If E₁, E₂, E₃ and E₄ are elementary events associated to a random experiment such that $P (E_{1}) = \frac{1}{10}, P (E_{2}) = \frac{1}{2}, P (E_{3}) = \frac{1}{10},$ then P(E₄) =____________.

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Solution

Since E₁, E₂, E₃ and E₄ are elementary events then P(E₁) + P(E₂) + P(E₃) + P(E₄) = 1
$i . e \frac{1}{10} + \frac{1}{2} + \frac{1}{10} + P (E_{4}) = 1$
$\begin{array}{rcl} i . e P (E_{4}) & = & 1 - \frac{1}{10} - \frac{1}{2} - \frac{1}{10} \\ = & \frac{10 - 1 - 5 - 1}{10} \\ ∴ P (E_{4}) & = & \frac{3}{10} \end{array}$

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Similar questions

If E1, E2, E3, ....., En are n elemantary events associated to a random experiment, then P(E1) + P(E2) + P(E3) + ........... + P(En) = ___

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

P (E)

E = E_{1} E_{2} E_{3} E_{4} E_{5}

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

P (E)

Events $E_{1}, E_{2}, E_{3}$ are the possible events of an experiment and their probabilities are recorded. Mark the possible correct answer

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