Probability & Statistics (21 of 62) Probability - 4 More Basic Examples

[@nikolaydudaev5076]

I find this video a little bit confusing. The multiplication of probabilities is introduced although it was not discussed before. And also some justification of why we use certain formula for certain example would help.

[@Peter_1986]

Here is an example that might give you some intuition: imagine that you want to select a 3-child family that has 1 girl. The _total_ number of arrangements of boys and girls is 2^3 = 8 (because the first choice has 2 outcomes, and then both those outcomes can themselves be followed by 2 possible outcomes, and then _they_ can be followed by 2 possible outcomes - you can see that the total number of outcomes is 8 if you draw a tree diagram), so the total sample space for all possible boy-girl arrangements is 8. And in this particular example, you want to choose a family with 1 girl. According to the Choose function, 1 girl in a 3-child family can be arranged in a total of 3 different ways (GBB, BGB and BBG), so this means that the _ratio_ between the number of ways to arrange 1 girl (3) and the _total possible_ number of arrangements (8) will equal 3/8. For the same reason, the probability of choosing a family where the girl is the _youngest_ will be 1/8, because there is only one way to arrange her so that she is the youngest (which, of course, is BBG).

[@justinseagull4278]

Laurelindo thank u!! U helped me a lot💚💚💚

[@jordanl9861]

You're really the best lecturer on youtube

[@6thHorseMan]

Thanks for making all these videos. This is helping me get a jump start on a probabilities course I will be taking in the future.

[@ahmedabdellatif781]

@MichelvanBiezen in the first example I understand why you multiplied instead of using the addition method as per the video # 14, which is the addition of the independent probabilities, and this because every time you pull a card it is considered as a new trial as per the sessions starting from #17, but because the denominator is changing every time you cannot use the Power option, but the question here: By using the multiplication method why don't we use the formula {1 - P(Not spade with denominator 52) x P(Not spade with denominator 51) x P(Not spade with denominator 50)), isn't this will match with your sessions starting from # 17?

[@vimvrown4264]

I think you've gotten ahead of yourself again...you haven't gone over the concept/logic of multiplying P's

[@Peter_1986]

The reason why you are multiplying probabilities is because each probability gives another certain number of ways for the next probability to happen. If you toss a coin then you have 2 possible events, and each of those 2 possible events have another 2 possible events that can follow after it, which gives you a total of 4 possible outcomes, and if you for example want 2 heads in a row then you only want the heads/heads event, in other words 1 outcome out of 4 possible outcomes, or 1/4. This is easy to see if you draw a tree diagram.

[@asmara5619]

I really enjoy all your lectures on physics and math.Thank you!!!

[@EP-rq3pn]

I'm starting to get kind of lost, unfortunately. Would you mind elaborating on when to use the n over k vs. the multiplication of each probability? Like between problems b and c. Is it strictly for the reason you stated that it simply depends on whether the order matters or not?

[@melontusk7358]

yes, that's correct, n over k is actually the combination, when the order doesn't matter. When the order does matter, try permutation.

[@samanthapikus3178]

Hi there. I really enjoy your lectures. I am struggling to understand this one and why you use two different formulas to work out the probabilities. c: The probability of finding a boy who is the youngest? How do you know that the boy is the youngest and we didn't take that into account. Question b: Not sure why you chose that formula, could you just not have said 1/2 * 1/2*1/2*1/2*1/2*1/2 to get the probability of selecting a six child family with one boy?

[@melontusk7358]

The boy here is the youngest of the family if ALL of his other siblings were born BEFORE him. (They're his sisters, this family only has a single boy with five girls)

[@umitsedgi]

thank you for this amazing video series! One little thing caught my attention. for b, total possibilities are 7. Tthe family can have 1b5g, 2b4g, 3b3g, 4b2g, 5b1g, 6b, 6g. So the probability of pulling 1 boy is 1/7. Am i wrong here?

