Question

Question:What is the probability that a player of a lottery wins the prize offered for correctly choosing five (but not six)numbers out of six integers chosen at random from the integers between 1 and 40, inclusive?

Short answer

Answer

The probability that a player of a lottery wins the prize offered for correctly choosing five (but not six) numbers out of six integers chosen at random from the integers between 1 and 40, inclusive is$P\left(E\right)=5.31\times {10}^{-5}$ .

Step-by-step solution

 Given  

The integers between 1 and 40.

The Concept of Probability

If$\mathit{S}$represents the sample space and$\mathit{E}$ represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

role="math" localid="1668501739834" $\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

Determine the probability

As per the problem we have been asked to find the probability that a player of a lottery wins the prize offered for correctly choosing five (but not six) numbers out of six integers chosen at random from the integers between 1 and 40, inclusive.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

So, the number of possible ways of select 6 numbers out of the first 40 positive integers is$n\left(s\right){=}^{40}{C}_6$

Total number of ways of selecting 5 correct and 1 incorrect number isrole="math" localid="1668501898037" $n\left(E\right){=}^6{C}_5{\times }^{34}{C}_1$.

There will be only one set of positive integers for which lottery can be drawn.

The probability that a player of a lottery wins the prize offered for correctly choosing five (but not six) numbers out of six integers chosen at random from the integers between 1 and 40, inclusive is as follows:

$$\begin{gathered} P\left(E\right)=\frac{{}^6{C}_5{\times }^{34}{C}_1}{40} \\ P\left(E\right)=\frac{\frac{6!}{5!1!}\times \frac{34!}{133!}}{\frac{40!}{6!34!}} \\ P\left(E\right)=5.31\times {10}^{-5} \end{gathered}$$