[{"Name":"Multiple Choice Questions for the Entire Course","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"1m 15s","ChapterTopicVideoID":13247,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/13247.jpeg","UploadDate":"2018-08-16T06:47:55.3730000","DurationForVideoObject":"PT1M15S","Description":null,"MetaTitle":"Exercise 1 - Multiple Choice Questions for the Entire Course: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Multiple Choice Questions for the Entire Course practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/probability/multiple-choice-questions-_-probability/multiple-choice-questions-for-the-entire-course/vid13735","VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:03.600","Text":"The following diagram, this thing right here,"},{"Start":"00:03.600 ","End":"00:06.540","Text":"displays 3 normal probability distributions of"},{"Start":"00:06.540 ","End":"00:10.920","Text":"3 different groups drawn on a system of coordinate axis."},{"Start":"00:10.920 ","End":"00:13.140","Text":"Now, the probability distributions have been"},{"Start":"00:13.140 ","End":"00:15.765","Text":"numbered in order to distinguish between them."},{"Start":"00:15.765 ","End":"00:19.200","Text":"Here, this is distribution number 1,"},{"Start":"00:19.200 ","End":"00:21.975","Text":"this is distribution number 2 right here,"},{"Start":"00:21.975 ","End":"00:25.665","Text":"and this is distribution number 3."},{"Start":"00:25.665 ","End":"00:27.880","Text":"In this question, we\u0027re asked,"},{"Start":"00:27.880 ","End":"00:31.470","Text":"which probability distribution has the highest average?"},{"Start":"00:31.470 ","End":"00:35.380","Text":"First of all, this is the x-axis right here."},{"Start":"00:35.380 ","End":"00:39.725","Text":"All 3 distributions are drawn on the same axis."},{"Start":"00:39.725 ","End":"00:42.800","Text":"In a normal probability distribution,"},{"Start":"00:42.800 ","End":"00:47.060","Text":"we know that the average equals the mode and that equals the medium,"},{"Start":"00:47.060 ","End":"00:50.825","Text":"and that\u0027s in the middle of the distribution."},{"Start":"00:50.825 ","End":"00:55.235","Text":"This line right here would be the average"},{"Start":"00:55.235 ","End":"00:59.590","Text":"of distribution number 1 and of distribution number 2."},{"Start":"00:59.590 ","End":"01:04.285","Text":"Here, that would be the average of distribution number 3."},{"Start":"01:04.285 ","End":"01:08.585","Text":"When we\u0027re asked which probability distribution has the highest average?"},{"Start":"01:08.585 ","End":"01:12.065","Text":"Distribution number 3 has the highest average."},{"Start":"01:12.065 ","End":"01:16.290","Text":"That means that c is the correct answer."}],"ID":13735},{"Watched":false,"Name":"Exercise 2","Duration":"1m 4s","ChapterTopicVideoID":13248,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.440","Text":"In this question we\u0027re asked which probability distribution has the highest mode."},{"Start":"00:04.440 ","End":"00:06.750","Text":"So here is our diagram again."},{"Start":"00:06.750 ","End":"00:08.910","Text":"This would be our x-axis,"},{"Start":"00:08.910 ","End":"00:12.060","Text":"that will be our variable, and here we have our 3 distributions."},{"Start":"00:12.060 ","End":"00:14.310","Text":"Now, we are asked about the mode."},{"Start":"00:14.310 ","End":"00:20.895","Text":"The mode is defined as the x value having the highest frequency."},{"Start":"00:20.895 ","End":"00:23.910","Text":"In a normal probability distribution,"},{"Start":"00:23.910 ","End":"00:27.705","Text":"the mode equals to the median and that equals to the average."