Probability without Replacement with Complement Events

YouTube player


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. 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 wrong 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 the top and bottom and simplifying to get 1 on 10, okay?

b! Find the probability that neither globe is faulty, so I want both globes that I pick not to be faulty. Three globes are not wrong, so five minus two is three. So three out of five are not faulty. So it’s going to be three out of five times one less on top and bottom having two out of four, okay? So there are 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 subtract one from both the numerator and denominator. That’s the idea.

c! Find the probability that the first globe is faulty and the second is not. So I want the first globe to be faulty and the second one to be not faulty. So I know there are two out of five faulty globes and three out of five non-faulty globes. But I’ve picked one from the first pick, which is two on five this one, the second one is 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 not. So it’s pretty much the same thing. But the order is different. So there are three out of five gloves are not faulty. So I want to start with three out of five because that’s the non-faulty globe, and the faulty globes, there are 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.

Twelve houses are in the street, and someone is 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 randomly, 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 going to be that if I pick five out of the twelve, one empty house because I’ll only have four 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 will be empty. So there will be 11 houses left to choose from, and four empty houses are left. And then I’ve already picked two. So there are three empty houses left out of the 10. So I’m subtracting from the numerator and denominator for the next two, 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 removed 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 are one out of the four shapes. So hearts is a quarter of the 52 cards, so there are 13 hearts in a pack. Because it will not be any replacement, it will 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 have 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.


Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume

Your email address will not be published. Required fields are marked *