Advance maths problem 71 please dont answer compulsory i...

Name: advance maths problem 71 please dont answer compulsory i will give dislike i have k different keys on my key ring when i come home at night i try one key after another until i find the key that fits m
Uploaded: 2021-10-17T19:55:11-08:00
Duration: 10 min 6 s
Channel: Gaurav Kalra
Description: advance maths problem 71 please dont answer compulsory i will give dislike i have k different keys on my key ring when i come home at night i try one key after another until i find the key that fits m

Advance Maths Problem 71 Please don’t answer compulsory I will give dislike I have k different keys on my key ring. When I come home at night I try one key after another until I find the key that fits my front door. What is the probability that I find the correct key, for the first time, on the nth attempt in each of the following three cases? (i) At each attempt, I choose a key that I have not tried before, each choice being equally likely. (ii) At each attempt, I choose a key from all my keys, each of the k choices being equally likely. (iii) At the first attempt, I choose from all my keys, each of the k choices being equally likely. At each subsequent attempt, I choose from the keys that I did not try at the previous attempt, each of the k ? 1 choices being equally likely. Comments This is very easy, and would really be far too easy if the answers were given. You should set out your argument clearly and concisely, because if you come up with the wrong answer and an inadequate explanation you will not get many marks. Even if you write down the correct answer you may not get the marks if your explanation is inadequate. You should of course run the usual checks on your answers. Do they lie in the range 0 ? p ? 1? If you sum over all outcomes, do you get 1? (The latter is a good and not difficult exercise; I insist that you try it.) You will probably be struck by the simplicity of the answer to part (i), after simplification. Is there an easy way of thinking about it (or, if you did it the easy way, is there a hard way)?

Transcript

00:02 Hi, as for the given statement, this is the banemal distribution we have here.

00:05 So we have 55 % of households from a 13 large region no longer have a landline and instead only have a cell phone service.

00:15 Suppose three households from the region are selected.

00:18 Three households, we have n equals three.

00:22 It is a banemal distribution.

00:24 What is the reality that all three have only cell phone service.

00:29 So 55 % of households no longer have a landline and instead only have a minimum.

00:33 Cell phone service so p is coming up to be 0 .55 p is 0 .55 you need to find out p of x equal to 3 all 3 have cell phone service formula we have is total 3 say 3 probability of success p to the part 3 and probability of failure q is coming out to be 1 negative 0 .55 that is 0 .45 so input the value here probability of success that is 0 .55 whole q, take it as here whole q times 0 .45 to the part 3 negative 3.

01:18 Let's a formula i can even give you the general formula we have for minimal distribution p of x equals r that is given as nc r probability of success to the power r q to the power n negative r general formula we have so if we just work on this and find out p of x equals 3 that's number 0 .1664 the next part is given as what is what is that at least one has only at least one has only cell phone service at least one so for at least one has one has cell phone service that means we'll have to go for p of x greater than equal to 1 at least one as cell phone service p of x greater than equal to 1 we can go for p of x equals 1 plus p of x equals 2 plus p of x equals 3 that is coming out to be 3 c1 we have probability of success that is given as 0 .55 to the power 1 0 .45 to the power 1 0 .45 to the power 3 negative 1 here you have a value of success to 0 .55 right plus for p x equal 2 that will be 3c2 0 .55 to the power 2 0 .45 to the power we have 3 negative 2 plus you have 3 c3 0 .55 to the power 3 0 .45 to the part 2 negative 3 so i can just give the values for this so so we are it as 0 .9089.

03:31 That is question number one.

03:33 Next come to question number two.

03:37 Recent polls is 2 % of residents from a certain large region took a vacation away from home in 2017.

03:44 Suppose two residents from the region are randomly selected.

03:46 So n equals to, again this is a banal distribution.

03:54 So what is the probability that both took a vacation away from home in 2017? so we have probability of success is p is 0 .62.

04:09 So both took a vacation away from homes, we'll find out p of x equals 2.

04:15 That's going to be 2c2, probability of success, 0 .62 with the power 2, probability of failure, 1 negative 0 .62, with the power we have 2 .2.

04:24 So let's calculate this.

04:28 Let's come out of 0 .384 .4.

04:34 Next we move on question number 3.

04:35 Roll are four fair six sided tie five times record the number of spots on top with the following sequences is more likely explain sequence a is five five five five so this is question number three sequence a is five five five five sequence b is two three six five five five five where the fairs decided die sequence is more likely because it is the probability of nothing.

05:15 No, actually the thing is that all these events are, if we just here, if you roll or face, they say it dive five times, right? so it is independent or four five times it is independent...

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