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A coin Is tossed 4 times and the faces obtained are observed Find the probabllity of the following events. Give your answer as a fraction in its lowest termslandin...

Question

A coin Is tossed 4 times and the faces obtained are observed Find the probabllity of the following events. Give your answer as a fraction in its lowest termslanding more heads than tailslanding at least two tails

A coin Is tossed 4 times and the faces obtained are observed Find the probabllity of the following events. Give your answer as a fraction in its lowest terms landing more heads than tails landing at least two tails



Answers

A coin is tossed four times and the number of heads that appear is counted. Fill in the following table listing the probabilities of obtaining various numbers of heads. What do you observe? Are all of these outcomes equally likely?

And this problem were asked if I'm the probability of getting exactly two heads are getting at least two tails in an experiment and wherever tossing four coins. To do this, we can use the union rule where event he's getting exactly two heads and Event F is getting at least two tails. But before we start computing these probabilities bracelets notice that the intersection between these two events is the same as Event E. That's because if both of these events are true, there's exactly two heads and at least two tails than 1/2 speech you that there's two heads and two tails, which is the same as Event E. So with this knowledge we can simplify our union room. I noticed that the first and third terms cancel out. So now all we need to do is find the probability of Event F getting at least two tails. This probability can be computed by dividing the number of elements in the event by the number of elements in the sample space. Since all of the outcomes in the samples face are equally likely, the number of the elements in the samples and in this event F is 11. We count that by looking at the, um, outcomes where there's four taels, three tails or two tails and there are 16 elements of the sample space. So the final probability of the game in between exactly two heads and getting at least you had tails is 11 over 16.

Okay, so we're asked upon the probability of tossing exactly two heads or exact or at least details soapy of two heads or at least details. Right? So what's the probable? You two heads won't help. Let's see here. Let's find the ones were too high. It's right here, right here. Here, here. That's equal to right. This one is a 123456 Yes, six out of 16. Plus, let's see. At least two tells actresses one down from Tonto details. But this a green. Just one to tell Carrie to tell here. Details, details, details, Details. 33 before us up T o 123456789 10 11 11 out of 16 and we need to subject our intersection of the two. That's this one right here, right here, Here, here, here and here. So that's minus six. Wait. 123456 and 26 over 16. Sorrow. Our probability of two heads, or at least two tails, is just 11 over it. 16

Three coins that are tossed. We want to know what is the probability of getting two heads and one tail so we can make a tree toe? Figure this question out our first flip, we could either get a head or a tail on our second flip. We could either get a head or a tail, and our third flip, we could get either a head or a tail. And now we need to find the options, um, or the paths that have two heads in one tail and all. Circle them in green. So this is head, head head had had tail. So this one right here is one head tail head is one head tail. Tail doesn't work. Tail had head works. Tail head tail does not tail tail head does not tell tale tale does not. So there are three, um, outcomes that meet the criteria out of 12345678 total outcomes. So the probability of getting two heads and one tail is 3/8

So we have an unfair coin. Biased coin and tails is twice as likely to occur heads. And that is what that the second equation represents that hails is twice as likely to occur His heads and our first equation is this basic, um, foundation of statistics. And that two thesis, um of the outcomes has to equal one right, probably of heads. Plus, probably tales as people won in a fair coin will be 50 50. But this coins unbiased. So it should be it is biased. So our probability is going to change a little bit and how we could fade it out is using substitution. So we can sub the probability of tea in our first equation with two times, probably of H. Because the second question says that they are equal. So when we write rewrite that we have two times the probability of age. Um, plus, the probability of age is going to be equal to what and now we have two times probably page plus the probably date itself, and that gives us three times the probably of age because we can treat the probability as a variable or as if we're combining like terms, So we have three times it, probably of heads equals one. So when we solve for the probably of heads, we divide by three and we get that the probability of heads is 1/3 and going back to our first equation where the sum has people one we know that the probability of tales has to be 2/3. So knowing that we can get on to solving our questions. So the 1st 1 wants to know what the probability of getting exactly two heads is if we tossed a coin by this this coin four times. So I listed all the waves that you could get exactly two heads. There are six of them now, if safe were to take this just this first scenario, we get two heads in a row and then two tails we would get 1/3 times 1/3 times, 2/3 times 2/3 right. We have probably of heads and then heads again, and then tales and tails. But if we look at the other five ways to get exactly two has the values, the probabilities are the same. They might just be a different order. Example. This next one our tails Tails has has would be 2/3 times, 2/3 times 1/3 time's one thing. So in reality were multiplying these same four probabilities just maybe in different orders. Maybe this is a year this is over there or what not. So if we're to take just this, probably we have here we have one times one times two times two, which is four. And then we have three times three, which is nine. And then three times three, which also nine. You would have nine times nine, which is just 81. So the probability of just one of these events appearing is four out of 81. But since there are six ways to get exactly two heads, we then have to multiply this by six. To get 24 out of 81 is our total probability of getting exactly two heads follow the same logic with our next probably which is we want at least three tails. So here all the ways that we can get at least retails. This last probability is all four taels. But that's OK because we want, at least so again, if we look at just this 1st 1 We have tails three times and then heads, so that would be 2/3 times 2/3 times 2/3 times 1/3. And that total probability is two times two is four times to his age. So we have a judge out of 81 and that's gonna happen four times. So we'll have to multiply this by four. And so we get 32 out of 81. But this last probability of all four taels is going to be a bit different because we have to factor in the extra tail. So instead of 2/3 2 3rd 2/3 and 1/3 it would be all four would be 2/3. That would change our probability to two times two times two times two, which is 16. So we have to add an additional probability of 16 out of 81. Then, when we add these, probably together we get a grand total of 48 out of 81 is the probability that we get at least read tales


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