Probability without Replacement with AND OR

Transcript

A box contains five light globes, two of which are faulty. Two globes are selected, one at a time without replacement.

Now, a! Find the probability that both globes are faulty. Now, two are faulty out of the five, so two out of five are faulty globes. Now we want, we know that there’s no replacement here. So when I’ve picked one I’m just going to keep it and I’m not going to replace it back. So there’s one less globe to choose from in the remaining pile. If both are faulty, I’ll have two out of five for the first pick. Now, for the second one, I know there’s only one faulty globe left, so it’s going to subtract two, subtract one from two to get one, and then out of the five I have one less globe to choose from, so I have four. So I’m simply subtracting one from both top and bottom and simplify to get 1 on 10, okay?

b! Find the probability that neither globe is faulty, so I want both of the globes that I pick to not be faulty. Now, three of the globes are not faulty, so five minus two is three, isn’t it? So three out of five is not faulty. So it’s gonna be three out of five times one less on top and bottom having two out of four, okay? So there’s only two non-faulty globes left after I’ve picked one, so there’s two on top and out of five I only have four left to choose from, so it’s out of four, okay? And multiplies it. You should get 3 on 10. So do you get the idea? So usually we just subtract one from both numerator and denominator. That’s the idea.

c! Find the probability that the first globe is faulty and the second one is not. So I want the first globe to be faulty and the second one to be not faulty. So I know that there’s two out of five for faulty globes and three out of five for non-faulty globes. But I’ve picked one from the first pick which is two on five this one, the second one, they’re asking you to find the non-faulty one. So make sure it’s three out of one less than five which is four. So there’s two out of five probability for the faulty globe, so it’s two out of five, and uh the non-faulty globes are three out of five but because I’ve picked one, I only have four left to choose from. So it’s 3 out of 4, okay? And then simplify that you should get 3 on 10.

d! Find the probability that the second globe is faulty and the first one is not. So I want the second one to be faulty and the first one to be not faulty. So it’s pretty much the same thing. But the order is different. So there’s three out of five gloves that are not faulty. So I want to start off with three out of five because that’s the non-faulty globe and the faulty globes there’s two out of five but because I’ve picked one already from my first pick. I’ll only have four left to choose from, so it’s two out of the four, okay? Just simplified also to 3 on 10.

There are twelve houses in the street and there is someone home at seven of them. Okay, So seven out of 12, there will always be someone. That’s the probability of having someone inside a house. Find the probability that if three houses are visited at random they are all empty. 7 out of 12 is the probability that there’s someone in the house and therefore the remaining 5 out of 12 will be the probability that they are all empty which means it’s going to be 5 out of 12 times one less because we want all three to be empty. It’s gonna be that if I pick five out of the twelve which is one empty house because I’ll only have four left empty houses. It will be four out of the eleven to choose from. So after I’ve picked one empty house I want to know whether the next house is going to be empty or not. So there’s going to be 11 houses left to choose from and there’s four empty houses left. And then I’ve picked already two. So there’s three empty houses left out of the 10. So I’m subtracting from the numerator and denominator for the next two as well and then you calculate that to get 1 over 22. So that’s the answer.

Susan draws three cards in succession from a pack of 52 cards. What is the probability that they are all hearts if three cards are drawn without replacement? So again we’re not going to replace it and we want all three cards to be hearts and as I said before, guys, hearts is one out of the four shapes. So hearts is a quarter of the 52 cards, so there are exactly 13 hearts in a pack. Because it’s not going to be any replacement. It’s simply going to be 13 out of 52 times. One less on top and bottom, so 12 out of 51 times 11 out of 50. So I hope I don’t have to explain why you see why because I’m having one less card to choose from, so I’ll have one less each time on the denominator and I’ll have one less heart. So I’ll keep having one subtracted from the numerator and calculate it to get that, okay? Just use your calculator and that’s it, guys.

 

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