Gaming machine : The probability of winning

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SUMMARY

The probability of winning on a gaming machine is established at 25%. When a player engages in 20 rounds, the probability of winning exactly five times can be calculated using the binomial distribution formula: $$P(5\mid 20)=\binom{20}{5}0.25^5\cdot 0.75^{15}$$. For ten rounds, the probabilities of winning five and ten times are given by $$P(5\mid 10)=\binom{10}{5}0.25^5\cdot 0.75^{5}$$ and $$P(10\mid 10)=\binom{10}{10}0.25^{10}\cdot 0.75^{0}$$ respectively. Additionally, the probability of winning zero times in twenty tries is calculated as $$P(0\mid 20)=\binom{20}{0}0.25^0\cdot 0.75^{20}$$, assuming the machine operates fairly and independently.

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mathmari
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Hey! :o

At a gaming machine the probability of winning is 25%. A player plays 20 rounds.

  1. Which is the probability that he wins exactly five times?
  2. Which is the probability that in ten rounds he wins five times?
  3. Which is the probability that in ten rounds he wins ten times?
  4. Which is the probability that in twenty tries he wins zero times?

Do we use at each case the binomial distribution? (Wondering)

  1. $$P(5\mid 20)=\binom{20}{5}0.25^5\cdot 0.75^{15}$$
  2. $$P(5\mid 10)=\binom{10}{5}0.25^5\cdot 0.75^{5}$$
  3. $$P(10\mid 10)=\binom{10}{10}0.25^{10}\cdot 0.75^{0}$$
  4. $$P(0\mid 20)=\binom{20}{0}0.25^0\cdot 0.75^{20}$$
 
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We are making the assumption that the gaming machine is fair that is that the the goes are independent. If this is the case then your results are indeed correct. Well done
 
IanCg said:
We are making the assumption that the gaming machine is fair that is that the the goes are independent. If this is the case then your results are indeed correct. Well done

Thank you very much! (Smile)
 

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