# X is taking

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.

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