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# probability corbett maths

P30 Frequency Treeâs Mathster; Corbett Maths A probability is a number that tells you how likely (probable) something is to happen. P10 Pie Chartsâ¦Drawing Share: Share on Facebook Share on Twitter Share on Linkedin Share on Google Share by email. $$S = \{ (H H H) , \color{red}{(H H T)} , \color{red}{(H T H)} , (H T T) , \color{red}{(T H H)} , (T H T) , (T T H) , (T T T) \}$$Event $$E$$ of getting 2 heads out of 3 tosses is given by the set$$E = \{ \color{red}{(H H T)} , \color{red}{(H T H)} , \color{red}{(T H H)} \}$$In one trial ( or one toss), the probability of getting a head is$$P(H) = p = 1/2$$and the probability of getting a tail is$$P(T) = 1 - p = 1/2$$The outcomes of each toss are independent, hence the probability $$P (H H T)$$ is given by the product:$$P (H H T) = P(H) \cdot P(H) \cdot P(T) \\ Venn Diagrams - (Corbettmaths) Print page. If it rains, the probability of a bus being late is 0.4. Einstein's Relativistic Train in a Tunnel Paradox: Special Relativity. = p^2 (1-p)$$In a similar way we get$$P (H T H) = p \cdot (1-p) \cdot p = p^2 (1-p)$$$$P (T H H) = (1-p) \cdot p \cdot p = p^2 (1-p)$$$$P( E ) = P ( \; (H H T) \; or \; (H T H) \; or \; (T H H) \;)$$Use the sum rule knowing that $$(H H T) , (H T H)$$ and $$(T H H)$$ are mutually exclusive$$P( E ) = P( (H H T) + P(H T H) + P(T H H) )$$Substitute$$P( E ) = p^2 (1-p) + p^2 (1-p) + p^2 (1-p) = 3 p^2 (1-p)$$All elements in the set $$E$$ are equally likely with probability $$p^2 (1-p)$$ and the factor $$3$$ comes from the number of ways 2 heads $$(H)$$ are within 3 trials and that is given by the formula for combinations written as follows:$$3 = \displaystyle {3\choose 2}$$$$P(E)$$ may be written as$$\displaystyle {P(E) = {3\choose 2} p^2 (1-p)^1 = {3\choose 2} p^2 (1-p)^1 = {3\choose 2} p^2 (1-p)^{3-2}}$$Hence, the general formula for binomial probabilities is given by   Hegerty Maths Youtube. Each question has five possible answers with one correct answer per question. Contents. P14 Introducing Sampling Methods Kes Maths. Ratio and Proportion: 33 lessons Teachit Maths - Drawing Pin Probability Experiment ♥ (2) Student carry out an experiment in which they calculate the experimental probability of the pin landing 'pin up' or 'pin down'. Conditional probability: Test your knowledge. A card is drawn from a deck of 52 cards at random, its color noted and then replaced back into the deck, 10 times. arrow_back Back to Probability of Combined Events Probability of Combined Events: Worksheets with Answers. P26 Relative Frequency and Experimental Probability, P27 Two-Way Tables the probability of getting a red card in one trial is $$p = 26/52 = 1/2$$The event A = "getting at least 3 red cards" is complementary to the event B = "getting at most 2 red cards"; hence$$P(A) = 1 - P(B)$$$$P(A) = P(3)+P(4) + P(5)+P(6) + P(7)+P(8) + P(9) + P(10)$$$$P(B) = P(0) + P(1) + P(2)$$The computation of $$P(A)$$ needs much more operations compared to the calculations of $$P(B)$$, therefore it is more efficient to calculate $$P(B)$$ and use the formula for complement events: $$P(A) = 1 - P(B)$$.$$P(B) = \displaystyle {10\choose 0} 0.5^0 (1-0.5)^{10-0} + {10\choose 1} 0.5^1 (1-0.5)^{10-1} + {10\choose 2} 0.5^2 (1-0.5)^{10-2} \\ = 0.00098 + 0.00977 + 0.04395 = 0.0547$$$$P(\text{getting at least 3 red cards}) = P(A) = 1 - P(B) = 0.9453$$. 