Week 27 - (27 April 2020) Ratio. A powerpoint introduction to Probability. Kes Maths. This video is a guide to probability. eval(ez_write_tag([[728,90],'analyzemath_com-banner-1','ezslot_9',367,'0','0']));Example 2A fair coin is tossed 5 times.What is the probability that exactly 3 heads are obtained?Solution to Example 2The coin is tossed 5 times, hence the number of trials is \( n = 5\).The coin being a fair one, the outcome of a head in one toss has a probability \( p = 0.5 \) and an outcome of a tail in one toss has a probability \( 1 - p = 0.5 \)The probability of having 3 heads in 5 trials is given by the formula for binomial probabilities above with \( n = 5 \), \( k = 3 \) and \( p = 0.5\)\( \displaystyle P(3 \; \text{heads in 5 trials}) = {5\choose 3} (0.5)^3 (1-0.5)^{5-3} \\ = \displaystyle {5\choose 3} (0.5)^3 (0.5)^{2} \)Use formula for combinations to calculate\( \displaystyle {5\choose 3} = \dfrac{5!}{3!(5-3)!} 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 \( k \).\( \displaystyle P( \text{at least 5 heads} ) = {7\choose 5} (0.5)^5 (1-0.5)^{7-5} + {7\choose 6} (0.5)^6 (1-0.5)^{7-6} + {7\choose 7} (0.5)^7 (1-0.5)^{7-7} \\ = 0.16406 + 0.05469 + 0.00781 = 0.22656 \). Mr Bartons Maths. come with answers. P8 Time Series Graphs (1) (2) 14 S (2) Sian thinks of two different numbers Videos, worksheets, 5-a-day and much more arrow_back Back to Probability with Venn Diagrams Probability with Venn Diagrams: Videos. come with answers. Number: 62 lessons Corbettmaths - 304 Followers, 31 Following, 592 pins | Corbettmaths - Home to video tutorials, 5-a-day, practice questions and much more. Khan Academy video on conditional probabilities and combinations. The Corbettmaths Textbook Exercise on Conditional Probability. Corbett Maths offers outstanding, original exam style questions on any topic, as well as videos, past papers and 5-a-day. P19 Averages From A Grouped Frequency Table P9 Pictograms Maths revision video and notes on the topic of probability. Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. Maths Genie ¦ Corbett Maths ¦ Mr Barton Maths Takeaway ¦ Mr Barton Maths Topic Search ¦ Just Maths 10.1 Combined Events - Use the product rule for finding the number of outcomes for two or more events P22 Probability Scale My name is Billy, a maths teachers working in the North West of the UK. Key Words List by Unit. All lessons introduce each topic with examples that are all fully annotated. This is the same for all topics eg, Algebra will cost Â£10 for all lessons. Corbett Maths. At the end of each presentation there are summary questions with answers attached. 20 Maths Working Wall Displays (Mostly Editable). We believe every child can do maths. = 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\( \) \( \) \( \) Welcome to Corbettmaths! We explain what probability / chance / likelihood means, how children are taught about probability from Year 5 and the kinds of mathematical problems involving probability … How to - do Simple Probability with number lines. Contents. P13 Scattergraphs - Line of Best Fit registered in England (Company No 02017289) with its registered office at 26 Red Lion Algebra: 81 lessons Mrs B Maths. Week 26 - (20 April 2020) Algebra. The probability of something not happening is 1 minus the probability that it will happen. Venn Diagrams - (Corbettmaths) Print page. In a binomial experiment, you have a number \( n \) of independent trials and each trial has two possible outcomes or several outcomes that may be reduced to two outcomes.The properties of a binomial experiment are:1) The number of trials \( n \) is constant.2) Each trial has 2 outcomes (or that can be reduced to 2 outcomes) only: "success" or "failure" , "true" or "false", "head" or "tail", ...3) The probability \( p \) of a success in each trial must be constant.4) The outcomes of the trials must be independent of each other.Examples of binomial experiments1) Toss a coin \( n = 10 \) times and get \( k = 6 \) heads (success) and \( n - k \) tails (failure).2) Roll a die \( n = 5\) times and get \( 3 \) "6" (success) and \( n - k \) "no 6" (failure).3) Out of \( n = 10 \) tools, where each tool has a probability \( p \) of being "in good working order" (success), select 6 at random and get 4 "in good working order" and 2 "not in working order" (failure).4) A newly developed drug has probability \( p \) of being effective.Select \( n \) people who took the drug and get \( k \) "successful treatment" (success) and \( n - k \) "not successful treatment" (failure). 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