Question

A box contains 24 transistors 5 of witch are defective if 5 are selected at random find the probability that none are defective. please help

Answer 1

This type of question normally requires using the binomial probability, but since you're looking to find 0 defective transistors (as opposed to 1,2,3, or 4), you can represent the problem a different way.

Rather than thinking about this as drawing 5 transistors out of 24, instead treat the situation as drawing one good transistor out of 24, one out of 23, etc. Then you can multiply all the probabilities together as you want each event to happen. Here's how it looks:
\[(19/24)*(18/23)*(17/22)*(16/21)*(15/20)\]
The first probability accounts for the number of good resistors out of the initial 24, then each subsequent probability accounts for taking one good resistor out each time. The product of these is the probability of drawing 5 good resistors:
\[(19/24)*(18/23)*(17/22)*(16/21)*(15/20)=969/3542\]

Answer 2

ok but how did you come up with the 19,18,17,16,15 not understanding that part.

Answer 3

There are 24 total transistors. Since 5 of them are defective, the rest are good. So, 24-5=19, which we use in the first probability.

After that, the other numbers come from us removing one transistor each time. So, the second pull has 18 good transistors left out of the new total 23, and the other three trials continue this way.

Answer 4

ok thank you