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In a certain​ lottery, an urn contains balls numbered 1 to 34. from this​ urn, 6 balls are chosen​ randomly, without replacement. for a​ $1 bet, a player chooses one set of six numbers. to​ win, all six numbers must match those chosen from the urn. the order in which the balls are selected does not matter. what is the probability of winning this lottery with one​ ticket? g

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konrad509

[tex] \displaystyle
|\Omega|=\binom{34}{6}=\dfrac{34!}{6!28!}=\dfrac{29\cdot30\cdot31\cdot32\cdot33\cdot34}{2\cdot3\cdot4\cdot5\cdot6}=1344904\\
|A|=1\\\\
P(A)=\dfrac{1}{1344904} [/tex]

Using the combination formula, it is found that there is a [tex]p = \frac{1}{1344904}[/tex] probability of winning the lottery with one ticket.

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  • A probability is the number of desired outcomes divided by the number of total outcomes.
  • In this problem, the order in which the balls are chosen is not important, thus, the combination formula is used to find the number of total outcomes.
  • You win the lottery with only one outcome.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem:

  • The total numbers of outcomes is 6 numbers from a set of 34, thus:

[tex]C_{34,6} = \frac{34!}{6!28!} = 1344904[/tex]

Thus, the probability of winning with one ticket is:

[tex]p = \frac{1}{1344904}[/tex]

A similar problem is given at https://brainly.com/question/24650047

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