مثالهای عمومی قانون AND
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Examples of Generalized AND Rule
Now we’ll talk about some examples of the generalized AND rule. So, first of all, recall the generalized AND rule says, the probability of A and B is the probability of A times the probability of B given A. Or the probability of A and B is the probability B times the probability of A given B, these two expressions here, B given A or A given B, these are conditional probabilities which we discussed in the last video.
So these formulas are used when A and B are not independent, of course if A and B are independent, much simpler, we just multiply the two probabilities directly. So, when do we use this on the test? Well, in practice, most of the cases on the test in which you will have to use this involve selection without replacement.
Well, let’s recall what this means exactly. Without replacement means each choice changes the probability of subsequent choices, so the choices are not independent. So, how does this play out in practice? Suppose I’m picking cards out of a deck. So I start with a nice shuffled full deck of 52 cards.
I pick one card, I put it aside, I don’t put it back into the deck, I don’t replace it. I put it aside. Now, the second card I’m going to pick, I’m picking this now from my deck of 51 cards. And when I pick that, I put it aside.
The third choice will be from a deck of 50 cards. The fourth choice will be from a deck with 49 cards and so forth. Each time I pick, I’m picking from a smaller deck. And so the number in the deck, the number in the selection deck, changes every time which means all the ratios change which means all the probabilities change.
In particular, suppose I pick a heart on my first card and then I pick a heart on my second card, well now it’s much less likely that I’m going to pick a heart on my third card, because there are two hearts missing from the deck. That’s a deck now, it’s a full deck of cards minus two hearts. So there are fewer hearts in that deck, and so that means that it’s less likely to pick a third heart, given that I’ve already picked two.
So that’s an example of how earlier choices change probabilities of later choices in a scenario when you’re picking without replacement. So, let’s see this played out in a calculation. A box has five green balls and seven red balls. Assume that all the balls in the box are equally likely and that the balls are picked without replacement.
So we have that magic phrase there, without replacement. What is the probability that the first two balls picked are both green? So, let’s think about this. I’m briefly gonna draw this out with five green balls, and seven red balls. That’s our box. So, the probability that the first choice is green, well that’s easy, there are 12 balls in the box, 5 of the 12 are green.
So, that’s 5 over 12. That part is easy. Now notice, once we pick that, that ball is missing. So, we no longer can think of this as 12 balls. Now there are 11 balls in the box. So, now when I wanna know what’s the probability that my second choice is green, given that the first one is green.
Now I’m considering this new scenario, 11 balls in the box, and 4 of the 11 are green. So that would be 4 11th, and so the total probability, I would multiply these two. Now, of course, if we’re clever about fractions, we realize the worst thing in the world we could possible do is multiply 5 times 4 and multiply 12 times 11.
That would be a really bad idea. You always want to cancel before you multiply. Why get big numbers if you don’t need to get big numbers. Cancel before you multiply. So I’m just gonna cancel the 4 and the 12, so those go down to 1 and 3. In the numerator I get 5, in the denominator I get 3 times 11, which is 33 and that is the answer.
Let’s look at another problem. From a standard shuffled deck of 52 cards, what’s the probability of picking three hearts on the first three cards drawn, if the cards are selected without replacement? So again, the magic phrase, without replacement. So, probability that the first is a heart.
Well, as you may know, in a standard deck of cards there are 4 suits, and each suit has exactly 13 cards. The four suits are hearts, diamonds, clubs and spades. And so, when I’m starting out with a 52 card deck, there are, is an equal number of cards in each suit, so every suit is equally likely, the four suits are equally likely, so the probability of picking a heart is one quarter.
Now starts to get a little tricky. The second pick. What’s the probability that the second pick is a heart given that the first pick was a heart. Well, so now we’re dealing with a deck of 51 cards, not 52. So I’m gonna put 51 in the denominator.
And now, instead of having 13 hearts, 1 has been taken out already so now we’re only left with 12 hearts. So 12 of those 51 cards are hearts. So that’s a probability of 12 over 51. So notice, it’s a different probability from the first one because we’re selecting without replacement, these probabilities are not independent, and what happens on the first outcome changes the probability on the second outcome.
So now we have to think about a third choice, so this third choice. Given that the first is heats and the second is hearts. Well now, I’m picking from a deck of 50 cards. Two cards have been removed. And both cards removed, according to this condition, both cards removed have been hearts.
So now, instead of 13 hearts, I’m down to 11 hearts. So 11 of those 50 cards are hearts. And we get this. Let’s see. Incidentally, the second fraction I’m gonna simplify a little bit. I can cancel a 3 in the numerator and denominator so that’s gonna get on to 4 over 17.
Now, I’m just gonna multiply these three. Multiply the original probability times the conditional probabilities. 4 over 17 times 11 over 50. Again, the worst thing in the world you could do, which is multiply straight across before canceling. We can cancel the 4’s right away, things get much easier.
In the numerator we just have 1 times time 1 times 11. In the denominator we get 17 times 50. Well 17 times 5 is 85 so 17 times 50 is 850, this is the probability of picking 3 cards and having all 3 of them be hearts.
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