SOLUTION: Winning the jackpot in a particular lottery requires that you select the correct three numbers between 1 and 64 ​and, in a separate​ drawing, you must also select the correct s

Algebra ->  Permutations -> SOLUTION: Winning the jackpot in a particular lottery requires that you select the correct three numbers between 1 and 64 ​and, in a separate​ drawing, you must also select the correct s      Log On


   



Question 1206897: Winning the jackpot in a particular lottery requires that you select the correct three numbers between 1 and 64 ​and, in a separate​ drawing, you must also select the correct single number between 1 and 27. Find the probability of winning the jackpot.

Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!

there are 64 numbers between 1 and 64.
there are 27 numbers between 1 and 27.

the probability of getting 3 correct numbers between 1 and 64 is 1/64 * 1/64 * 1/64.

the probability of getting 1 correct number between 1 and 27 is 1/27.

the probability of getting 3 correct numbers between 1 and 64 and 1 correct number between 1 and 27 is 1/64 * 1/64 * 1/64 * 1/27.

that probability is equal to 1 / 7,077,888.

that's what it is, if i analyzed it correctly.















Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

In Math problems on lotteries, it is traditionally assumed that 

    (1)  the numbers on a ticket are different (no repeating),

and

    (2)  the order of numbers on a ticket does not matter to determine winning.


If follow to the tradition in this current problem, it should be stated in the 
problem, that 


    +---------------------------------------------------------------------+
    |   tree integer numbers from 1 to 64 on a ticket are different       |
    |   (no repeating) and the order of the three numbers on the ticket   | 
    |                    does not matter.                                 |
    +---------------------------------------------------------------------+


Below is my solution to the problem in this formulation.


There are  C%5B64%5D%5E3 = %2864%2A63%2A62%29%2F%281%2A2%2A3%29 = 41664 different possible triples of numbers.


Combining it with the separate integer number from 1 to 27, we have, in all,

     41554*27 = 1121958  possible outcomes (or tickets).


Only one of these tickets wins. So, the probability of winning is

    P = 1/1,121,958 = 1%2F1121958 = 8.91299E-07.


It is a standard pattern of analysing/solving traditional Math problerms on lotteries.

As you see, the analysis and my answer are significantly different from that by @Theo.

Solved.


===================


My opinion is that, as a Math problem, this post must be worded differently,
to reflect all features of a lottery in explicit form.