Math Question: Many states in U.S.A have a lottery game, usually called a Pick-4, in which you pick a four-digit number such as 7359. During the lottery drawing, there are four bins, each containing balls numbered 0 through 9. One ball is drawn from each bin to form the four-digit winning number. a. You purchase one ticket with one four-digit number. What is the probability that you will win this lottery game? (2 marks) b. There are many variations of this game. The primary variation allows you to win if the four digits in your number are selected in any order as long as they are the same four digits as obtained by the lottery agency. For example, if you pick four digits making the number 1265, then you will win if 1265, 2615, 5216, 6521, and so forth, are drawn. The variations of the lottery game depend on how many unique digits are in your number. Consider the following four different versions of this game. Find the probability that you will win this lottery in each of these four situations. i. All four digits are unique (e.g., 1234) ii. Exactly one of the digits appears twice (e.g., 1223 or 9095) iii. Two digits each appear twice (e.g., 2121 or 5588) I have the answers but I do not have the working. Can you please show the working. a.) .01% b.) i.) 0.0048 ii.) 0.0028 iii.) 0.0222

Asked By Avishan On 10/20/2020 22:48

Math Answers

Answer: a. Since this drawing is randomly happen, so assume the probability of winning the game is P(A). The first figure of drawing to win is 1/10. Because this is randomly happen, without considering the que, so the probability of second figure to win is 1/10 as well. So, as the third and the forth figure. Therefore, the probability of total 4 figures to win is P(A)=1/10*1/10*1/10*1/10=1/1000=1/10000=0.01% b. i) Assume the probability of winning of drawing is P(b). Under the condition of selected, P(b) has to consider the que. As the figure is selected, assume they are W X YZ. As they are unique, so each figure can not be the same. So, P(b)=P(w)*P(x)*P(Y)*P(Z). P(w)=4/10( 4 figures selected by the agency, the pool of drawing is 10 figures from 0-10). P(X)=3/9( 3 figures left to be drawn from agency, the pool has drawn 1 figure for P(w), so only 9 figures in pool). P(Y)=2/8 P(Z)=1/7 So, P(b)=4/10*3/9*2/8*1/7=0.00476 approximately to 0.0048=0.48%
Answered On 10/28/2022 08:34

Eric
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