A box contains 40 red, 10 blue, and 10 white cards. Two are withdrawn, one after the other without a replacement. So there’s no replacement here. Find the probability that the first card is withdrawn. A first card is that is withdrawn is blue. Okay. So you can see that 40 plus 10 plus 10, 60, there’s 60 cards in total and for the blue, it’s going to be 10 out of the 60. That’s the probability, isn’t it? So I want to find the problem that the first card withdrawn is blue. So the second one could be red or the second one could be white. So it could be 10 out of 60 which is simply 1 out of 6 which is just this guys. So the first card, the next card could be red it, could be white but we don’t have to worry about the next card. The first card doesn’t depend on what the second card is. The second one is in the future, so we don’t have to worry about what the second card is. So it’s simply just 10 over 60.
b! Find the probability that a white card is not withdrawn. Okay. The white is not withdrawn which means it’s going to be the rest which is 50 because white is 10. So the remaining of 40 plus 10 which is 50 out of 60. And the second one I don’t want it to be a white as well. So it’s going to be 49 because 1 less than 50 is 49 because I’ve already picked one out of that 50. I’m going to have 49 left and out of the 60 cards in total, I’ll only have 59 to choose from, okay? So see how I’m excluding that 10 because I don’t want white and I just multiply those together to get that. So I’ve just subtracted one from top and bottom.
c! Find the probability that the second card withdrawn is red. So we want the second one to be red. Now this one because it’s the second card it is dependent of the first card, so depending on what the first card is the second card the probability will change. If the second card is withdrawn as red it could be two cases. One which is red red or the second which is a non-red and then a red. So the second one could be a blue or a white. right? So the first one, when the first one is red. There’s 40 red out of the 60, so it’s 40 out of 60 for the first pick and the second one will be one less from top and bottom which is 39 out of 59. Or if the first card could be a blue or a white. So there’s 10 blue and 10 white. So 20 out of 60 for blue or white, so it’s 20 out of 60 but we want the second one to be red. So there’s 40 of red. But it’s going to be out of 59 because I’ve already picked one for my first pick. So it’ll be only 59 left which is one less than 60. Calculate that you add them because it’s this or this. So we add it and then you can simplify it to 2 on 3. Okay? Do you get the idea?
d! Find the probability that a white and a red are withdrawn in that order. So we must have white first and red next because that’s what that their order wants to be. So the right than the red is white there’s 10, so 10 out of 60 for the white pick. Now there’s 40 red. So 40 out of 59 because it’s one less. We’ve already picked once there’s 59 left. Okay, so make sure the denominator is different and calculate it to get 20 over 177 and that’s it, guys. Okay?
Now with e now! Find the probability that both cards are red which means we want red red which is 40 out of 60 times one less on top and bottom, so 39 over 59 because we’ll have one less red card out of one less of the total number of cards to choose from. And that’s 26 on 59. Very very simple question.
An urn contains 5 white and 5 black marbles. If three marbles are drawn from the urn at random without replacement, so without replacement, find the probability that two of the three are black. Okay, I want two of the three to be black. So there’s different cases you’re going to think about. Black black white, black white black or white black black. All three cases have two blacks. So all three cases are possible. But because there is no replacement, we must consider the orders, okay? We must consider all different cases and even if there is replacement you must consider the different cases because the order is different.
We know that there’s ten marbles in total, so there’s five white and five black. There’s five out of ten for black and five out of ten for white. So we do the same thing. Black black white, we’ll have five out of ten. And there’s one less black marble. So we’ll have four out of the nine left because there’s nine left to choose from. And then here white is five out of ten but because I’ve already picked two. There’s eight left to choose from. So five out of eight for the first one here, okay?
And then the same thing guys, black white black there’s five out of ten black, white, that’s five out of ten but because we’ve already picked one, we’ll have five out of nine, and for black, because we’ve already picked a black, we’ll have four left uh four blacks left out of eight because I’ve already picked two. I’ll have eight left to choose from. So this is the second one and white black black, there’s five out of ten whites and five out of ten blacks for the second one but we’ve already picked one. So it’s gonna be five out of nine. One less than ten. And for the next black, we’ll have four left blacks four blacks left out of eight because we’ve already picked two. So we’ve eight left to choose from. So I don’t have to really explain that too much right? That’s pretty much it. You guys can calculate it all. It will be simplified to 5 on 12. So you get the idea, guys? And make sure you add it because it’s this or this or this. So make sure you remember to add not multiply.
Algebra Algebraic Fractions Arc Binomial Expansion Capacity Chain Rule 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 Proof Pythagoras Theorem Quadratic Quadratic Factorise Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume