Question

A restaurant hands out a scratch-off game ticket with prizes being worth purchases at the restaurant. The back of the ticket lists the odds of winning each dollar value: 0.05 for $10, 0.04 for $25, 0.01 for $50, and 0.003 for $100. What are the odds that the ticket is worth at least $25?

Answer 1

Answer: 0.05412

Step-by-step explanation:

Formula : Odds of having an event is given by  $o=\dfrac{p}{1-p}$, where p = probability that event happens.

In terms to find p , we use $p=\dfrac{o}{1+o}$

Given, he back of the ticket lists the odds of winning each dollar value: 0.05 for $10, 0.04 for $25, 0.01 for $50, and 0.003 for $100.

Let X be the worth of ticket.

Then, the probability that the ticket is worth at least $25 =

$P(X\geq 25)=P(X=25)+P(X=50)+P(X=100)$

$=\dfrac{0.04}{1+0.04}+\dfrac{0.01}{1+0.01}+\dfrac{0.003}{1+0.003}\\\\=0.05135$

The odds that the ticket is worth at least $25 = $\dfrac{0.05135}{1-0.05135}$

=0.05412

hence, the odds that the ticket is worth at least $25 is 0.05412 .