Ohio Pick 4 In Ohio’s Pick 4 game, you pay $1 to select a sequence of four digits, such as 7709. If you buy only one ticket and win, your prize is $5000 and your net gain is $4999.
a. If you buy one ticket, what is the probability of winning?
b. Construct a table describing the probability distribution corresponding to the purchase of one Pick 4 ticket.
c. If you play this game once every day, find the mean number of wins in years with exactly 365 days.
d. If you play this game once every day, find the probability of winning exactly once in 365 days.
e. Find the expected value for the purchase of one ticket.
Answer :
Step 1 of 1
Given
Ohio Pick 4 In Ohio’s Pick 4 game, you pay $1 to select a sequence of four digits, such as 7709. If you buy only one ticket and win, your prize is $5000 and your net gain is $4999.
a)
If you buy one ticket, the probability of winning is.
Let we are altered to pick a 0 for the first digit.
Number of four digit sequence or
So, Number of four digit = 10,000
Now if we buy one ticket =
= 0.0001
Therefore the probability of winning is 0.0001
b)
Let the table describing the probability distribution corresponding to the purchase of one Pick 4 ticket.
Don’t get dollar back to that we used to buy ticket with 4999
Then the probability table is given below.
x 
p(x) 
x p(x) 

Win 
4999 
1/10000 
4999/10000 = 0.4999 
Lose 
1 
9999/10000 
9999/10000 = 0.9999 
Sum 
0.5 
c)
If you play this game once every day,the mean number of wins in years with exactly 365 days.
So we know that the mean number of wins in years with exactly 365 days.
Then Therefore the probability of winning is or 0.0001
= 365
= 365 (0.0001)
= 0.0365
Therefore the mean number of wins in years with exactly 365 days is 0.0365.
d)
If you play this game once every day.
We have to find the probability of winning exactly once in 365 days.
= 365
= 0.0352
Therefore the probability of winning exactly once in 365 days = 0.0352
e)
Let the expected value for the purchase of one ticket.
The expected value formula is given by
Where
x =an outcome
μ = mean
) = probability of that out come
E(x) = expected value
x p(x) 
4999/10000 = 0.4999 
9999/10000 = 0.9999 
0.5 
From the above table we know that
Therefore the expected value for the purchase of one ticket = 0.5