Question

Samsung, Panasonic, and LG are producing Single Board Computers (SBCs) for hobbyists. Samsung’s SBCs take up 40% of the market, Panasonic’s SBCs take up 25% of the market, and LG’s SBCs take up the rest. 1% of all Samsung and Panasonic’s SBCs are defective, whereas 2% of all LG SBCs are defective. If the SBC you bought was defective, what is the probability that it is a Panasonic SBC?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Samsung, Panasonic, and LG are producing Single Board Computers (SBCs) for hobbyists. Samsung’s SBCs take up 40% of the market, Panasonic’s SBCs take up 25% of the market and LG take rest.

Let assume that

$\rm \: E_1 : \: manufactured \: by \: Samsung$

$\rm \: E_2 : \: manufactured \: by \: Panasonic$

$\rm \: E_3 : \: manufactured \: by \: LG$

$\rm \: A : getting \: defective \:Single Board \: Computers\:(SBCs)$

So,

$\rm \: P(E_1) = \dfrac{40}{100}$

$\rm \: P(E_2) = \dfrac{25}{100}$

$\rm \: P(E_3) = \dfrac{35}{100}$

Now, further given that 1% of all Samsung and Panasonic’s SBCs are defective, whereas 2% of all LG SBCs are defective.

$\rm \: P(A|E_1) = \dfrac{1}{100}$

$\rm \: P(A|E_2) = \dfrac{1}{100}$

$\rm \: P(A|E_3) = \dfrac{2}{100}$

Now, Required Probability the SBC you bought was defective, that it is a Panasonic SBC is evaluated using Baye's Theorem as

$\rm \: P(E_2|A) = \dfrac{P(E_2)P(A|E_2)}{\displaystyle\sum_{ i = 1}^3\rm P(E_i)P(A|E_i)}$

So, on substituting the values, we get

$\rm \: = \: \dfrac{\dfrac{25}{100} \times \dfrac{1}{100} }{\dfrac{40}{100} \times \dfrac{1}{100} + \dfrac{25}{100} \times \dfrac{1}{100} + \dfrac{35}{100} \times \dfrac{2}{100} }$

$\rm \: = \: \dfrac{25}{40 + 25 + 70}$

$\rm \: = \: \dfrac{25}{135}$

$\rm \: = \: \dfrac{5}{27}$

Hence,

$\bf\implies \:\boxed{ \rm{ \:\rm \: P(E_2|A) = \dfrac{5}{27} \: \: }}$