (at random) from ten red balls numbered from 11 to 20. When you buy a ticket, you choose three numbers from 1 to 10 and one number from 11 to 20.

If the numbers on your ticket match at least two of the white balls or match the red SuperBall, then you win a super prize. What is the probability that you win a super prize?

I tried this many times but I'm stumped. I would also like to have a step-by-step explanation of how you did it please. thanks for your help if you answered this!

Guest May 1, 2022

#1**0 **

On average, 1 out of 10 times you will draw the Superball (and this wins all by itself so we don't care what the other balls are) for a winning probability of 0.10.

Now, for the 9 out of ten times that you do not draw the Superball:

We will have to get a losing blue ball, I'll call that B (and the probability of getting that losing ball is 0.90).

We can win by getting two or more of the white balls. I'll classify a winning white ball (whose probability is 1/10 = 010)

as W and a losing white ball as L (whose probability is 9/10 = 0.90).

So, we can still win if we have this:

B x W x W x W = (0.90) x (0.10) x (0.10) x (0.10) = 0.0009

B x W x W x L = (0.90) x (0.10) x (0.90) x (0.10) = 0.0081

B x W x L x W = ...

B x L x W x W = ...

Since these are independent, add the above 5 results together: 0.10 + 0.009 + 0.0081 + ... + ... = 313/2500.

Guest May 1, 2022

#3**0 **

There are 3 cases to consider: 2 white balls, 3 white balls, and the Powerball

Powerball: There is a \(1 \over 10 \) chance of this happening, and it is a win automatically

2 white balls: There are \({10 \choose 3 } = 120\) ways to choose the 3 numbers. There are \({3 \choose 2} = 3\) ways to choose the 2 numbers that will be drawn. We also need to pick one of the 7 losing numbers. Thus, the probability of this happening is \({ 7 \times 3 \over 120} = {7 \over 40}\)

3 white balls: There are \(10 \choose 3\) ways to choose the 3 numbers. There is \({3 \choose 3 }= 1\) way to choose the 3 numbers. Thus, the probability of this is \(1 \over 120\)

Thus, the total probability is \({1 \over 10} + {7 \over 40} + { 1 \over 120} = \color{brown}\boxed{17 \over 60}\)

BuilderBoi May 2, 2022