My question is to get a better understanding of this. Basically, these two problems, I did it WITHOUT replacement. So I would multiply each defects and so forth. My question is, would I have to subtract the total from 1 in the second question? I am not exactly clear on when to subtract from 1 and not to...
#1) A sample of 4 different calculators is randomly selected from a group containing 20 that are defective and 31 that have no defects. What is the probability that at least one of the calculators is defective?
(A) 0.863 (B) 0.874 (C) 0.200 (D) 0.126
My work:
(31/51)(30/50)(29/49)(28/48) = .1259
1 - .1259 = .8741
#2)A sample of 4 different calculators is randomly selected from a group containing 49 that are defective and 26 that have no defects. What is the probability that all four of the calculators selected are defective?
(A) 0.1822 (B) 0.0793 (C) 14.1724 (D) 0.1743
My work:
(49/75)(48/74)(47/73)(46/72) = .1743
#1. The probability that NONE are defective is 31/51 *30/50 * 29/49 * 28/48 = 0.1259
Probability of at least 1 = 1 - .1259 = .8741 . Correct
#2 Also correct
To answer your question about when to subtract from 1, let's first understand the concept of probability.
Probability represents the likelihood of an event occurring. It is often expressed as a value between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
In your first problem, you correctly calculated the probability of none of the calculators being defective as 0.1259. The probability of at least one calculator being defective is the complement of this event, which means the probability of it not happening.
To find the probability of at least one calculator being defective, you subtract the probability of none being defective from 1. This is because in the sample of calculators, the event "at least one defective" and the event "none defective" are mutually exclusive and exhaustive. In other words, if the probability of none being defective is 0.1259, then the probability of at least one being defective is 1 - 0.1259 = 0.8741.
In your second problem, you correctly calculated the probability of all four calculators being defective as 0.1743. In this case, there is no need to subtract from 1 because you are already calculating the probability of a specific event happening (i.e., all four calculators being defective).
So, to summarize, you subtract from 1 when you want to find the probability of the complementary event (event not happening) to the one you are considering. In other cases, you directly calculate the probability of the given event.