35 students only study French 2 students do not study French or German. Student Assessment Sheets. Maths Genie ¦ Corbett Maths ¦ Mr Barton Maths Takeaway ¦ Mr Barton Maths Topic Search ¦ Just Maths 13.1 Calculating Probability - Calculate simple probabilities from equally likely events P33 Probability Tree Diagrams â Conditional arrow_back Back to Probability with Venn Diagrams Probability with Venn Diagrams: Videos. All lessons introduce each topic with examples that are all fully annotated. (1) (2) 14 S (2) Sian thinks of two different numbers A little bit of maths each day After a particularly difficult year, we need to now focus on getting our students ready for exams in the summer of 2021 despite them having a four month gap in their education at the end of year 10. There are 80 students in year 11. Week 25 - (30 March 2020) Numbers. The Corbettmaths Textbook Exercise on Probability. How to tell your life story in your college application essays. P39 (H) Histograms â Drawing Corbett Maths; Maths Genie; Hegarty Maths; Other Videos Not my own work, merely a collection of other resources, BBC Bitesize, CGP and other web resources. The probability of rain in the village is 0.3. Corbettmaths - This video explains the probability scale and goes through some typical exam questions. What is the probability of-- so I once again, I have a deck of 52 cards, I shuffled it, randomly pick a card from that deck-- what is the probability that that card that I … https://corbettmaths.com/2019/09/02/probability-practice-questions At the end of each presentation there are summary questions with answers attached. Hence Corbettmaths - 304 Followers, 31 Following, 592 pins | Corbettmaths - Home to video tutorials, 5-a-day, practice questions and much more. We believe every child can do maths. c: oqq unwpst unwp6L ;usu ecsls pe10M' 01 10110,wua pseu wseŒq ou e!x-aqsq qgce (p) hon 1011 s unwpek ou su otq!usù, elx aqsq Week 26 - (20 April 2020) Algebra. Samples of 1000 tools are selected at random and tested.a) Find the mean and give it a practical interpretation.b) Find the standard deviation of the number of tools in good working order in these samples.Solution to Example 4When a tool is selected, it is either in good working order with a probability of 0.98 or not in working order with a probability of 1 - 0.98 = 0.02.When selecting a sample of 1000 tools at random, 1000 may be considered as the number of trials in a binomial experiment and therefore we are dealing with a binomial probability problem.a) mean: $$\mu = n p = 1000 \times 0.98 = 980$$In a sample of 1000 tools, we would expect that 980 tools are in good working order .b) standard deviation: $$\sigma = \sqrt{ n \times p \times (1-p)} = \sqrt{ 1000 \times 0.98 \times (1-0.98)} = 4.43$$, Example 5Find the probability that at least 5 heads show up when a fair coin is tossed 7 times.Solution to Example 5The number of trials is $$n = 7$$.The coin being a fair one, the outcome of a head in one toss has a probability $$p = 0.5$$.Obtaining at least 5 heads; is equivalent to showing : 5, 6 or 7 heads and therefore the probability of showing at least 5 heads is given by$$P( \text{at least 5}) = P(\text{5 or 6 or 7})$$Using the addition rule with outcomes mutually exclusive, we have$$P( \text{at least 5 heads}) = P(5) + P(6) + P(7)$$where $$P(5)$$ , $$P(6)$$ and $$P(7)$$ are given by the formula for binomial probabilities with same number of trial $$n$$, same probability $$p$$ but different values of $$k$$.$$\displaystyle P( \text{at least 5 heads} ) = {7\choose 5} (0.5)^5 (1-0.5)^{7-5} + {7\choose 6} (0.5)^6 (1-0.5)^{7-6} + {7\choose 7} (0.5)^7 (1-0.5)^{7-7} \\ = 0.16406 + 0.05469 + 0.00781 = 0.22656$$. 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