Step by step solution to "When an experiment is performed, one and only..."

High School
Calculus and Analysis
Precalculus
Probability and combinatorics

hexacordoK

Answered question

2021-09-03

When an experiment is performed, one and only one of the events

A_{1}, A_{2}

and

A_{3}

will occur. Find

P (A_{1}), P (A_{2})

and

P (A_{3})

under each of the following assumptions: a)

P (A_{1}) = P (A_{2}) = P (A_{3})

P (A_{1}) = P (A_{2}) and P (A_{3}) = \frac{1}{2}

P (A_{1}) = 2 P (A_{2}) = 3 P (A_{3})

Answer & Explanation

Sadie Eaton

Skilled2021-09-04Added 104 answers

Here these events are mutually exclusive events..

a) $P (A_{1}) = P (A_{2}) = P (A_{3})$
so, $P (A_{1}) = P (A_{2}) = P (A_{3}) = \frac{1}{3}$
b) $P (A_{1}) + P (A_{2}) = 1 - P (A_{3}) = 1 - \frac{1}{2} = \frac{1}{2}$
so, $P (A_{1}) = P (A_{2}) = \frac{1}{4}$
c) $P (A_{1}) = 2 P (A_{2}) = 3 P (A_{3})$
we can write, $P (A_{2}) = \frac{P (A_{1})}{2}$
$P (A_{3}) = \frac{P (A_{1})}{3}$
now, $P (A_{1}) + \frac{P (A_{1})}{2} + \frac{P (A_{1})}{3} = 1$
$\Rightarrow \frac{11}{6} \cdot P (A_{1}) = 1$
$\Rightarrow P (A_{1}) = \frac{6}{11}$
using value of $P (A_{1})$ we find... $P (A_{2}) = \frac{3}{11}$
$P (A_{3}) = \frac{2}{11}$

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