Question

Consider the case of three boxes containing resistors of different values according to the table shown below. Now, consider an experiment where you first randomly choose a box then you randomly select a resistor from that box.

Resistor Values in Ohms	Box 1	Box 2	Box 3	Totals
1Ω	20	95	25	140
10 Ω	55	35	75	165
100 Ω	70	80	145	295
Totals	145	210	245	600

a. Calculate the probability of selecting 1 Ohm resistor given that box 2 is chosen.

b. If a 1 Ohm resistor is selected, calculate the probability it came from the third boc

Verify if PX(k) = {2k/n(n+1), 1<= k <= n

{0, otherwise

can be a valid PMF for the random ?, where ? =>1 is an integer.

Answer 1

a.

P(1 ohm | box 2) = (no. of 1 ohm resistor in box 2)/(total resisitor in box 2)

= 95 / 210

= 0.4524

b.

P(box 3 | 1 ohm) = (no. of 1 ohm resistor in box 3)/(total 1 ohm resisitor)

= 25 / 140

= 0.1786

Verify if PX(k) = {2k/n(n+1), 1<= k <= n

   {0, otherwise

can be a valid PMF for the random ?, where ? =>1 is an integer.

for it to be valid PMF PX(k) should be 1 for all k>n {since the distribution lies in 1 to n}

but for k > n it is given 0 which shows it cannot be a valid PMF

(please UPVOTE)