Q. 245.0( 4 Votes )

# X is taking up subjects – Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets

i. Grade A in all subjects

ii. Grade A in no subjects

iii. Grade A in two subjects

Answer :

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.

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