},{"Start":"00:27.705 ","End":"00:29.888","Text":"Now, distribution 1 then,"},{"Start":"00:29.888 ","End":"00:31.830","Text":"the mode would be right here,"},{"Start":"00:31.830 ","End":"00:36.140","Text":"and that would be the same for distribution number 2."},{"Start":"00:36.140 ","End":"00:38.660","Text":"Mode for distribution 1,"},{"Start":"00:38.660 ","End":"00:41.600","Text":"that equals to the mode for distribution number 2."},{"Start":"00:41.600 ","End":"00:46.430","Text":"Now here, that will be the mode for distribution number 3."},{"Start":"00:46.430 ","End":"00:53.270","Text":"This is the x value of distribution number 3 that has the highest value."},{"Start":"00:53.270 ","End":"00:56.450","Text":"We see that the mode for distribution number 3 is"},{"Start":"00:56.450 ","End":"00:59.870","Text":"greater than the mode of distributions 1 and 2,"},{"Start":"00:59.870 ","End":"01:04.320","Text":"so C would be the correct answer here."}],"ID":13736},{"Watched":false,"Name":"Exercise 3","Duration":"1m 32s","ChapterTopicVideoID":13249,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"In a right asymmetric probability distribution,"},{"Start":"00:02.610 ","End":"00:08.085","Text":"the standard deviation is greater than in a left asymmetric probability distribution."},{"Start":"00:08.085 ","End":"00:10.770","Text":"Now, is this statement always correct,"},{"Start":"00:10.770 ","End":"00:15.780","Text":"never correct, or the information is insufficient to determine the answer?"},{"Start":"00:15.780 ","End":"00:20.770","Text":"Well, first of all, let\u0027s take a look at asymmetric distributions."},{"Start":"00:21.170 ","End":"00:26.540","Text":"Here we have a right asymmetric probability distribution."},{"Start":"00:26.540 ","End":"00:27.860","Text":"It has a standard deviation,"},{"Start":"00:27.860 ","End":"00:32.560","Text":"we\u0027ll call this SDR for right asymmetry."},{"Start":"00:32.560 ","End":"00:35.990","Text":"What I did here is I took this distribution,"},{"Start":"00:35.990 ","End":"00:38.450","Text":"I took the mirror image of this distribution to"},{"Start":"00:38.450 ","End":"00:42.310","Text":"create a left asymmetric probability distribution."},{"Start":"00:42.310 ","End":"00:45.348","Text":"Now here, this has a standard deviation,"},{"Start":"00:45.348 ","End":"00:48.680","Text":"we\u0027ll call this SDL for left asymmetry."},{"Start":"00:48.680 ","End":"00:52.610","Text":"Now we see that the 2 standard deviations are equal to each"},{"Start":"00:52.610 ","End":"00:57.365","Text":"other so saying that the statement is always correct, that\u0027s wrong."},{"Start":"00:57.365 ","End":"01:01.979","Text":"Now, what about saying that the statement is never correct?"},{"Start":"01:02.390 ","End":"01:07.850","Text":"Here I have a left asymmetric probability distribution"},{"Start":"01:07.850 ","End":"01:12.260","Text":"and I can obviously see that the standard deviation here,"},{"Start":"01:12.260 ","End":"01:16.230","Text":"SDL, is smaller than"},{"Start":"01:16.230 ","End":"01:20.275","Text":"this standard deviation right here for the right asymmetric distribution,"},{"Start":"01:20.275 ","End":"01:23.420","Text":"so by saying that the statement is never correct,"},{"Start":"01:23.420 ","End":"01:25.415","Text":"well, that\u0027s wrong as well."},{"Start":"01:25.415 ","End":"01:28.520","Text":"In essence, C is the correct answer by saying"},{"Start":"01:28.520 ","End":"01:32.970","Text":"that the information is sufficient to determine the answer."}],"ID":13737},{"Watched":false,"Name":"Exercise 4","Duration":"1m 4s","ChapterTopicVideoID":13250,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.140","Text":"On the numbers axis,"},{"Start":"00:01.140 ","End":"00:05.760","Text":"most of the values in a right asymmetrical probability distribution are high values,"},{"Start":"00:05.760 ","End":"00:08.595","Text":"equally distributed between high and low values,"},{"Start":"00:08.595 ","End":"00:13.515","Text":"low values, or the information is insufficient to determine the answer."