Question 596214: A die is rolled 15 times. What is the probability of at least two 5's appearing?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Use the binomial probability distribution to compute the probability of getting zero fives (ie P(X = 0)) and the probability of getting one five (ie P(X = 1)) .
P(X = k) = (n ncr k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (15 ncr 0)*(0.1667)^(0)*(1-0.1667)^(15-0)
P(X = 0) = (15 ncr 0)*(0.1667)^(0)*(0.8333)^(15-0)
P(X = 0) = (1)*(0.1667)^(0)*(0.8333)^15
P(X = 0) = (1)*(1)*(0.064866)
P(X = 0) = 0.064866
P(X = k) = (n ncr k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (15 ncr 1)*(0.1667)^(1)*(1-0.1667)^(15-1)
P(X = 1) = (15 ncr 1)*(0.1667)^(1)*(0.8333)^(15-1)
P(X = 1) = (15)*(0.1667)^(1)*(0.8333)^14
P(X = 1) = (15)*(0.1667)*(0.077843)
P(X = 1) = 0.194646
In short,
P(X = 0)= 0.064866
P(X = 1)= 0.194646
Now add the two probabilities
P(X = 0)+P(X = 1) = 0.064866 + 0.194646
P(X = 0)+P(X = 1) = 0.259512
So P(X <= 1) = 0.259512
Since the binomial distribution is a discrete distribution, this means that
P(X >= 2) = 1 - P(X <= 1)
So this means...
P(X >= 2) = 1 - P(X <= 1)
P(X >= 2) = 1 - 0.259512
P(X >= 2) = 0.740488
So the probability of at least two 5's appearing is approximately 0.740488
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