Probability of Simple Events - Experiments, Outcome, Sample Space and Event @MathTeacherGon

He will also demonstrate how to find the probability of simple events using the formula.
Permutation - Fundamental Counting Principle (Probability and Combinatorics)
Fundamental Counting Principle (Tree Diagram)
Distinguishable Permutations - Permutations and Combinations
Combinations - (Permutation and Combination)
Probability of Simple Events - Experiments, Outcome, Sample Space and Event

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[MathTeacherGon]

Hi guyz, pwede ko bang malaman kung saang school kayo nagaaral.😊

[maicho]

i don't speak your language but thank you so much for this! it helped me understand the questions in my exam >< may god bless you, mister!

[lazzada]

Much better than 1 hour lecture of my prof. U deserve my subscribe!

[ichiiro]

Thank you Sir! Natulungan ako para mas maunawaan ang topic na ito. More power po

[sorinasoriano]

Thank you so much sir.. it really helped me in my exam, it is really frustrating to have low grades basically in math😪♥️

[soundeffectandbackground]

Thank youuu po! It helped me so much to understand kasi hindi ako pumapasok sa sub namin na business statistics bc of my religion, nanonood nalang ako ng videos mo po 😭🫶🏻✨❤

[juvelleannflores]

Thank you so much sir!!!very helpful po itong video lalo na di ko gets sa school ang bilis ng paliwanag ng teacher namin 😅😅😅di pa po tapos yung video gitna palang pero gets ko na poo, thank you sir!!❤❤

[patriznicoledemandante]

Thank you po sir
D ko na talaga alam isasagot buti nalang meron ka pong video tapos tagalog pa

[margielynmarthaferrer]

I just subscribed. Nag karoon na ko ng idea pano ko ba mahahanap yung probability of winning ng isang laro for school project. Thank youu sir💖

[mariajoyalaisa]

Thank you po sir..mas madali q po maiuturo sa anak q😊

[RozelynRenoballesTV]

thank you sir for this video. dahil sa inyo naintindihan ko itong topic namin

[archanalawande]

Sir please answer 🙏
Q. Write down the sample space for the following experiments.Also write the nature of the sample space.


1) Ten seeds are planted and total number of seeds germinated are recorded after a week.

2) A point is randomly placed inside a circle with radius 10 cm . Its distance from the centre is measured.

[chloesantos]

Tysm :))
It really helped me understand my math module TnT
(Since my teacher doesn't give us online classes :/)
And you explain more clearly than my teacher! Tysm!

[angelitosabulaanofficial]

Thank you po. This is very helpful for me. BTW I'm studying at St. John Paul II College of Davao.

[岡安一壽-gy]

Thank you sir.
Many people get the following question wrong, so please explain.
[Question]
Find the probabilities that the following events will occur when two dice are rolled.
①Both show odd numbers
②One shows odd number and another shows evenn number
③Both show even number

P(E):the probability that event E occurs.

【Distinguishable dice A and B】
(the number that dice A shows, the number dice B shows):event
If we only judge whether the numbers indicated by the dice are odd or even, the following four events will occur.
　　(odd, odd) (odd, even)

　　(even, odd) (even, even)
If each event occurs with equal probability, the probability is 1/4.
Therefore, P(①)=1/4, P(②)=1/2, P(③)=1/4 (1)
If we judge the number(1, 2, ···, 6) indicated by the dice, the following 36 events will occur.
(1, 1), (1, 3), (1, 5) (1, 2), (1, 4), (1, 6)
(3, 1), (3, 3), (3, 5) (3, 2), (3, 4), (3, 6)
(5, 1), (5, 3), (5, 5) (5, 2), (5, 4), (5, 6)

(2, 1), (2, 3), (2, 5) (2, 2), (2, 4), (2, 6)
(4, 1), (4, 3), (4, 5) (4, 2), (4, 4), (4, 6)
(6, 1), (6, 3), (6, 5) (6, 2), (6, 4), (6, 6)
If each event occurs with equal probability, the probability is 1/36.
Therfore
P(①)=9×(1/36)=1/4, P(②)=18×(1/36)=1/2, P(③)=9×(1/36)=1/4 (2)
(2) matches (1).

【Indistinguishable two dice】
(odd, even)is the same event as(even, odd).
Thererore, if we only judge whether the numbers indicated by the dice are odd or even, the following three events will occur.
　　(odd, odd)

　　(even, odd) (even, even)
If each event occurs with equal probability, the probability is 1/3.
Therefore
P(①)=1/3, P(②)=1/3, P(③)=1/3 (3)
(1, 3) is the same event as (3, 1).
If we judge the number(1, 2, ···, 6) indicated by the dice, the following 21 events will occur.
(1, 1)
(3, 1), (3, 3)
(5, 1), (5, 3), (5, 5)

(2, 1), (2, 3), (2, 5) (2, 2)
(4, 1), (4, 3), (4, 5) (4, 2), (4, 4)
(6, 1), (6, 3), (6, 5) (6, 2), (6, 4), (6, 6)
If each event occurs with equal probability, the probability is 1/21.
Therefore
P(①)=6×(1/21)=2/7, P(②)=9×(1/21)=3/7, P(③)=6×(1/21)=2/7 (4)
(4) contradicts (3).

[archanalawande]

New subscriber here

Best channel ❤👍
Love your teaching sir 👍🙏
Thankyou so much for this video
It helped me a lot.

I am so lucky to get this chance to learn from a best teacher 💓 💖

[cleoarganda]

Thank you so much sir your vedio lessons helped me a lot💖🙃

[josekinzd.sanbuenaventura]

Salamat po sa tulong maka anser sa modules ko po I am from colegion del santisimo rosario in bicol

[joshuagabriels.maninang]

Salamat sir, minsan mahina ang net kaya hindi ko maintindihan yung lesson namin putol putol, kaya dito ako sa yt nanonood ng lesson. Eto po yung performance task namin salamat po😁

[jessasiervo]

Actually Math subject sakin ang pinakamahirap😥but don't give up to learn Math...