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 are the odds of one play winning the jackpot?

1/30 = 0.03 or 3%

3/(30+30+30) = 3/90 = 0.33 or 33%

3/(30×30×30) = 3/27000 = 0.0001 or 0.01%

1/(30×30×30) = 1/27000 = 0.00003 or 0.003%

The odds of one play winning the jackpot such that all 3 dials are in the jackpot position is 1/27000 =  0.00003 or 0.003%

Number of dias = 3

Number of positions per dial = 30

Of the 30 positions, 1 is jackpot

To win, all 3 dials must be in the jackpot position:

Since, the probability of Event is independent (The probability of occurence of one event has no effect on the probability of another) ;

The probability of the 3 dials being in the jackpot position can be defined as :

P(jackpot 1 ) × P(Jackpot 2) × P(Jackpot 3)

Recall:

Probability = required outcome / Total possible outcomes

Required outcome = jackpot position = 1

Total possible outcomes = total number of positions per dial = 30

Hence,

P(jackpot 1 ) × P(Jackpot 2) × P(Jackpot 3) = (1/30 × 1/30 × 1/30) = 1 / (30 × 30 × 30) = 1 / 27000 = 0.00003 or 0.003%

Learn more : link

(1/30)^3, or 3.7·10^(-5)

Step-by-step explanation:

The probability of each dial reaching the jackpot position is 1/30, as each dial has 30 positions.  The behaviors of the three dials are independent, and so the probability of reaching the jackpot on each is (1/30)^3; we merely multiply the three probabilities together.  Winning the jackpot consists of all three dials being at 'jackpot' position.

the answer for this question is 38

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.