A radio station runs a promotion at an auto show with a money box with 15 $50 tickets, 11 $25 tickets, and 15 $5 tickets. The box contains an additional 20 "dummy" tickets with no value. Three tickets are randomly drawn. Find the probability that all three tickets have no value.

0.0316

Step-by-step explanation:

Given:

A radio station runs a promotion at an auto show with a money box with 15 $50 tickets, 11 $25 tickets, and 15 $5 tickets.

The box contains an additional 20 "dummy" tickets with no value.

Thus, total number of tickets = 15 + 11 + 15 + 20 = 61

Three tickets are randomly drawn.

Question asked:

Find the probability that all three tickets have no value.

Solution

As we know:

First of all we will find favorable outcome for drawing 3 dummy tickets.

Favorable outcome for drawing 3 dummy tickets out of 15 $50 tickets =

Favorable outcome for drawing 3 dummy tickets out of 11 $25 tickets =

Favorable outcome for drawing 3 dummy tickets out of 15 $5 tickets =

Favorable outcome for drawing 3 dummy tickets out of 20 dummy ticket =

Thus, total Favorable outcome for drawing 3  dummy tickets =

Total outcome for drawing 3  dummy tickets out of 61 tickets =

Now,

Thus, the probability of three tickets that are randomly drawn are dummy tickets is 0.0316

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0

2500

step-by-step explanation:

2500 multiplied by 0.05 = 125