AN IB MATH SL INTERNAL ASSESSMENT ABOUT THE PROBABILITY AND LOTTERIES
Application of Birthday Paradox
With regards to the investigation of the application of birthday paradox, it becomes vital to follow some basic rules that relates with lottery1. In this case, consider the situation of a ticket SuperCash. In this case, make a choice of six various numbers that have a range from 1 to 39 with the objective of marking them on the panels. In this regard, it is possible to calculate the odds for winning the jackpot of a value of $350,000 through the use of the combinatorics simply because the order of the numbers does not have an effect on the outcomes. ...view middle of the document...
For instance, the first person does not see the significance of his birthday though the second guy his birthday may find it possible in getting his birthday by finding the probability that the same person does not share his birthday with the first person. Hence, this instance should give the total days that are possible less one. On the other hand, when the third person finds his way into the room, in this case, the probability that this very person will not share his birthday with the initial person in this case is the total number of possible days less two. In consequence, through the simple multiplication of the numbers, it then leads to the attaining of the probability4. Likewise, this pattern should progress until the attainment of the probability of the 23rd person altogether.
3Huck, Schuyler W. 2012. Reading Statistics And Research. Boston: Pearson.56
365365× 364365×363365×362365…343365=probablity of not having someone with the same birthday≅0.493≅49.3%
(Assuming of all the options of days is 365 and that the probability for each day is similar)
Elsewhere, to obtain the complementary value, it is then just the simple subtraction of 1 form the value5. For example, it signifies that the probability for at least two people that have the same birthdays is equivalent to 1 less the probability of not being with someone that has the same birthday; 1-0.4893- 0.507=50.7%. As much as this scenario may appear odd for just mere number consisting of 23 people that have a 50% chance of someone that has similar birthday, in other cases it may appear that different. Consequently it may call for looking for the paired numbers consisting of 23 people with the capability to make with their birthdays where in this case it will be the number that leads to the creation of a more realistic fact or idea altogether. For this reason, by use of the combinatorics;
∅k= ∅!k!∅-k!=23!2!21!=253 pairs
Consequently, it is then possible to deduce the general patters from the birthday paradox as;
5Wazir, Ibrahim, Tim Garry, Peter Ashbourne, Paul Barclay, Peter Flynn, Kevin Frederick, and Mike Wakeford. 2012. Mathematics 2012 Edition - Higher Level. London: Pearson.54
Where n ∈R, n ≤ ∅; n=number of people; ∅=Total number of sets of value;c = probability of not having a repeating birthday
Where z = probability of having a repeating birthday
In this regard, then the formula assumes;
By applying the Wisconsin SuperCash! Formula, there indeed a lot of changes such as the need by a person to search for set of 6 numbers where that can lead to the purchase of the whole lottery as a single entity6. With respect to the same, being that there are 322623 variations that may lead to being a winner for the lottery such as various sets of numbers, also there are 3262623 numbers of combinations such as (∅=3262623). On the same note, n is the representation of the combinations that constitute...