Given:

⇒ P(M_A) = P(getting A in mathematics)

⇒ P(M_A) = 0.2

⇒ P(M_N) = P(not getting A in mathematics)

⇒ P(M_N) = 1-0.2

⇒ P(M_N) = 0.8

⇒ P(P_A) = P(getting A in physics)

⇒ P(P_A) = 0.3

⇒ P(P_N) = P(not getting A in physics)

⇒ P(P_N) = 1-0.7

⇒ P(P_N) = 0.3

⇒ P(C_A) = P(getting A in Chemistry)

⇒ P(C_A) = 0.5

⇒ P(C_N) = P(not getting A in chemistry)

⇒ P(C_N) = 1-0.5

⇒ P(C_N) = 0.5

We need to find the probability that:

i. X gets A in all subjects

ii. X gets A in no subjects

iii. X gets A in two subjects

⇒ P(X_all) = P(getting A in all subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_all) = (P(M_A)P(P_A)P(C_A))

⇒ P(X_all) = 0.2×0.3×0.5

⇒ P(X_all) = 0.03

⇒ P(X_none) = P(getting A in no subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_none) = (P(M_N)P(P_N)P(C_N))

⇒ P(X_none) = 0.8×0.7×0.5

⇒ P(X_none) = 0.28

⇒ P(X_two) = P(getting A in any two subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_two) = (P(M_A)P(P_A)P(C_N)) + (P(M_A)P(P_N)P(C_A)) + (P(M_N)P(P_A)P(C_A))

⇒ P(X_two) = (0.2×0.3×0.5) + (0.2×0.7×0.5) + (0.8×0.3×0.5)

⇒ P(X_two) = 0.03 + 0.07 + 0.12

⇒ P(X_two) = 0.22

∴ The required probabilities are 0.03, 0.28, 0.22.