Kashvi, Kalyani and Sara of class 12 are given a problem in Accounts whose respective probabilities of solving are $\frac{2}{5},\frac{1}{4}$ and $\frac{1}{6}$. They were asked to solve it independently.

The probability that the problem is solved is:

$\frac{5}{8}$

Let $P(A)=\frac{2}{5},P(B)=\frac{1}{4},P(C)=\frac{1}{6}$

so total probability of solving = $P(A∩B∩C)$

$=P(A)+P(B)+P(C)-P(A∩B)-P(B∩C)-P(A∩C)+P(A∩B∩C)$

$⇒P(A)+P(B)+P(C)-P(A)P(B)-P(B)P(C)-P(A)P(C)+P(A)P(B)P(C)$

$⇒\frac{2}{5}+\frac{1}{4}+\frac{1}{6}-\frac{2×1}{5×4}-\frac{1×1}{4×6}-\frac{2×1}{5×6}+\frac{2}{5×4×6}$

since all are independent events 

$P(A∩B∩C)=P(A)P(B)P(C)$

$\frac{48+30+20-12-5-8+2}{120}=\frac{100-25}{120}=\frac{75}{120}=\frac{5}{8}$