S={e1,e2,e3,e4}S=\{e_1,e_2,e_3,e_4\}S={e1,e2,e3,e4}
Given,
P(e2)=1/5P(e3)=1/3P(e_2)=1/5\\P(e_3)=1/3P(e2)=1/5P(e3)=1/3
and we know that
P(e1)+P(e2)+P(e3)+P(e4)=1 ⟹ P(e1)+15+13+P(e4)=1 ⟹ P(e1)+P(e4)=1−15−13=715P(e_1)+P(e_2)+P(e_3)+P(e_4)=1\\\ \\\implies P(e_1)+\dfrac{1}{5}+\dfrac{1}{3}+P(e_4)=1\\\ \\\implies P(e_1)+P(e_4)=1-\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{7}{15}P(e1)+P(e2)+P(e3)+P(e4)=1 ⟹P(e1)+51+31+P(e4)=1 ⟹P(e1)+P(e4)=1−51−31=157
Let P(e4)=115P(e_4)=\dfrac{1}{15}P(e4)=151
So, P(e1)=715−115=615P(e_1)=\dfrac{7}{15}-\dfrac{1}{15}=\dfrac{6}{15}P(e1)=157−151=156
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