A committee consists of 9 experts taken from three institutions A, B and C, of which 2 are from A, 3 from B and 4 from C. If three experts resign from the committee, then the probability of exactly two of the resigned experts being from the same institution, is equal to

$\frac{55}{84}$

These are three mutually exclusive cases:

(i) 2 are from A and another is from either from B or C.

Corresponding probability

$=\frac{{ }^2 C_2 \cdot\left({ }^3 C_1+{ }^4 C_1\right)}{{ }^9 C_3}=\frac{7}{84}$

(ii) 2 are from B and another is either from A or C.

Corresponding probability

$=\frac{{ }^3 C_2 \cdot\left({ }^2 C_1+{ }^4 C_1\right)}{{ }^9 C_3}=\frac{18}{84}$

(iii) 2 are from C and another is either from A or B.

Corresponding probability

$=\frac{{ }^4 C_2 \cdot\left({ }^2 C_1+{ }^3 C_1\right)}{{ }^9 C_3}=\frac{30}{84}$

Thus, required probability $=\frac{7+18+30}{84}$

$=\frac{55}{84}$