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Question Find the probability that all 3 randomly drawn tickets from a money box with 15 50tickets,1050 tickets, 10 25 tickets, and 10 $5 tickets, plus 20 dummy tickets, have no monetary value.
The probability that all 3 tickets drawn have no money value is 2055⋅1954⋅1853\frac{20}{55} \cdot \frac{19}{54} \cdot \frac{18}{53}.

Studdy Solution

STEP 1

Assumptions1. The box contains 15$5015 \$50 tickets, 10$2510 \$25 tickets, 10$510 \$5 tickets, and20 "dummy" tickets with no value. . Three tickets are randomly drawn from the box.
3. We need to find the probability that all three tickets drawn have no money value.

STEP 2

First, we need to find the total number of tickets in the box. We can do this by adding the number of each type of ticket.
Total tickets=$50 tickets+$25 tickets+$5 tickets+Dummy ticketsTotal\, tickets = \$50\, tickets + \$25\, tickets + \$5\, tickets + Dummy\, tickets

STEP 3

Now, plug in the given values for each type of ticket to calculate the total number of tickets.
Total tickets=15+10+10+20Total\, tickets =15 +10 +10 +20

STEP 4

Calculate the total number of tickets in the box.
Total tickets=15+10+10+20=55Total\, tickets =15 +10 +10 +20 =55

STEP 5

Now, we need to find the probability of drawing a "dummy" ticket. The probability of an event is the number of ways that event can occur divided by the total number of outcomes. In this case, the event is drawing a "dummy" ticket and the total number of outcomes is the total number of tickets.
(Dummy)=Number of dummy ticketsTotal number of tickets(Dummy) = \frac{Number\, of\, dummy\, tickets}{Total\, number\, of\, tickets}

STEP 6

Plug in the values for the number of dummy tickets and the total number of tickets to calculate the probability.
(Dummy)=2055(Dummy) = \frac{20}{55}

STEP 7

Now, we need to find the probability of drawing three "dummy" tickets in a row. Since the tickets are drawn without replacement, the probability changes each time a ticket is drawn. The probability of drawing three "dummy" tickets in a row is the product of the probability of drawing a "dummy" ticket on the first draw, the probability of drawing a "dummy" ticket on the second draw (given that a "dummy" ticket was drawn on the first draw), and the probability of drawing a "dummy" ticket on the third draw (given that "dummy" tickets were drawn on the first two draws).
(3 Dummies)=(Dummy on first draw)times(Dummy on second draw ∣ Dummy on first draw)times(Dummy on third draw ∣ Dummies on first and second draws)(3\, Dummies) =(Dummy\, on\, first\, draw) \\times(Dummy\, on\, second\, draw\, |\, Dummy\, on\, first\, draw) \\times(Dummy\, on\, third\, draw\, |\, Dummies\, on\, first\, and\, second\, draws)

STEP 8

Plug in the values to calculate the probability.
(3 Dummies)=2055times1954times1853(3\, Dummies) = \frac{20}{55} \\times \frac{19}{54} \\times \frac{18}{53}

STEP 9

Calculate the probability that all three tickets drawn have no money value.
(3 Dummies)=2055times1954times1853=.0244(3\, Dummies) = \frac{20}{55} \\times \frac{19}{54} \\times \frac{18}{53} =.0244The probability that all three tickets drawn have no money value is .0244.0244.

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