The probability that player B wins the game using binomial probability concept is 0.0329
For a throw of 2 - Fair dice :
Total required outcome = Sample space = 6² = 36
X = sum of the 2 rolled dice
For player A :
To win ; X>5
Required outcome = 26 (Number of sum greater than 5).
For player B :
To win ; X≤5
Required outcome = 10 (Number of sum less than or equal to 5).
The probability of player B winning any game :
- Recall : Probability = required outcome / Total possible outcomes
- P(B winning a round ) = 10 / 36
For player B to win the game, he must win atleast 6 rounds :
Using binomial probability :
.
where :
p = probability of winning
q = probability of not winning = 1 - p
x = number of winning or success
n = number of trials
Using a binomial probability calculator :
- P(x ≥ 6) = p(x=6) + p(x=7) + p(x=8) + p(x=9) + p(x=10)
- P(x ≥ 6) = 0.0263 + 0.0057 + 0.00083 + 0.00000711 + 0.000000273
- P(x ≥ 6) = 0.0329
Therefore, the probability of player B winning is 0.0329
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