# The state runs a lottery once every week in which six number

The state runs a lottery once every week in which six numbers are randomly selected from 16 without replacement. A player chooses six numbers before the state’s sample is selected. The player wins if all 6 numbers match. If a player enters one lottery each week, what is the probability that he will win at least once in the next 200 weeks? Report answer to 3 decimal places.
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Neil Dismukes
Step 1
In this question, there is a lottery in which a person needs to select the 6 numbers if those 6 numbers are in state’s sample then he/she win the lottery. We need to find the probability that he will win at least once in next 200 weeks.
Step 2
Let’s divide the numbers in two group,
Group 1, Those numbers which are in state’s sample
Group 2, Those numbers which are not in state’s sample
The player selects the 6 numbers he wins when all 6 numbers are in state’s sample,
Therefore, Probability of winning a lottery;
Calculations have shown below;
Probability of winning $=\frac{6{C}_{6}×10{C}_{0}}{16{C}_{6}}$
$=\frac{1}{8008}$
Step 3
Now, probability of in next 200 lotteries a person will wins at least once;
We will use the binomial distribution to get the required probability,
$P\left(x\ge 1\right)=1-P\left(x=0\right)$
$=1-200{C}_{0}×{\left(\frac{1}{8008}\right)}^{0}×{\left(1-\frac{1}{8008}\right)}^{200}$
$=0.025$
###### Not exactly what you’re looking for?
Thomas Lynn
Step 1
Hyper geometric distribution:
A hyper geometric distribution is a discrete probability distribution that determines the probability of getting k successes in n draws (without replacement) from a finite population of size N that contains exactly K success states.
Denote the total number of successes as k,
Denote the total number of objects that are drawn without replacement as n,
Denote the population size as N,
Denote the total number of success states in the population as K.
The probability distribution of k is a hyper geometric distribution with parameters (N, K, n) and the probability mass function (pmf) of k is given as:
$P\left(k\right)=\frac{\left(\begin{array}{c}K\\ k\end{array}\right)×\left(\begin{array}{c}N-K\\ n-k\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
Step 2
Parameters of hyper geometric distribution for the given situation:
In a state’s lottery, 6 numbers are randomly selected from a total of 17 numbers without replacement.
Here, the population size is $N=17$,
The number of draws is $n=6$.
Furthermore, it is given that, a player chooses 6 numbers before the state’s sample is selected and the player wins, if all the 6 numbers match with the state’s sample.
Thus, the number of success states in the population is $K=6$.
The probability distribution of k successes is a hyper geometric distribution with parameters ($N=17,K=6,n=6$) and the probability mass function (pmf) of k is given as:
$P\left(k\right)=\frac{\left(\begin{array}{c}6\\ k\end{array}\right)×\left(\begin{array}{c}17-6\\ 6-k\end{array}\right)}{\left(\begin{array}{c}17\\ 6\end{array}\right)}$
Step 3
Probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample:
It is given that, 3 of the 6 chosen numbers have to appear in the state’s lottery.
That is, the number of successes is $k=3$.
The probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample is obtained as 0.267 from the calculation given below:
$P\left(k=3\right)=\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)×\left(\begin{array}{c}17-6\\ 6-3\end{array}\right)}{\left(\begin{array}{c}17\\ 6\end{array}\right)}$
$=\frac{20×165}{12,376}$
$=\frac{3,300}{12,376}$
$=0.2666$
$=0.267$ (Rounded to 3 decimal places)
Step 4
The probability that exactly 3 of the 6 numbers chosen by a player appear in the state’s sample is 0.267.