Given: P(e1) = 0.1, P(e2) = 0.5, P(e3) = 0.1We know that,Sum of all probabilities = 1∴ P(e1) + P(e2) + P(e3) + P(e4) = 1⇒ 0.1 + 0.5 + 0.1 + P(e4) = 1⇒ P(e4) = 1 – 0.7⇒ P(e4) = 0.3Ans. If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) =.1, P(e2) = .5, P (e3) = .1, then the probability of e4 is 0.3

Q. 38

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Class 11th Mathematics - Exemplar 16. Probability

Answer :

Given: P(e₁) = 0.1, P(e₂) = 0.5, P(e₃) = 0.1

We know that,

Sum of all probabilities = 1

∴ P(e₁) + P(e₂) + P(e₃) + P(e₄) = 1

⇒ 0.1 + 0.5 + 0.1 + P(e₄) = 1

⇒ P(e₄) = 1 – 0.7

⇒ P(e₄) = 0.3

Ans. If e₁, e₂, e₃, e₄ are the four elementary outcomes in a sample space and P(e₁) =.1, P(e₂) = .5, P (e₃) = .1, then the probability of e₄ is 0.3

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