[@diaconescutiberiu7535]

If you rewatch video 10 with the Pascal triangle you would remember that we are interested in the total number of outcomes which here would be 1 event with 1 element= 6B; 1 event with 6 elements = 1B5G (that is BGGGGG, GBGGGG, GGBGGG...so on); 1 event with 15 elements for 2B4G ... so on so forth ... for a total of 64 elements/outcomes. For the numerator you have the 6 elements for 1B5G (see my parenthesis)

[@Javlini]

Thanks for you videos. Just a note- I think it would remove confusions by wording what's inside the paranthesises a tiny bit better. I.e (Selecting 1 boy who is the youngest within a 6-child family) for c) and (selecting 5 boys and 3 girls within a 8-child family) for d). Right now it sounds a bit like something else. Thanks again

[@84toanpham]

for a. we can use different method to find: there is 13 spades in 52 cards, we pull 3 out of it. so 13C3 divide by52C3= same answer note: 13 combination 3, cuz we take 3 out of 13 spades 52 combination 3 cuz we take 3 out of 52 cards.

[@84toanpham]

b. just =6C1/2^6 D. =8C5/2^8

[@ascnir]

I think the confusion is in the English, selecting a 6-child family means to select a family that has 6 children out of say a n= 100 families so and in this case the n is not given. And there the confusion arises. For b) if he meant to select a family with 1 boy in a sample size of 6 families then it makes a better sense. Can somebody please check if I am writing it correctly please? So, b) Get the probability of finding 1 boy in 6 children (= H or T (1 of 2 options) in 6 coins) of a family that has children. and c) Get the probability of finding 5 boys in 8 children ((= 5 H in 8 coins)). Again the 3 girls is an extra and unnecessary detail that gives the idea of an union or an "at least scenario" of 5B+3G in 256 combination which creates more confusion. Hope this helps.

[@MichelvanBiezen]

No, that is not what it means. Pick a family of 6 children, (only 1 family and only 6 children). What is the probability of having 1 boy and 5 girls, or 2 boys and 4 girls, or that the oldest 3 are boys and the youngest 3 are girls, etc.

[@ascnir]

@@MichelvanBiezen Wow the master himself..Thank you for the clarification dear professor. It makes perfect sense. By the way, I have become a big fan of your teaching; short, simple, sweet and too the point. Teaachers like you are a treasure to the society. I caanot express my gratitude enough. God bless your kind heart for this selfless service to the students around the world. You cleared my doubts in youtube videos that the University couldn't. By university I meant at Masters lvl. We were just cramming up formulae and let SAS or SPSS do the work. Those lecturer should be eshamed and learn from teachers like you. Again a big heartfelt thank for making me 're-love' my subject. God bless you.

[@rupayanhalder5048]

Sir thanks for the wonderful video , sir please tell me the difference between 2nd problem and 3rd problem

[@sairam.haridasu]

for the question a,c we are using formula like P(1st SP) or P(1st G). but for b,d questions we are using formula [i.e. (n k)/2 power n] which was stated in the previous videos. why cant we same formula for b,c questions and also can we use [(n k)/2 power n] for a,c as well?

[@AK-ty9wl]

Sairam Haridasu in the coin questions, the experiment is done multiple times without reducing the possible outcomes, however in cards, the experiment is repeated, but every time you draw a card, the possible outcomes (sample space) are reduced, therefore the ( k n ) method won’t work on cards.

[@FadilAidid]

we can use binomial(n k) method with a little bit modified the formula, it becomes like this: [ binomial(13, 3)*binomial(39, 0) ] / binomial(52, 3) the result is same : 0.012941

[@ehsansahil7366]

Hallo, tnx for the videos. i don' get it, You mention groups of families but never mention how many families are in that group on timeframe 4:40 p=3/32 but you don't mention how much is the number of the group in which many families are included? that group of families is mentioned in c and d also but you don't mention have much big is the group? And also at the end you mention group of families 7/32 but how mucht families set in that group? 32 families and out of thouse 32 families 7 families would have (5B3G) comination?

[@ravivenkatesh984]

Sir, for the b sum you used ( n k)/2^n. But 2^n in the denominator is total outcomes for flipping n coins. how it could be used here for different case.hope you would clarify my doubt.

[@MichelvanBiezen]

Sometime it is better to think through the problem one step at a time, instead of using a general equation.

[@ravivenkatesh984]

i cant get you. would you please be more clear

[@dasarrajukrishnamaraju6412]

latest news tv 9

[@annabelyan2506]

either boy or girl

[@amrhafez3212]

thanks for the video !! but how come for c, the denominators are all 2's?

[@MichelvanBiezen]

Since it can only be a boy or a girl, there is a binary probability that it is one or the other.

