Three dice are rolled together. What is the probability as getting at least one '4'
total number of ways:
n=63=216n=6^3=216n=63=216
number of ways to get one '4':
n1=3⋅62=108n_1=3\cdot6^2=108n1=3⋅62=108
number of ways to get two '4':
n2=3⋅6=18n_2=3\cdot6=18n2=3⋅6=18
number of ways to get three '4':
n3=1n_3=1n3=1
the probability of getting at least one '4'
p=n1+n2+n2n=108+18+1216=127216p=\frac{n_1+n_2+n_2}{n}=\frac{108+18+1}{216}=\frac{127}{216}p=nn1+n2+n2=216108+18+1=216127
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