A carton contains 20 bulbs, 5 of which are defective. The probability that, if a sample of 3 bulbs is chosen at random (with replacement) from the carton that exactly 2 are defective is.
Hint: Think the question as there are 3 different bags of bulbs, and each contains 5 defective bulbs and 15 working bulbs. We have to pick 1 bulb from each of the three bags so that out of the three bulbs we pick 2 are defective, try to find the probability.
Complete step by step answer: Before moving to the question, let us talk about probability. Probability, in simple words, is the possibility of an event to occur. Probability can be mathematically defined as . Now, let’s move to the solution to the above question. The question can be interpreted differently as there are three bags A, B, and C, and each bag contains 20 bulbs, out of which 5 are defective. Now we have to pick one bulb from each such that out of the three bulbs, 2 are defective. First, let us focus on bag A. Therefore, the probability of getting out a defective bulb from the bag is:
Also, the probability of getting out a working bulb from bag A is:
We know that all three bags A, B and, C are identical bags with the same number of total bulbs, and the same proportion of bulbs are defective in each bag. So, we can say that the probability of picking a defective bulb from each bag is , and the probability of picking a working bulb from each bag is . The possible cases for which we can get 2 defective bulbs can be: a working bulb that is picked from A, and defective bulbs are picked from the rest of the two bags. And similarly, other two bags can be selected to pick defective bulbs forming two more similar cases. So, the total number of possible cases is three. Therefore, the probability of picking 2 defective bulbs out three is:
Hence, we can conclude that the probability of selecting 3 bulbs out of which 2 are defective is .
Note: It is preferred that while solving a question related to probability, always cross-check the possibilities, as there is a high chance you might miss some or have included some extra or repeated outcomes. Also, when a large number of outcomes are to be analysed then permutations and combinations play a very important role as we see in the above solution.