},{"Start":"00:13.515 ","End":"00:19.230","Text":"First of all, let\u0027s take a look at a right asymmetric probability distribution."},{"Start":"00:19.230 ","End":"00:24.435","Text":"This is a distribution and we see that we have a tail going off to the right."},{"Start":"00:24.435 ","End":"00:26.810","Text":"Now, this is our x-axis."},{"Start":"00:26.810 ","End":"00:31.280","Text":"Now, what happens to most of the values in this type of distribution?"},{"Start":"00:31.280 ","End":"00:32.380","Text":"Where are they situated?"},{"Start":"00:32.380 ","End":"00:33.740","Text":"Where they\u0027re located?"},{"Start":"00:33.740 ","End":"00:39.530","Text":"We see that the mass of observations or most of"},{"Start":"00:39.530 ","End":"00:46.505","Text":"the area under the graph right here falls in the lower spectrum of values for x."},{"Start":"00:46.505 ","End":"00:50.040","Text":"That means that C is the right answer."},{"Start":"00:50.040 ","End":"00:55.685","Text":"The perfect example of this would be salaries where we have most of the people"},{"Start":"00:55.685 ","End":"00:59.480","Text":"earning salaries in this range and less and"},{"Start":"00:59.480 ","End":"01:04.290","Text":"less people earning a higher and higher salary."}],"ID":13738},{"Watched":false,"Name":"Exercise 5","Duration":"1m 8s","ChapterTopicVideoID":13251,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"In this question, we\u0027re told that Kathy, a famous statistician,"},{"Start":"00:03.060 ","End":"00:07.545","Text":"claims that when events E and F are mutually exclusive,"},{"Start":"00:07.545 ","End":"00:09.690","Text":"it can be said that the probability that"},{"Start":"00:09.690 ","End":"00:13.140","Text":"both events will occur equals to the product of the probability."},{"Start":"00:13.140 ","End":"00:17.260","Text":"In mathematical terms, the probability of the intersect of E and F,"},{"Start":"00:17.260 ","End":"00:20.730","Text":"that equals to the multiplication of the probabilities."},{"Start":"00:20.730 ","End":"00:22.575","Text":"Now, is this claim true?"},{"Start":"00:22.575 ","End":"00:25.440","Text":"The answer to this is no,"},{"Start":"00:25.440 ","End":"00:27.540","Text":"that it\u0027s not. Why is that?"},{"Start":"00:27.540 ","End":"00:30.840","Text":"If E and F are mutually exclusive,"},{"Start":"00:30.840 ","End":"00:34.740","Text":"then the probability of E intersect F,"},{"Start":"00:34.740 ","End":"00:36.840","Text":"that would be equal to 0."},{"Start":"00:36.840 ","End":"00:39.710","Text":"In a Venn diagram,"},{"Start":"00:39.710 ","End":"00:41.495","Text":"we\u0027d look at it like this."},{"Start":"00:41.495 ","End":"00:43.490","Text":"This would be event E,"},{"Start":"00:43.490 ","End":"00:48.290","Text":"and here, that would be event F. They\u0027re mutually exclusive."},{"Start":"00:48.290 ","End":"00:50.945","Text":"Now, you could have been confused with"},{"Start":"00:50.945 ","End":"00:54.320","Text":"the concept of mutually exclusive and independence."},{"Start":"00:54.320 ","End":"00:58.010","Text":"Now, if E and F would be independent,"},{"Start":"00:58.010 ","End":"01:00.740","Text":"then this equation would be true."},{"Start":"01:00.740 ","End":"01:02.135","Text":"But we weren\u0027t given that."},{"Start":"01:02.135 ","End":"01:04.610","Text":"We\u0027re given that they are mutually exclusive,"},{"Start":"01:04.610 ","End":"01:08.970","Text":"so this equation right here is not true."