Question

The local lottery is found by randomly selecting 6 ping pong balls from a container in order without replacement. There are 30 ping-pong balls in the container numbered 0 through 29. 1.) What is the probability you hold the winning ticket? 2.) How many tickets should you buy to give yourself a 1% chance of winning the lottery? A 10% chance? A 25% chance? A 50% chance? 3.) What is the probability all 6 numbers are even? At least one number is odd? 4.) What is the probability each number will be a single digit? 5.) What is the probability your birthday will be the winning ticket? (mm,dd,yy) 6.) Complete parts 1-5 again under the assumption the order of the digits is irrelevant. How are your chances of winning affected? Better? Worse? Explain

I ONLY NEED HELP WITH PART 6

Answer #1

Number of tickets for 1% chance = 0.01 / 0.000001684 = 5938.24 = 5939.

**6)** Since the order of the digits is irrelevant
we need to calculate combinations instead of permutations.

Probability of holding a winning ticket = 1 / 30C6 = 1 / 593775 = 0.000001684.

Since this value is bigger than the earlier value, the chances of winning are increased.

Number of tickets for 1% chance = 0.01 / 0.000001684 = 5938.24 = 5939.

Number of tickets for 10% chance = 5938.24 *10 = 59382.4 = 59383.

Number of tickets for 25% chance = 0.25 / 0.000001684 = 148456.06 = 148457.

Number of tickets for 50% chance = 0.5 / 0.000001684 = 296912.11 = 296913.

In all cases above, since fewer tickets need to be bought, the chances of winning are increased.

There are 15C6 ways to choose even numbers out of 30C6.

Probability of all even numbers = 15C6 / 30C6 = 5005 / 593775 = 0.0084

Probability that atleast one is odd = 1 - 0.0084 = 0.9916

There are 10C6 ways to choose single digit numbers out of 30C6.

Probability of all single digit numbers = 10C6 / 30C6 = 210 / 593775 = 0.00035.

**Probability of six digit winning ticket = 1 / 210 =
0.0047.**

**Since this value is greater than the earlier value, the chance of winning is increased.**

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