Probability of intersection

Three students have a probability of 0.7,0.5 and 0.4 to graduated from university respectively. What is the probability that at least one of them will be graduated?

Result

p =  0.91

Solution:

p1=0.7 p2=0.5 p3=0.4  p=p1+p2+p3p1 p2p2 p3p3 p1+p1 p2 p3=0.7+0.5+0.40.7 0.50.5 0.40.4 0.7+0.7 0.5 0.4=91100=0.91p_{1}=0.7 \ \\ p_{2}=0.5 \ \\ p_{3}=0.4 \ \\ \ \\ p=p_{1}+p_{2}+p_{3} - p_{1} \cdot \ p_{2}-p_{2} \cdot \ p_{3}-p_{3} \cdot \ p_{1} + p_{1} \cdot \ p_{2} \cdot \ p_{3}=0.7+0.5+0.4 - 0.7 \cdot \ 0.5-0.5 \cdot \ 0.4-0.4 \cdot \ 0.7 + 0.7 \cdot \ 0.5 \cdot \ 0.4=\dfrac{ 91 }{ 100 }=0.91



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