SOLUTION: Hi, please help me out with this question. It is very confusing and I don't know how to approach it. The tickets for a raffle have four different numbers chosen from 3- 13, wit

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Question 1189691: Hi, please help me out with this question. It is very confusing and I don't know how to approach it.
The tickets for a raffle have four different numbers chosen from 3- 13, with order not important. Determine the probability of winning first prize, which consists of matching all four numbers?
a. 0.3322%
b. 0.3%
c. 29%
d. 0.3030%
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Hi, please help me out with this question. It is very confusing and I don't know how to approach it.
The tickets for a raffle have four different numbers chosen from 3- 13, with order not important.
Determine the probability of winning first prize, which consists of matching all four numbers?
a. 0.3322%
b. 0.3%
c. 29%
d. 0.3030%
~~~~~~~~~~~~~~~~~ 

There are 11 numbers  3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.


In all, there are   =  = 330 different combinations of 11 items, taken 4 at a time
(330 quadruples).


Only one such combination (quadruple of numbers) provides winning the first prize.


So, the probability is   = 0.00303 = 0.3030%  (rounded).


ANSWER.  Option (d).

Solved.

--------------------

This problem is on COMBINATIONS.

On Combinations,  see introductory lessons
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - OVERVIEW of lessons on Permutations and Combinations
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.



Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
there are 11 numbers from 3 to 13.
you draw 4 times.
the probability of getting one of the 4 numbers on the first try is 4/11.
the probability of getting one of the remaining 3 numbers on the second try is 3/10.
the probability of getting one of the remaining 2 numbers on the third try is 2/9.
the probability of getting the remaining 1 number on the fourth try is 1/8.
the combined probabiity is 4/11 * 3/10 * 2/9 * 1/8 = .003030303.....
that is equal to .30303030.....%
round that to 4 decimal places to get .3030%.
that would be selection d.


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