}],"ID":13739},{"Watched":false,"Name":"Exercise 6","Duration":"1m 14s","ChapterTopicVideoID":13252,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.209","Text":"In this question, there are 5 shares in an investment portfolio."},{"Start":"00:03.209 ","End":"00:04.785","Text":"Now we define an event,"},{"Start":"00:04.785 ","End":"00:08.190","Text":"none of the shares in the portfolio increase tomorrow."},{"Start":"00:08.190 ","End":"00:11.910","Text":"Assume that a share can only increase or decrease,"},{"Start":"00:11.910 ","End":"00:14.760","Text":"so what would be the complimentary event here?"},{"Start":"00:14.760 ","End":"00:17.820","Text":"First of all, let\u0027s define the event,"},{"Start":"00:17.820 ","End":"00:22.245","Text":"none of the shares in the portfolio increase, that\u0027ll be A."},{"Start":"00:22.245 ","End":"00:27.555","Text":"Here we have A as no share increases,"},{"Start":"00:27.555 ","End":"00:30.510","Text":"none of the shares in the portfolio increase tomorrow."},{"Start":"00:30.510 ","End":"00:34.530","Text":"Now, what would be the complimentary event here?"},{"Start":"00:34.530 ","End":"00:38.320","Text":"The complimentary event is written as A bar,"},{"Start":"00:38.320 ","End":"00:46.760","Text":"and that would be all the options in the sample space other than the event in question,"},{"Start":"00:46.760 ","End":"00:49.200","Text":"that means other than A."},{"Start":"00:49.340 ","End":"00:53.990","Text":"That means that we\u0027re looking at all the options in"},{"Start":"00:53.990 ","End":"00:57.995","Text":"the sample space other than no share increases."},{"Start":"00:57.995 ","End":"01:08.650","Text":"Now, that would be that at least 1 share, that would increase."},{"Start":"01:08.840 ","End":"01:10.925","Text":"Now, where\u0027s that here?"},{"Start":"01:10.925 ","End":"01:15.180","Text":"This is A, so A is the correct answer."}],"ID":13740},{"Watched":false,"Name":"Exercise 7","Duration":"2m 19s","ChapterTopicVideoID":13253,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.475","Text":"In this question, we\u0027re given 2 events, A and B,"},{"Start":"00:02.475 ","End":"00:05.535","Text":"where the probability of A equals 0.45,"},{"Start":"00:05.535 ","End":"00:08.670","Text":"and the probability of B equals to 0.5,"},{"Start":"00:08.670 ","End":"00:11.610","Text":"and the probability of the union of A and B, well,"},{"Start":"00:11.610 ","End":"00:13.380","Text":"that equals to 0.95,"},{"Start":"00:13.380 ","End":"00:15.870","Text":"and we\u0027re asked, which statement is true?"},{"Start":"00:15.870 ","End":"00:19.475","Text":"Well, A says that the events are independent."},{"Start":"00:19.475 ","End":"00:22.265","Text":"Well, in order for A and B to be independent,"},{"Start":"00:22.265 ","End":"00:25.430","Text":"then the probability of A intersect B,"},{"Start":"00:25.430 ","End":"00:30.940","Text":"that has to be equal to the probability of A times the probability of B."},{"Start":"00:30.940 ","End":"00:34.130","Text":"We have the probabilities of both events,"},{"Start":"00:34.130 ","End":"00:36.520","Text":"but we don\u0027t have the probability of the intercepts."},{"Start":"00:36.520 ","End":"00:38.045","Text":"Let\u0027s calculate that."},{"Start":"00:38.045 ","End":"00:42.860","Text":"Well, we know that the probability of A union B,"},{"Start":"00:42.860 ","End":"00:46.595","Text":"that has to be equal to the probability of A plus"},{"Start":"00:46.595 ","End":"00:52.310","Text":"the probability of B minus the probability of A intersect B."},{"Start":"00:52.310 ","End":"00:54.245","Text":"Let\u0027s just plug in the numbers."},{"Start":"00:54.245 ","End":"00:56.945","Text":"Here, that\u0027ll be 0.95."},{"Start":"00:56.945 ","End":"01:02.000","Text":"That equals to 0.45"},{"Start":"01:02.000 ","End":"01:07.865","Text":"plus 0.5 minus the probability of A intersect B."