[@shimalyounus]

thanks for all your lectures but i I could not understand from where you get first example (13/52)(12/51)(11/50) can you responsible about it with my best regards

[@josephmusonda7548]

sir i did video 1 to 20 ,can you please help me understand what concept in the previous videos did you use to answer question a and c? Thank you!

[@MichelvanBiezen]

When you are trying to find the probability of multiple events (A and B) then the probability that both events will happen is P(A) x P(B).

[@kylecatman7738]

Doing God's work my man!

[@MichelvanBiezen]

Thank you. Most of us contibutes to the well being of our community and the world beyond. Glad we have the opportunity to do our small part.

[@lena9724]

I can't understand why you used [(n k)/2^n] for part b. Could you please clarify it further sir. Thank you

[@MichelvanBiezen]

That is the equation used for such an example. Note that if you find all the probabilities of 0, 1, 2, 3, 4, 5, and 6 boys in a family of 6 you get: (1/64), (6/64), (15/64), (20/64), (15/64), (6/64), (1/64) using this formula

[@Peter_1986]

@@MichelvanBiezen I usually solve a problem like "b)" by first calculating the probability of the boy taking up any one individual position - which is 1/64 - and then I multiply that value by 6 to take all of his possible available positions into account, which gives me 6/64 = 3/32.

[@dirasadirasa2499]

in 05:50 why not 1/6 ?

[@panoschristopoulos4542]

It may be the probability of selecting a 6 child family with 1 boy multiplied with 1/6 (boy being the youngest) so that gives us 3/32*1/6=1/64 the same mr Michel gets

[@tsoojbaterdene7793]

P(selecting an 8-child family with 5B and 3G)=P(selecting an 8-child family with 5G and 3B) Is it correct,Sir?

[@wilsonkurien369]

This is precious.

[@JaroslawSchaller]

Thank you for this video, Mr. van Biezen, I really appreciate it! but I struggle to understand b) and d). Whats wrong with the following approach: let's say Omega={6B, 5B1G, 4B2G, 3B3G, 2B4G, 1B5G, 6G} meaning the group of 6 children consists of either 6 boys or 6 girls or any configuration in between. So the probability for one boy must be all relevant events over all possible, so P(1B5G)=1/7, right? I can't see my error...

[@MichelvanBiezen]

For part b there can only be 1 boy. Since that boy could have been born 1st, 2nd, 3rd, etc. then there are 6 possible births out of 2^6 possibilities.

[@JaroslawSchaller]

Ah ok, I got it! The question is just put a little bit tricky but I see. Thx

[@tinaphiri2026]

Hey sir I'm lost where did you get 13 over 52 ,12 over 51 and 11 over 50😢 and why we have used different formulas for q1 and q2

[@MichelvanBiezen]

There are 13 of each of the 4 types of cards (hearts, spades, etc.) There are 52 cards in one deck of cards. Once you pulled 1 card from the deck, you only have 12 left of that type and 51 total cards remain.

[@tinaphiri2026]

@@MichelvanBiezen I get it now thank you very much sir

[@ahmedabdellatif781]

@@MichelvanBiezen in the first example I understand why you multiplied instead of using the addition method as per the video # 14, which is the addition of the independent probabilities, and this because every time you pull a card it is considered as a new trial as per the sessions starting from #17, but because the denominator is changing every time you cannot use the Power option, but the question here: By using the multiplication method why don't we use the formula {1 - P(Not spade with denominator 52) x P(Not spade with denominator 51) x P(Not spade with denominator 50)), isn't this will match with your sessions starting from # 17?

[@MrWinter2]

Lovely examples! Thank you

[@cadeaulazare1427]

Thanks sir for the video and it really helped alot ❤❤❤, but I'm still failling to differentiate between Q.b and Q.c, cuz they look to be simple but we have used different formulas, would you please help and cralify on that,🙏🙏🙏🙏

[@jyosthnamarati1458]

Hello sir, i couldn't understand the problem c and d. Looks alike but why are we using the different formula for each of them

[@MichelvanBiezen]

c) What is the probability of having a family with 6 children where the oldest 5 children are girls and the younges child is a boy. d) what is the probaility of having a family with 8 children where 5 are boys and 3 are girls.

