4 months ago
Q:

A slot machine has 3 dials. Each dial has 40 positions, two of which are "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? 1/20 * 1/20 * 1/20 = 1/8000 =0.000125=0.0125% B 1/40 * 1/40 * 1/40 = 1/64000 =0.0000156=0.00156% 1/3 * 1/3 * 1/3 = 1/27 =0.037=3.7% frac 3 * frac 3 * frac 3=frac 9=0.00014=0.014%

A slot machine has 3 dials. Each dial has 40 positions, two of which are "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? 1/20 * 1/20 * 1/20 = 1/8000 =0.000125=0.0125% B 1/40 * 1/40 * 1/40 = 1/64000 =0.0000156=0.00156% 1/3 * 1/3 * 1/3 = 1/27 =0.037=3.7% frac 3 * frac 3 * frac 3=frac 9=0.00014=0.014%

Accepted Solution

A:
Answer: there is a 1.2% of winningStep-by-step explanation: there is a 1.2% chance of winning. A slot machine has 3 dials. Each dial has 40 positions, two of which are "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? 1/20 * 1/20 * 1/20 = 1/8000 =0.000125=0.0125% B 1/40 * 1/40 * 1/40 = 1/64000 =0.0000156=0.00156% 1/3 * 1/3 * 1/3 = 1/27 =0.037=3.7% frac 3 * frac 3 * frac 3=frac 9=0.00014=0.014% 65048252cb3e8.webp