},{"Start":"01:07.865 ","End":"01:10.400","Text":"Now, after moving things around,"},{"Start":"01:10.400 ","End":"01:16.025","Text":"we see that the probability of A intersect B has to be equal to 0."},{"Start":"01:16.025 ","End":"01:20.775","Text":"As such, this condition isn\u0027t met."},{"Start":"01:20.775 ","End":"01:23.945","Text":"That means that A is wrong."},{"Start":"01:23.945 ","End":"01:27.395","Text":"But we\u0027re giving it an insight here."},{"Start":"01:27.395 ","End":"01:31.880","Text":"If the probability of A intersect B is equal to 0,"},{"Start":"01:31.880 ","End":"01:35.855","Text":"then that means that A and B are mutually exclusive."},{"Start":"01:35.855 ","End":"01:38.905","Text":"So B is the correct answer."},{"Start":"01:38.905 ","End":"01:40.690","Text":"Now, what about D?"},{"Start":"01:40.690 ","End":"01:43.700","Text":"D says that the events are complimentary."},{"Start":"01:43.700 ","End":"01:46.849","Text":"Well, in order for the events to be complimentary,"},{"Start":"01:46.849 ","End":"01:53.300","Text":"then the probability of A plus the probability of B has to be equal to 1."},{"Start":"01:53.300 ","End":"01:56.920","Text":"We see that that\u0027s just not true, so D is wrong."},{"Start":"01:56.920 ","End":"02:01.165","Text":"What about C? Event B contains event A."},{"Start":"02:01.165 ","End":"02:04.325","Text":"Well, let\u0027s just take a look at here."},{"Start":"02:04.325 ","End":"02:05.930","Text":"This is A,"},{"Start":"02:05.930 ","End":"02:09.755","Text":"and since they\u0027re mutually exclusive, this is B."},{"Start":"02:09.755 ","End":"02:14.240","Text":"There\u0027s no way that B contains A,"},{"Start":"02:14.240 ","End":"02:15.440","Text":"so C is wrong,"},{"Start":"02:15.440 ","End":"02:19.350","Text":"and therefore B is the only correct answer."}],"ID":13741},{"Watched":false,"Name":"Exercise 8","Duration":"1m 14s","ChapterTopicVideoID":13254,"CourseChapterTopicPlaylistID":245058,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.990","Text":"In this question, we want to calculate the chances of the union of 2 events."},{"Start":"00:03.990 ","End":"00:05.730","Text":"Now, in order to do so,"},{"Start":"00:05.730 ","End":"00:11.190","Text":"the probabilities of the 2 events may be added only if what happens?"},{"Start":"00:11.190 ","End":"00:17.070","Text":"Let\u0027s define A and B as events in a sample space."},{"Start":"00:17.070 ","End":"00:20.670","Text":"Now we\u0027re looking at the probability of A union B,"},{"Start":"00:20.670 ","End":"00:26.310","Text":"and we want to know when can this equal to the sum of the probabilities."},{"Start":"00:26.310 ","End":"00:30.450","Text":"Probability of A plus the probability of B."},{"Start":"00:30.450 ","End":"00:34.240","Text":"Now, we know that the probability of A union B,"},{"Start":"00:34.240 ","End":"00:38.510","Text":"that equals to the probability of A plus the probability of"},{"Start":"00:38.510 ","End":"00:43.625","Text":"B minus the probability of A intersect B."},{"Start":"00:43.625 ","End":"00:46.040","Text":"Now, in order for this to happen,"},{"Start":"00:46.040 ","End":"00:51.180","Text":"then the probability of A intersect B that has to equal to 0,"},{"Start":"00:51.180 ","End":"00:54.560","Text":"but this is the definition of events that are"},{"Start":"00:54.560 ","End":"00:58.700","Text":"mutually exclusive so A is the correct answer."},{"Start":"00:58.700 ","End":"01:04.340","Text":"So again, if the probability of A intersect B equals 0,"},{"Start":"01:04.340 ","End":"01:14.100","Text":"then the probability of the union of the events is equal to the sum of the probabilities."}],"ID":13742}],"Thumbnail":null,"ID":245058}]
