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More Chance of Being Trampled By a Herd of Elephants!

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#37648
Fun:*** (2.23)
Difficulty:*** (2.35)
Submitted By:leftclickAau*****!
Corrected By:nerdyiscool




Leftclick Lotteries Inc. runs a lottery in which 5 balls are drawn from a barrel of numbered white balls (so that each white ball may be drawn once only), and 1 ball is drawn from a different barrel of numbered black balls. There are 45 white balls and 45 black balls. Players must correctly select the numbers of all 5 white balls, as well as that of the black ball in order to win the main prize.

What is the probability of winning the main prize with any single given entry?


The probability of winning the main prize is 1 in 54,979,155.

The probability of winning is the probability of guessing each ball correctly. For the black ball, this is 1/45. For each white ball, it is 5/45 (there are 5 correct balls out of 45 possible), 4/44 (there is 1 less correct ball and 1 less to choose from), 3/43, 2/42 and 1/41. This gives a probability of winning of 1 in 45*(45!)/(40!)*(5!), or 1 in 54,979,155.

The title is an expression that describes an event as being highly unlikely. Of course, the actual chance of being trampled by a herd of elephants is very difficult to measure and varies widely!


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Aug 04, 2007

I don't believe this answer is correct, unless the order in which the balls are drawn must also be matched. (45*44*43*42*41) represents all permutations of drawing 5 white balls. However, it is the combinations that matter, not the permutations, therefore this should be divided by 5! (or 5*4*3*2*1). The odds are still very, very, very much against winning however, which is why I don't buy lottery tickets.
Aug 04, 2007

I agree with gizzer. It was not stipulated that order matters, so divide by 120. (45!)/((40!)(5!))
Aug 04, 2007

I don't... understand =D
Aug 05, 2007

at first I thougt I had it right, bt after reading the comments, I see I was wrong. Good catch guys.
Aug 05, 2007

Yes it is wrong, I will submit a correction...

The first ball has a 5 in 45 chance of being one of the numbers picked, the second ball has a 4 in 45 chance, and so on... My answer says that they all have only a 1 in 45 chance.

Jan 21, 2008

I arn't very good at this probability stuff. 51! means as much mathematical sence to me as 6435.46!
Mar 19, 2008

i didn't get that 1 correct
Apr 21, 2008

Would be 1 chance in : 45[45!/(40!x 5!)] or 1/54,979,155.

There are C(45,5) ways of choosing the white balls and hence 45 x C(45,5) of choosing all of the balls. The desired probability is 1/[(45 x C(45,5) ].

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