Question

Question:What is the probability that six consecutive integers will be chosen as the winning numbers in a lottery where each number chosen is an integer between 1 and 40 (inclusive)?

Short answer

Answer

The resultant answer is\(\frac{1}{{109,668}} \approx 9.1184 \times {10^{ - 6}}\).

Step-by-step solution

Given data

The given expression is six consecutive integers

Concept of Permutation 

Definition permutation (order is important):\(P(n,r) = \frac{{n!}}{{(n - r)!}}\)

Definition combination (order is not important): \(C(n,r) = \left( {\begin{aligned}{{}{}}n\\r\end{aligned}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

with \(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\).

Simplify the expression 

There are 40 numbers between 1 and 40 (inclusive):

\(\begin{aligned}{}n = 40\\r = 6\end{aligned}\)

The order of the numbers does not matter (since a different order result in the same numbers being selected), thus we need to use the definition of a combination.

Repetition of the numbers is not allowed:

\(\begin{aligned}{}\# {\rm{ of possible outcomes }} &= C(40,6)\\\# {\rm{ of possible outcomes }} &= \frac{{40!}}{{6!(40 - 6)!}}\\\# {\rm{ of possible outcomes }} &= \frac{{40!}}{{6!34!}}\\\# {\rm{ of possible outcomes }} &= 3,838,380.\end{aligned}\)

There are 35 possible combinations of 6 numbers from 1 to 40 such that the numbers are consecutive \((1 - 6,2 - 7,3 - \)\(8, \ldots ,35 - 40)\).

\(\# \)of favorable outcomes \( = 35\)

We then have 1 chance in 3,838,380 to select the winning combination:

\(\begin{aligned}{}P({\rm{ win }}) &= \frac{{\# {\rm{ of favorable outcomes }}}}{{\# {\rm{ of possible outcomes }}}}\\P({\rm{ win }}) &= \frac{{35}}{{3,838,380}}\\P({\rm{ win }}) &= \frac{1}{{109,668}}\\P({\rm{ win }}) \approx 9.1184 \times {10^{ - 6}}\end{aligned}\)