Answer:
[tex]Pr =\frac{1}{38955840}[/tex]
Step-by-step explanation:
Given
[tex]n = 35[/tex] ---- 1 to 35
[tex]r = 5[/tex] -- selection
Required
The probability of winning
The probability of getting the first number correctly is:
[tex]P(1) = \frac{1}{35}[/tex]
At this point, the remaining numbers are 34
So, the second selection has the following probability
[tex]P(2) = \frac{1}{34}[/tex]
Following the above sequence, we have:
[tex]P(3) = \frac{1}{33}[/tex]
[tex]P(4) = \frac{1}{32}[/tex]
[tex]P(5) = \frac{1}{31}[/tex]
So, the required probability is:
[tex]Pr =P(1) * P(2) * P(3) * P(4) * P(5)[/tex]
[tex]Pr =\frac{1}{35} *\frac{1}{34}*\frac{1}{33}*\frac{1}{32}*\frac{1}{31}[/tex]
[tex]Pr =\frac{1}{35*34*33*32*31}[/tex]
[tex]Pr =\frac{1}{38955840}[/tex]
