Question

question 25
a slot machine has 3 dials. each dial has 30 positions, one of which is jackpot. to win the jackpot, all three dials must be in the jackpot position. assuming each play spins the dials and stops each independently and randomly, what is the probability of one play winning the jackpot?
a 1/(30×30×30)=1/27000 = 0.00003 or 0.003%
b 3/(30×30×30)=3/27000 = 0.0001 or 0.01%
c 3/(30+30+30)=3/90 = 0.33 or 33%
d 1/30=0.03 or 3%
5 points

Explanation:

Step1: Calculate total number of possible outcomes

Since each of the 3 dials has 30 positions, by the multiplication - principle, the total number of possible combinations is $30\times30\times30 = 27000$.

Step2: Calculate probability of winning

There is only 1 winning combination (all dials in jackpot position). The probability $P$ of an event is the number of favorable outcomes divided by the number of total outcomes. So the probability of winning is $\frac{1}{30\times30\times30}=\frac{1}{27000}\approx0.000037$ or $0.0037\%$.

Answer:

A. $1/(30\times30\times30)=1/27000 = 0.00003$ or $0.003\%$