[@jyosthnamarati1458]

Thank you sir. Understood 😊

[@graymatter6810]

I am sorry.. but this video became unclear for problems b,c, and d.. for exa, in B, you talk about the probability of selecting a 6-child family with 1 boy.. But what is the sample space that we are selecting from? I understand the binary possibility here, but why is n = 6? Are you talking about picking 1 boy from only ONE family of 6 children or about picking 6-child families with 1 boy? If its the latter, then we do not know n isn't it? In your video, you say you are pulling 6-child family from a group, but then you take n=6, meaning that the sample space is 6 children and not a group of families. So, this has all become very confusing. Kindly clarify. Thx in advance for your response..

[@MichelvanBiezen]

For b the sample space is 6 children. (therefore n = 6) [picking from a 6 child family where one of the children is a boy]

[@graymatter6810]

@@MichelvanBiezen thank you for ur reply.. so does this mean that you are calculating the probability of picking a boy from any 6-child family given that the arrangement could be GGGGGB, BGGGGG, GGBGG, and so on?

[@MichelvanBiezen]

What is the probabiliy of selecting a 6-child family that has 5 girls and 1 boy out of a random group of families that have 6 children.

[@graymatter6810]

@@MichelvanBiezen got it. Cuz i was beating my head over what was the sample space itself.. but i see it better now.. thank you for taking the time to reply. Appreciate it :)

[@chikh2000]

I appreciate your lectures, thank you for your excellent work. in case (c) is that mean we have the same probability for having the boy first or second or ...

[@MichelvanBiezen]

Interestingly enough, yes.

[@diaconescutiberiu7535]

I dont understand a) You said that we do not return the card to the deck after each draw. That means my 2nd (and 3rd) draw will be influenced by 1st draw. Thus the problem has dependent events and we should go for P(A1)*P(A2|A1)*P(A3| A2 interesected A1) ???

[@manjunathg3]

Try it this way. Look at the deck in isolation after every draw. Originally the deck has 13 spades out of 52 cards. P(1st) = 13/52 If we have gotten a spade in the first draw, then the deck has one card less, and one spade less. Therefore spades = 12, and cards = 51, P(2nd) = 12/52 Take the same 51 card deck and draw one more spade from the set of 12 spades. Now Spades = 11 and cards = 50, P(3rd) = 11/50

[@diaconescutiberiu7535]

@@manjunathg3 I kinda get that, but now makes me wonder when do i actually need to use the conditional formula

[@manjunathg3]

@@diaconescutiberiu7535 Well i guess it takes a bit of practice to recognize the pattern of the problem. All I can suggest is to practice as many problems as possible. There is no other way to learn math. Practice enough that it becomes a second nature to your mind. Eventually, it will make more sense and one day, it will all click.

[@Peter_1986]

Wahoo! I solved all of them. :)

[@zounio]

On d) shouldn't we make the probability of the boys then of the girls and add them all up?

[@Clifffffffffford]

Thanks!

[@XxXRapidoSxXx]

Hello I didn't understand about the c) one ...

[@MichelvanBiezen]

You take the probability of each child and multiply the probabilities together. Since each can only be a boy or a girl, it is like flipping a coin. What is the probability with 6 tosses that the first five are heads and the last one is tails? It will be the same probability.

[@XxXRapidoSxXx]

+Michel van Biezen oh yeah ! thank you teacher.

[@worldacademy1706]

Why frist girl probability is not 1/6 so that I need only one girl out of six sample space ?

[@MichelvanBiezen]

You are not picking one child out of six children. That question is: what is the probability that a family with 6 children will have 1 boy and 5 girls?

[@clarittana]

As much as I loved your videos, I disliked this one. I did not understand a word. Neither the questions neither why did we use each of the formulas were clear or explained.

[@MichelvanBiezen]

We appreciate the feedback.

[@mhlurishivambu]

This probability thing is confusing now😢😢😢

[@MichelvanBiezen]

Stay with it and it'll become clear.

[@worldacademy1706]

Why frist girl probability is not 1/6 so that I need only one girl out of six sample space ?

[@MichelvanBiezen]

You are not picking one child out of six children. That question is: what is the probability that a family with 6 children will have 1 boy and 5 girls?

[@worldacademy1706]

@@MichelvanBiezen So does it mean that there is one boy and five girls in that family?

[@MichelvanBiezen]

the question asks: "what is the probablity that a family of 6 children will have one boy and 5 girls?"

[@worldacademy1706]

Why frist girl probability is not 1/6 so that I need only one girl out of six sample space ?

[@MichelvanBiezen]

You are not picking one child out of six children. That question is: what is the probability that a family with 6 children will have 1 boy and 5 girls?