Question

A chairman is biased so that he selects his relatives for a job 3 times as likely as others. If
there are 3 posts for a job. Find the probability distribution for selection of persons other than
their relatives.
If the chairman is biased than which value of life will be demolished?

Answer 1

challenging question

Answer 2

.25 i guess

Answer 3

do we need a relative quantification of the number of relatives in the probability pool?

Answer 4

For 0 relatives, it is 100%
For 3 relatives, it is 0%

Answer 5

Let the probability of selecting an other person = x
Then probability of selecting a relative = 3x
\[P(0\ others)=\left(\begin{matrix}3 \\ 0\end{matrix}\right)x ^{0}(1-x)^{3}=(1-x)^{3}\]
\[P(3\ relatives)=\left(\begin{matrix}3 \\ 3\end{matrix}\right)(3x)^{3}(1-3x)^{0}=(3x)^{3}\]
\[P(0\ others)+P(3\ relatives)=1\]
\[(1-x)^{3}+(3x)^{3}=1\ .......................(1)\]
The non-zero, positive solution of equation (1) is x=0.2868
Using the value of 0.2868 for the probability of selecting a person other than a relative the binomial distribution can be used to find P(0 others), P(1 other), P(2 others) and P(3 others).

Answer 6

also how ur 0.2868 comes, plse explain

Answer 7

If you are doing Bernoulli trial, I would suggest 1:3 converted to \(p=1/4,q=3/4\)

Answer 8

\[
\Pr(X=0) = \binom{3}{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^3 = \frac{27}{64}\\
\Pr(X=1) = \binom{3}{1}\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^2 = \frac{27}{64}\\
\Pr(X=2) = \binom{3}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^1 = \frac{9}{64}\\
\Pr(X=3) = \binom{3}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^0 = \frac{1}{64}
\]

Answer 9

But if you use 0.25 for the probability of choosing a relative and 0.75 for the probability of choosing other than a relative the sum of Pr(X = 0) + Po(X = 3) does not equal 1.
Also how did you find the value p = 1/4 ?

Answer 10

To find \(p\) I used a system of equations. \[
p = 3q \\
p = 1-q
\]

Answer 11

The sum \(\Pr(X=0) + \Pr(X=3)\) should not equal \(1\), because there is nonzero probability that \(X=1,X=2\)

Answer 12

My bad. The sum is
Po(X = 0) + Pr(X = 3)
where Po is the probability of other than a relative being selected and Pr is the probability of a relative being selected. If Po = 1/4 and Pr = 3/4 this sum does not equal 1. Therefore your values are not a valid solution.

Answer 13

When you say 'other than a relative being select' and 'a relative being selected'...
You're not talking about mutually exclusive events, so due to double counting, they should be greater than 1 when added.

Answer 14

In my model, 'other than a relative being select' and 'a relative being selected' would be represented by \(\Pr(X\gt 0)\) and \(\Pr(X\lt 3)\) respectively.

Answer 15

I agree that my reasoning is flawed :(

Answer 16

@wio
the concept of 4 is not understood to me
plse explain

Answer 17

Is this advanced probability? College level?

Answer 18

Let the probability that a person other than a relative is selected = x
Then the probability that a relative is selected = 3x
The sum of the above probabilities must equal one.
Therefore x + 3x = 1
4x = 1
x = 1/4 = 0.25
Then use the formula for the binomial distribution with n = 3 and p = 0.25 to find
P(X = 0), P(X = 1), P(X = 2), P(X = 3).

Answer 19

@msingh Can you be more specific about what you don't understand?
Do you not understand the Binomial Distribution? Do you not understand how I found \(p\) and \(q\)?

Answer 20

@wio
i know binomial theorem and also understood the concept of 4
thank u once again