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Suppose that coin I has probability $0.7$ of coming up heads, and coin 2 has probability $0.6$ of coming up heads. If the coin flipped today comes up heads, then we...

Question

Suppose that coin I has probability $0.7$ of coming up heads, and coin 2 has probability $0.6$ of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or $operatorname{coin} 2$, then what is the probability that the coin flipped on the third day after the initial flip is coin $1 ?$

Suppose that coin I has probability $0.7$ of coming up heads, and coin 2 has probability $0.6$ of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or $operatorname{coin} 2$, then what is the probability that the coin flipped on the third day after the initial flip is coin $1 ?$



Answers

Suppose that coin I has probability $0.7$ of coming up heads, and coin 2 has probability $0.6$ of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or $operatorname{coin} 2$, then what is the probability that the coin flipped on the third day after the initial flip is coin $1 ?$

Okay, so you're looking at the expected value of a biased coin. So the expected value is that the coin flip is mutually independent. We can do the number of trials, times the probability of event occurring for getting the number of heads. Okay, He were given 10 trials. I was gonna be 10 times, But instead of it being a 50% chance, it's going to be 60% chance. So we're gonna multiply that by a 600.6 instead of the one a half, And that's gonna result in the expected number of heads being 6 10 10.6 is six.

Okay, so he was trying to figure out we flip a fair coin, the number of flips that would be attained, he said. A number of flips that would be obtained either until two tails are achieved or until the coin has been flipped six times. It's well lit xB the number of queen flips. And then today, fair coin. The probability of getting a heads is equal to the probability of getting eight Hales, which is 1/2. So don't do this in stages, right, because we don't have too many cases to look at. The number of folks is a minimum of two, and it goes up to six. So we'll look at what happens when X equals two when X equals 345 and six and then to get out from there. So the first case, it's gonna be the probability of when X is equal to two eso This assumes and that our 1st 2 flips for both tails and that occurs with the probability of it entails times the probability of getting tails. Or we could just square it. This would be 1/2 times 1/2 or point 25 Okay, so then The second case is going to be where we have three coin flips. And so the thing is, in this instance, we know the third football details. But one of the 1st 2 is the heads. The order does not matter. And since order doesn't matter, we will use a combination. So a combination, uh, to choose one because we're choosing one of the 1st 2 to be heads multiplying that, you know that there are in these three flips there's two tails. So the probability of tales times probably tails and there is one head, so we want to multiply it by the probability of getting heads. You'll see that the experience here there is a two and a one is equal to the number of flips, right? Yes. Once we types that into our calculator but ends up being to choose one. This these problems are the same. It's really getting 10.5 to the third. Karen's gonna be equal to 0.25 master a new page here. So for the probability that we have four flips, he's gonna follow this same formula. We're gonna have to choose a combination again or isn't matter. But two of the 1st 3 foots were heads or choosing to from three to be our heads. Two of these four is going details. We're stopping at two tails. Okay, so you want the probability of Tales Square? But the other two are going to be heads really wanted ticket, probably a head squid. And again you see the exponents two and two is equal to four that this will be equal to three Choose to That was 1/2 to the fourth and this ends up being equal to we put it in the calculator 0.187 five. So there's two more cases. Nest is where we have five flips and following the same procedure in yet choosing three from Forbes. We're choosing three of the first four flips to be heads, knowing that the 5th 1 was tails And then since we have two tails will multiply that by the problem of getting entails squared. But then we also have three heads. So I want to point out that it probably getting our heads three times. Where to the third power? Yes, this is it being for choose three times 1/2 to the fifth, which is equal to point 12 five. Yes, and we have one case left to look at. And that's going to be where we have six flips, right? So regardless of what happens on the sixth flip, we start flipping. Okay, but four or five of the 1st 5 flips has to contain heads. Eso basically there could be at most one. And so too encompass that we have to add these two cases together. So we have the case where we're choosing five. Choose four with one has one tails and the foreheads. Or there's the case we're choosing 55 Where's all heads and no Kale's. Okay, so there's either case, whether it was one tails just this first part or the case where there was no tales in the second part. And so we add that up together. So get five shoes for okay times 1/2 to the fifth and leaving it at that to five. Choose five, 1/2 to the fifth. And, of course, whether the last one is tails or not, we are stopping. So it it is a relevant okay, and this is equal to 0.18 seven fine and cement. We figured out the artillery for each case, we can move on to find the expected value. Okay, so the expected value then is going to be equal to the number off each. Tom had tails occurrence times the probability of that currents. So when we have the case where X is equal to two for two flips, that probability was, uh, 0.25 I can't allow that to. When we had three flips thin, the probability was also 0.25 for four flips got more interesting. It was 0.1. Eat 75 for five. Flips are probability was 0.1 to five. And then, lastly, to get six flips, probably of that occurring was 60.18 75 Okay, so once we add all that up, we'll find that the expected value is about three point 75 So you'd expect to get between three and four flips to get your two tails, but closer to four

So if we flip a coin six times, what is the probability that each of the flips gives us head well for each of the slips, we have two possible outcomes and these are heads entails. And we are multiplying this six times since we're doing the same action six times. So we have to to the six, which is 64 possible outcomes by the product rule. Well, one of those 64 possible outcomes, we have equal probability of getting each and every one of them. Since this is a random action that we're doing well, there's only one way to get heads every single time. And that is if we get heads, heads, heads, heads, heads and hats which we can only get it once. Therefore, the probability that we get heads every single time we flip a coin. ISS won over 64

So we want to know the conditional probability that, exactly foreheads appear when a fair cornice slipped five times. Given that the first flip first flip is Ah, heads well, let me be the events that exactly four heads appear or hiss that as be the event that the first flip comes up heads first Prince. Well, we know that the probability of F is equal to two to the fourth power over to to the fifth, two to the fifth. And that is because there are two to the fifth possible outcomes. Since we're flipping five times and we can have either heads or tails for each outcome and for the first for the two to go for it, it is because we know that the first outcome is ahead. And then for the fourth other outcomes, we can either have heads or tails, and this is equal to 1/2. So the probability that the first flip is ahead is one house Now, for the probability of we know that the condition of probability is found. So the probability that we get four heads when the first flip is ahead is equal to the probability of e intersection f we were the probability of us. We have found the probability of F and then we want to find the probability off P intersection of intersection right here we have four heads and we have a new initial H. Therefore, the soul T can only be one of the 2nd 3rd or four said one of the 2nd 3rd 4th or fifth flips, says the first been is supposed to be ahead because we have right here that the first minutes ahead and we have four heads total. Therefore, the T the only tail can only be in the 2nd 3rd 4th or fifth flip. Therefore, there are four ways to get to get four heads and the first had be the first Libby head. And then the possible outcomes is to to the fifth. Therefore, the probability that we get foreheads. Given that the first flip was ahead is Eagle two after fight clips and this is equal to four over to to the fifth over 1/2 and that is equal to won over fourth. And that is a probability that we flip a coin five times. We get foreheads. And given that the first flip was


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