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?

Correct result:

p =  0.91

Solution:

$$\begin{gathered} p_1=0.7 \\ {p}_2=0.5 \\ {p}_3=0.4 \\ p=p_1+p_2+p_3-p_1\cdot p_2-p_2\cdot p_3-p_3\cdot p_1+p_1\cdot p_2\cdot p_3=0.7+0.5+0.4-0.7\cdot 0.5-0.5\cdot 0.4-0.4\cdot 0.7+0.7\cdot 0.5\cdot 0.4={\displaystyle \frac{91}{100}}=0.91 \end{gathered}$$

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