 We're going to do an example and make our own probability distribution or table from scratch a Supplier of aircraft parts has 12 altimeters that are correctly calibrated and two that are not Three altimeters are randomly selected without replacement without replacement Let X represent the number that aren't correctly calibrated create a probability distribution table So just from reading this question I was able to pick up on a few things and the first thing is that if I have 12 Altimeters that are correctly calibrated and to that or not means I have 14 Altimeters total so that's how many we had to pick from in our first trial 14 Next I am selecting three without replacement Which means that as I go to collect my three and select them, I will have one less every single time So these trials are dependent on each other Well first and foremost if you are considering Altimeters that aren't correctly calibrated and you have two that are not It is impossible for you to get three altimeters that aren't correctly calibrated So the probability that three altimeters are not correctly calibrated would be zero. It's not possible There's only two Altimeters that are not correctly calibrated Next the probability you get zero altimeters that are not correctly calibrated so For probability That X equals zero so you get zero altimeters Not correctly calibrated would be the probability that you would Get an altimeter that's calculated in the first trial probability you get an altimeter in the second Trial that's correctly calibrated and also the probability you get a correctly calibrated altimeter in the third trial So in the first trial the out of 14 altimeters How many would be correctly calibrated what are the favorable outcomes there would be 12 Second trial since you've already picked one that was correctly calibrated in the first trial You now have 13 altimeters to pick from and there are 11 that are correctly calibrated Third trial you now only have 12 altimeters to correctly that are correctly calibrated and you actually only have 10 That are correctly calibrated. That's 12 altimeters total 10 of them are correctly calibrated. So multiply 12 times 11 times 10 14 times 13 times 12 and I believe you'll get 1320 over 2184 You can write this as a decimal to For decimal places you get point six zero four four So the probability you get zero Altimeters not correctly calibrated is point six zero four four Alright, let's do the probability you get one One altimeter correctly calibrated not correctly calibrated So that means You want one altimeter that's not calibrated not correctly calibrated In the other two trials you do want the altimeters correctly calibrated So we have 14 altimeters total how many are not calibrated that would be two All right now, let's do the probability That in our second trial that we get a altimeter that is correctly calibrated out of 13 altimeters there would be 12 All 12 are still correctly calibrated because in my first trial I pulled out one that was not correctly calibrated Third trial out of my 12 total left 11 are correctly calibrated So two times 12 times 11 14 times 13 times 12. Excuse me 264 out of 2184 This is not the answer there Because how many ways are there for you to get one altimeter not correctly calibrated in three trials? Well The first one drawn Could be not calibrated The rest could be calibrated the second one drawn could be not correctly calibrated or Even the third one drawn could be not calibrated So there's three different ways total so take this probability 264 out of 2184 and multiply it by three So multiply it by three When you do that You will get point three six two six. I'm rounding everything to four decimal places We're applicable So we got point three six two six Next let's find the probability we get two altimeters that are not correctly calibrated or just not calibrated at all So probability that the altimeter on the first trial is not calibrated Probability that the altimeter we pick on the second trial is not calibrated And on the third trial probability that the altimeter we pick is calibrated So first trial out of my 14 altimeters how many are not calibrated that would be two Second trial out of the 13 remaining altimeters how many are not calibrated That would be one because I picked one out in the first trial And then third trial out of 12 altimeters how many are calibrated? All 12 are still in the mix So this is going to give me two times one times 12 which is 24 And then 14 times 13 times 12 That's just 2184 This is just one of a few ways that you can get two altimeters that are not calibrated So You could get two altimeters right off the bat that are not calibrated and the third one's calibrated Your not calibrated altimeters might be the first center of the third pool Or They might even be the last pool the last two pooled So there's three different ways you could obtain two altimeters that are not calibrated So take your answer three times 24 over 2184 Multiply together The answer that you get is actually 0.033 And if it makes you feel better you can add the zero at the end so everything's the four decimal places So we just created a probability distribution Let's now find the expected value or the mean of a probability distribution you could use technology But rather than having to pull up the technology that we could use Or pulling out my calculator. I can actually find the expected value or the mean Just by looking at my table Because of the formula for expected value or mean of a probability distribution I literally just multiply each Possible value by its probability so multiply across the rows of your probability distribution I got zero One times point three six two six is point three six two six On two times point zero three three Is going to be Point zero six six Three times zero zero So what i'm going to do here is I'm going to add this last column All of my products from multiplying the rows together. I'm going to add these results. So my expected value represented by the notation given here Or my mean they mean the same thing Be zero plus point three six two six plus point zero six six plus zero Which gives me an expected value of Point four two eight six Make sure you round to the appropriate number of decimal places Based on what the question states So once again the expected value or mean as point four two Eight six