1
$\begingroup$

I have an exercise in probability theory, which I can't solve:

There are 3 factories A B C, which produce 3 types of light bulbs. Factory A / B / C makes 40 / 20 / 40 percent of whole light bulbs. The probability of manufacturing first type of light bulb in factory A / B / C is 0.6 / 0.3 / 0.5. The probability of manufacturing second type of light bulb in factory A/ B / C is 0.3/0.4/0.2. The probability of manufacturing third type of light bulb in factory A / B / C is 0.1/0.3/0.3.

We buy 6 light bulbs. What is the most probable set of light bulbs (number of each type)? What is its probability?

So I figured out it's multinomial distribution with parameteres 6 and probabilities $p_1=0.5$, $p_2=0,22$, $p_3=0.28$, where $p_i$ is probability of buying a light bulb of $i-th$ type.

Let $X_i$ denote a number of light bulbs of $i-th$ type in the set of 6 light bulbs we bought. To solve first problem we want to maximaze $P(X_1=k_1, X_2=k_2, X_3=k_3) = \frac{6!}{k_1!\cdot k_2! \cdot k_3!}0.5^{k_1} \cdot 0.22^{k_2} \cdot 0.28^{k_3}$

So we can write a function $f(x,y,z)=\frac{6!}{x!\cdot y! \cdot z!} 0.5^{x}\cdot 0.22^{y}\cdot 0.28^{z}$. Since sum of all variables must be 6, we can reduce one variable. Then we want to maximaze function $g(x,y)=\frac{6!}{x!\cdot y!\cdot (6-x-y)!}0.5^{x}\cdot 0.22^{y}\cdot 0.28^{6-x-y}$. I don't how to do it because of the factorials. And also $k_i$ are natural numbers so i doubt that this aproach is good. I think you can examine this formula with sequences, but there is a lot of work to do then. I think there should be a theorem, which I don't know, that would help, because it was for an exam on some not mathematical (economy sth) studies and they have a small amount of math classes there.

$\endgroup$
2
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Aug 4 '18 at 10:49
  • $\begingroup$ You may take a look at the mode of Binomial first: en.wikipedia.org/wiki/Binomial_distribution#Mode The mode of a multinomial distribution should be also similar, close to the mean - so you may first search around $(6\times 0.5, 6 \times 0.22, 6 \times 0.28) = (3, 1.32, 1.68)$ - the first guess could be $(3, 1, 2)$. $\endgroup$ – BGM Aug 4 '18 at 15:14
0
$\begingroup$

Max $P = \dfrac{6!}{x_1!.x_2!.x_3!.x_4!.x_5!.x_6!.x_7!.x_8!.x_9!}.24^{x_1}.12^{x_2}.04^{x_3}.06^{x_4}.08^{x_5}.06^{x_6}.2^{x_7}.08^{x_8}.12^{x_9}$ with

$x_1$ - Type 1 made in Factory A - probability $= .6*.4 = 0.24$

$x_2$ - Type 2 made in Factory A -probability $= .3*.4 = 0.12$

$x_3$ - Type 3 made in Factory A probability $= .1*.4 = 0.04$

$x_4$ - Type 1 made in Factory B probability $= .3*.2 = 0.06$

$x_5$ - Type 2 made in Factory B probability $= .4*.2 = 0.08$

$x_6$ - Type 3 made in Factory B probability $= .3*.2 = 0.06$

$x_7$ - Type 1 made in Factory C probability $= .5*.4 = 0.20$

$x_8$ - Type 2 made in Factory C probability $= .2*.4 = 0.08$

$x_9$ - Type 3 made in Factory C probability $= .3*.4 = 0.12$

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9=6$ and $x_i \ge 0$

Ran a simulation and found out that the number and itemtypes are

$x_1 = 2, x_2 = 1, x_7 = 2$ and $x_9 = 1$

Most probable with maximum probability $= 0.005971$

$\endgroup$
3
  • $\begingroup$ Thank you! But this should be solved without a simulation. It was an exercise on an exam. $\endgroup$ – Derm Aug 4 '18 at 14:00
  • $\begingroup$ Do you know the answer for this? Let me know how you got your probabilities? Thanks $\endgroup$ – Satish Ramanathan Aug 4 '18 at 15:35
  • $\begingroup$ I don't have the answer. Probability of buying a light bulb of first type is 0.4*0.6 + 0.2*0.3 + 0.4*0.5 = 0.5 (x_1 + x_4 + x_7) $\endgroup$ – Derm Aug 5 '18 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.