Question

On September​ 11, 2002, a particular state​ lottery's daily number came up 9-1-1. Assume that no more than one digit is used to represent the first nine months. ​a) What is the probability that the winning three numbers match the date on any given​ day?​ b) What is the probability that a whole year passes without this​ happening? ​c) What is the probability that the date and winning lottery number match at least once during any​ year? ​d) If 27 states have a​ three-digit lottery, what is the probability that at least one of them will come up 1-1-4 on January 14​?

Answer 1

The probability that the winning three numbers match the date on any given​ day is the likelihood of the winning number

• The probability that the winning three numbers match the date on any given​ day is 0.001
• The probability that a whole year passes without this​ happening is 0.7407
• The probability that the date and winning lottery number match at least once during any​ year is 0.2593
• The probability that at least one of them will come up 1-1-4 on January 14​ is 0.0267

The probability that the winning three numbers match the date on any given​ day

Start by calculating the number of possible outcomes

Each of the three digits can take any of 0 - 9 i.e. 10 digits.

So, the number of possible outcomes is:

$n =10^3$

$n =1000$

There can be only one winning date.

So, the probability (p) is:

$p = \frac 1n$

This gives

$p = \frac 1{1000}$

$p = 0.001$

Hence, the probability that the winning three numbers match the date on any given​ day is 0.001

The probability that a whole year passes without this​ happening

In a calendar year, there are 65 days, where the whole year passes without matching the winning numbers.

The dates are:

• October 10- October 31 (22 days)
• November 10 - November 30 (21 days)
• December 10 - December 31 (22 days)

This means that the matching numbers can occur on the remaining 300 days.

The probability that the winning date does not occur on a date is calculated using the following complement rule

$p' = 1 - p$

So, we have:

$p' = 1 - 0.001$

$p' = 0.999$

For the 300 days, we have:

$p = 0.999^{300}$

$p = 0.7407$

Hence, the probability that a whole year passes without this​ happening is 0.7407

The probability that the date and winning lottery number match at least once during any​ year

This is calculated using the following complement rule

$p' = 1 - p$

So, we have:

$p' = 1 - 0.7407$

$p' = 0.2593$

Hence, the probability that the date and winning lottery number match at least once during any​ year is 0.2593

The probability that at least one of them will come up 1-1-4 on January 14​

In (b), we have:

$p' = 0.999$

So, the probability is calculated as:

$p =1 - (p')^n$

Where n is the number of states i.e. n = 27.

So, we have:

$p = 1 - 0.999^{27}$

$p = 0.0267$

Hence, the probability that at least one of them will come up 1-1-4 on January 14​ is 0.0267

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