Given:
⇒ P(MA) = P(getting A in mathematics)
⇒ P(MA) = 0.2
⇒ P(MN) = P(not getting A in mathematics)
⇒ P(MN) = 1 - 0.2
⇒ P(MN) = 0.8
⇒ P(PA) = P(getting A in physics)
⇒ P(PA) = 0.3
⇒ P(PN) = P(not getting A in physics)
⇒ P(PN) = 1 - 0.7
⇒ P(PN) = 0.3
⇒ P(CA) = P(getting A in Chemistry)
⇒ P(CA) = 0.5
⇒ P(CN) = P(not getting A in chemistry)
⇒ P(CN) = 1 - 0.5
⇒ P(CN) = 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(Xall) = P(getting A in all subjects)
Since getting A in different subjects is an independent event, their probabilities multiply each other
⇒ P(Xall) = (P(MA)P(PA)P(CA)
⇒ P(Xall) = 0.2 × 0.3 × 0.5
⇒ P(Xall) = 0.03
⇒ P(Xnone ) = P(getting A in no subjects)
Since getting A in different subjects is an independent event, their probabilities multiply each other
⇒ P(Xnone ) = (P(MN)P(PN)P(CN))
⇒ P(Xnone ) = 0.8 × 0.7 × 0.5
⇒ P(Xnone ) = 0.28
⇒ P(Xtwo ) = P(getting A in any two subjects)
Since getting A in different subjects is an independent event, their probabilities multiply each other
⇒ P(Xtwo ) = (P(MA)P(PA)P(CN)) + (P(MA)P(PN)P(CA)) + (P(MN)P(PA)P(CA))
⇒ P(Xtwo ) = (0.2 × 0.3 × 0.5) + (0.2 × 0.7 × 0.5) + (0.8 × 0.3 × 0.5)
⇒ P(Xtwo ) = 0.03 + 0.07 + 0.12
⇒ P(Xtwo ) = 0.22
∴ The required probabilities are 0.03, 0.28, 0.22.