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To calculate the probability that a number of transistors are defective, we will make an assumption/postulation that the transistors are randomly and uniformly distributed or placed in the box.
The probability that X number of transistors picked is calculated as follows:-
X/5 * 5/13
5/13 is the probability of picking a defective transistor.
1/5 is the probability of picking one defective transistor from the five defective transistors.
Hence X/5 is the probability of picking X defective transistors from the five transistors.
For example the probability that 3 of the transistors picked are defective is given by,
3/5 * 5/13 = 3/13 or 0.2307 or 23.07%
* stands for multiplication
Continuing from the explanation above;
a. if all are defective, the following are the events,
(D and D and D and D and D)
D stands for the probability of picking a defectivetransistor while ND stands for non defective
the probability of D is 5/13 for the first pick.
In probability "and" means multiplication since these are joint events.
Therefore the probability that all transistors picked are defective is given by 5/13*4/12*3/11*2/10*1/9 =120/154440 =0.0777%
b. The probability that none of the transistors picked are defective is the probability that all transistors picked are non defective.
(ND and ND and ND and ND and ND)
=(8/13*7/12*6/11*5/10*4/9) = 6720/154440 = 4.3512%
Note: I have changed the formula and method of calculation since your follow up question implies a method where the transistors are picked without replacement.
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