A ball is chosen at random and it is noted whether it is red. Solution: (a) The repeated tossing of the coin is an example of a Bernoulli trial. Example 2: For the same question given above, find the probability of: Solution: P (at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1). × (½)2× (½)3 P(x=2) = 5/16 (b) For at least four heads, x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5) Hence, P(x = 4) = 5C4 p4 q5-4 = 5!/4! Each question has four possible answers with one correct answer per question. Hence. There are two possible outcomes: true or false, success or failure, yes or no. n = 10, p=0.15, q=0.85, x=4. The 0.7 is the probability of each choice we want, call it p. The 2 is the number of choices we want, call it k. And we have (so far): = p k × 0.3 1. 6 times, a ball is selected at random, the color noted and then replaced in the box.What is the probability that the red color shows at least twice?Solution to Example 7The event "the red color shows at least twice" is the complement of the event "the red color shows once or does not show"; hence using the complement probability formula, we writeP("the red color shows at least twice") = 1 - P("the red color shows at most 1") = 1 - P("the red color shows once" or "the red color does not show")Using the addition ruleP("the red color shows at least twice") = 1 - P("the red color shows once") + P("the red color does not show")Although there are more than two outcomes (3 different colors) we are interested in the red color only.The total number of balls is 10 and there are 3 red, hence each time a ball is selected, the probability of getting a red ball is \( p = 3/10 = 0.3\) and hence we can use the formula for binomial probabilities to findP("the red color shows once") = \( \displaystyle{6\choose 1} \cdot 0.3^1 \cdot (1-0.3)^{6-1} = 0.30253 \)P("the red color does not show") = \( \displaystyle{6\choose 0} \cdot 0.3^0 \cdot (1-0.3)^{6-0} = 0.11765 \)P("the red color shows at least twice") = 1 - 0.11765 - 0.30253 = 0.57982. eval(ez_write_tag([[300,250],'analyzemath_com-large-mobile-banner-2','ezslot_13',701,'0','0']));Example 880% of the people in a city have a home insurance with "MyInsurance" company.a) If 10 people are selected at random from this city, what is the probability that at least 8 of them have a home insurance with "MyInsurance"?b) If 500 people are selected at random, how many are expected to have a home insurance with "MyInsurance"?Solution to Example 8a)If we assume that we select these people, at random one, at the time, the probability that a selected person to have home insurance with "MyInsurance" is 0.8.This is a binomial experiment with \( n = 10 \) and p = 0.8. Hence if you When we replace in the formula: Interpretation: The probability that exactly 4 candies in a box are pink is 0.04. Example 1A fair coin is tossed 3 times. There are two parameters n and p used here in a binomial distribution. According to the problem: Number of trials: n=5 Probability of head: p= 1/2 and hence the probability of tail, q =1/2 For exactly two heads: x=2 P(x=2) = 5C2 p2 q5-2 = 5! 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: \( P(A) = 1 - P(B) \).\( P(B) = \displaystyle {10\choose 0} 0.5^0 (1-0.5)^{10-0} + {10\choose 1} 0.5^1 (1-0.5)^{10-1} + {10\choose 2} 0.5^2 (1-0.5)^{10-2} \\ = 0.00098 + 0.00977 + 0.04395 = 0.0547 \)\( P(\text{getting at least 3 red cards}) = P(A) = 1 - P(B) = 0.9453 \). When you throw the dice 10 times, you have a binomial distribution of n = 10 and p = ⅙. × (½)4× (½)1 = 5/32. \( S = \{ (H H H) , \color{red}{(H H T)} , \color{red}{(H T H)} , (H T T) , \color{red}{(T H H)} , (T H T) , (T T H) , (T T T) \} \)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) \\ Your email address will not be published. Each question has five possible answers with one correct answer per question. Here the number of failures is denoted by ‘r’. Hence Example 2 A fair coin is tossed 5 times. Finding the quantity of raw and used materials while making a product. "at least 8 of them have a home insurance with "MyInsurance" means 8 or 9 or 10 have a home insurance with "MyInsurance"The probability that at least 8 out of 10 have have home insurance with the "MyInsurance" is given by\( P( \text{at least 8}) = P( \text{8 or 9 or 10}) \)Use the addition rule\( = P(8)+ P(9) + P(10) \)Use binomial probability formula calling "have a home insurance with "MyInsurance" as a "success".\( = P(8 \; \text{successes in 10 trials}) + P(9 \; \text{successes in 10 trials}) + P(10 \; \text{successes in 10 trials}) \)\( = \displaystyle{10\choose 8} \cdot 0.8^8 \cdot (1-0.8)^{10-8} + \displaystyle{10\choose 9} \cdot 0.8^9 \cdot (1-0.8)^{10-9} + \displaystyle{10\choose 10} \cdot 0.8^10 \cdot (1-0.8)^{10-10} \)\( = 0.30199 + 0.26843 + 0.10737 = 0.67779 \)b)It is a binomial distribution problem with the number of trials is \( n = 500 \).The number of people out of the 500 expected to have a home insurance with "MyInsurance" is given by the mean of the binomial distribution with \( n = 500 \) and \( p = 0.8 \).\( \mu = n p = 500 \cdot 0.8 = 400 \)400 people out of the 500 selected at random from that city are expected to have a home insurance with "MyInsurance". The number of trials must be fixed. 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 \( k \).\( \displaystyle P( \text{at least 5 heads} ) = {7\choose 5} (0.5)^5 (1-0.5)^{7-5} + {7\choose 6} (0.5)^6 (1-0.5)^{7-6} + {7\choose 7} (0.5)^7 (1-0.5)^{7-7} \\ = 0.16406 + 0.05469 + 0.00781 = 0.22656 \). Therefore the probability of getting a correct answer in one trial is \( p = 1/5 = 0.2 \)It is a binomial experiment with \( n = 20 \) and \( p = 0.2 \).\( P(\text{student answers 15 or more}) = P( \text{student answers 15 or 16 or 17 or 18 or 19 or 20}) \\ = P(15) + P(16) + P(17) + P(18) + P(19) + P(20) \)Using the binomial probability formula\( P(\text{student answers 15 or more}) = \displaystyle{20\choose 15} 0.2^{15} (1-0.2)^{20-15} + {20\choose 16} 0.2^{16} (1-0.2)^{20-16} \\ \quad\quad\quad\quad\quad + \displaystyle {20\choose 17} 0.2^{17} (1-0.2)^{20-17} + {20\choose 18} 0.2^{18} (1-0.2)^{20-18} \\ \quad\quad\quad\quad\quad + \displaystyle {20\choose 19} 0.2^{19} (1-0.2)^{20-19} + {20\choose 20} 0.2^{20} (1-0.2)^{20-20} \)\( \quad\quad\quad\quad\quad \approx 0 \)Conclusion: Answering questions randomly by guessing gives no chance at all in passing a test.

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