Question #22549

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Question #22549

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Answer 1 · 2017-12-20T13:40:56

Please see the processed steps below;

Explanation:

Probability of $3$ $3$ posts selected is $\frac{3}{9} = \frac{1}{3}$ $3/9 = 1/3$

Probability of none selected is $\frac{9 - 3}{9} or (1 - \frac{1}{3})$ $(9-3)/9 or (1 - 1/3)$

$\frac{9 - 3}{9}$ $(9-3)/9$

$\frac{6}{9}$ $6/9$

$\frac{2}{3}$ $2/3$

Hence;

Probability of none selected is $= \frac{2}{3}$ $= 2/3$

Answer 2 · 2018-01-02T06:22:53

$\frac{2}{3}$ $2/3$

Explanation:

$9$ $9$ applicants , $3$ $3$ posts
so probability of anyone selected at any post is $\frac{3}{9} = \frac{1}{3}$ $3/9=1/3$
so, probability that the same person is not selected can be easily identified
$P (s) + P (n) = 1$ $P(s)+P(n)=1$ where $P (s)$ $P(s)$ means selected and $P (n)$ $P(n)$ means not selected )
$\frac{1}{3} + P (n) = 1$ $1/3+P(n)=1$
$P (n) = 1 - \frac{1}{3}$ $P(n)=1-1/3$
$P (n) = \frac{3 - 1}{3}$ $P(n)=(3-1)/3$
$P (n) = \frac{2}{3}$ $P(n)=2/3$
Cheers!
If the question is that none of them is selected , then we woupd require probability of selection of each which on an average would not be greater than $\frac{1}{3}$ $1/3$ .
My interpretion of question is that we need to find the probability of anyone not being selected .

It might require use of linear inequalities if the question isn't what i interpret .

for more please see my hand written script below
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Question #22549

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