Probability #13 - Questions and Answers

Question 1 Think about a binary code that uses 4 bits (0 or 1) per code word. A bit in every code word is zero with a probability of 0.6, unrelated to any other bit. (a) What is the probability of the code word 0101? (b) What is the probability that a code word contains exactly two ones? Answer Step 1 Given data: The probability of 0 bit is P(0) = 0.6. The probability of one bit is expressed as, P(1) = 1 - P(0) Replace the specified values in the aforementioned expression. P(1) = 1 - 0.6 = 0.4 a) The expression for the probability of code-word 0101 is, P(a) = P(0) * P(1) * P(0) * P(1) Substitute the given and above-calculated values in the expression. P(a) = (0.6) (0.4) (0.6) (0.4) = 0.0576 Thus, the probability of code-word 0101 is 0.0576. Step 2 b) The expression for the probability that the code-word contains exactly 2 ones is, P(b) = (4 choose 2) * (P(1))²(P(0))² Substitute the given values in the above expression. P(b) = (4 choose 2) (0.4)²(0.6)² = (4! / (2! * 2!)) * (0.4)² (0.6)² = 6 * (0.4)² (0.6)² = 0.3456 Thus, the probability that the code-word contains exactly 2 ones is 0.3456. Question 2 The next number in the series 38, 36, 30, 28, 22 is? Answer 1 There is a pattern to the terms in the given order: minus two, minus six, minus two, minus six, minus, and so on. So 38 - 2 = 36 36 - 6 = 30 30 - 2 = 28 28 - 6 = 22 22 - 2 = 20 Answer: the next number is 20. Answer 2 The number 20 comes after 38, 36, 30, 28, and 22 in the sequence. Since there are no inherent differences among the numbers, this is discovered by looking for patterns in the supplied series of numbers. The first number in the sequence is 38, and each number after that differs by two or six. Instead of advancing, the series is falling. The number preceding 22 is 28, and the final number is 22. There are six differences between them. As a result, the following number should be different by two. The following number in the given sequence of numbers is given by the equation 22 - 2 = 20. Answer 3 Write alternate numbers, we get two different series, 38, 30, 22, ... Common difference = 30 - 38 = 22 - 30 = -8 36, 28, ... common difference = 28 - 36 = -8 Next number in this series = 28 - 8 = 20 Answer 4 Explanation: To solve the series and find the next number, we can observe the pattern in the given sequence: The given series is: 38, 36, 30, 28, 22 We can see that each number in the series is decreasing by 2 and 6 alternately. So, the pattern can be represented as follows: First number: 38 Subtract 2: 36 Subtract 6: 30 Subtract 2: 28 Subtract 6: 22 Continuing with the pattern, the next number can be obtained by subtracting 2 from the previous number: Subtract 2: 22 - 2 = 20 Therefore, the next number in the series is 20. Question 3 A pair of dice is rolled. The outcomes are summarized in the following table: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5.1) (5,2) (5,3) (5.4) (5,5) (5,6) (6,1) (6,2) (6,3) (6.4) (6,5) (6,6) Find the probability, that the sum is 11 (adding the two numbers will equal to 11). Give your answer as a decimal, rounded to two decimal digits. Find the probability, that the sum is more than 10 (which means 11 or 12). Give your answer as a decimal, rounded to two decimal digits. Find the probability, that the sum is 6 OR the sum is more than 10. Give your answer as a decimal, rounded to two decimal digits. Answer 1 Obtain the probability that the sum is 11. The probability that the sum is 11 is obtained below as follows: Let S denotes the total outcomes. That is, N(S) = 36. Let E denote the event that the sum is 11. That is, N(E) = [(5,6). (6,5)] = 2 The required probability is, P(E) = 2 / 36 = 1/18 = 0.0556 ≈ 0.06 The probability that the sum is 11 is 0.06. Answer 2 Obtain the probability that sum is more than 10. The probability that the sum is more than 10 is obtained below as follows: Let E denote the event that the sum is more than 10. That is, N(E) = [(5,6) (6,5) (6,6)] = 3 The required probability is, P(E) = 3 / 36 = 1 / 12 = 0.0833 ≈ 0.08 The probability that the sum is more than 10 is 0.08. Answer 3 Obtain the probability that the sum is 6 OR the sum is more than 10. The probability that the sum is 6 OR the sum is more than 10 is obtained below as follows: Let A denote the event that sum is 6. That is, N (A) = [(1,5), (2, 4), (3, 3), (4, 2), (5, 1)] = 5 The required probability is, P(A or E) = P(A) + P(E) - P(A and E) = 5/36 + 3/36 - 0 = 8/36 = 0.2222 ≈ 0.22 The probability that the sum is 6 OR the sum is more than 10 is 0.22. Question 4 The mean of a normal probability distribution is 60; the standard deviation is 5. a. About what percent of the observations lie between 55 and 65? b. About what percent of the observations lie between 50 and 70? c. About what percent of the observations lie between 45 and 75? Answer 1 Mean of the normal distribution is μ = 60 and the standard deviation is σ = 5 a. Observations lie between 55 and 65 Suppose that area under the normal curve with these observations is within z standard deviation of the mean. ⇒ μ - zσ = 55 and μ + zσ = 65 ⇒ 60 - 5z = 55 and 60 + 5z = 65 ⇒ 5z = 60 - 55 and 5z = 65 - 60 ⇒ 5z = 5 and 5z = 5 ⇒ z = 1 and z = 1 As per the empirical rule, about 68% of the area under the normal curve is within one standard deviation of the mean. Answer 2 b. Observations lie between 50 and 70 Suppose that area under the normal curve with these observations is within z standard deviation of the mean. ⇒ μ - zσ = 50 and μ + zσ = 70 ⇒ 60 - 5z = 50 and 60 + 5z = 70 ⇒ 5z = 60 - 50 and 5z = 70 - 60 ⇒ 5z = 10 and 5z = 10 ⇒ z = 2 and z = 2 As per the empirical rule, about 95% of the area under the normal curve is within two standard deviation of the mean. Answer 3 c. Observations lie between 45 and 75 Suppose that area under the normal curve with these observations is within z standard deviation of the mean. ⇒ μ - zσ = 45 and μ + zσ = 75 ⇒ 60 - 5z = 45 and 60 + 5z = 75 ⇒ 5z = 60 - 45 and 5z = 75 - 60 ⇒ 5z = 15 and 5z = 15 ⇒ z = 3 and z = 3 As per the empirical rule, Practically all of the area under the normal curve is within three standard deviation of the mean. Answer 4 a. We want to calculate P(55 ≤ X ≤ 65), where x is a random variable following a normal distribution with mean μ and standard deviation σ. Using the standard normal distribution, we can standardize the values 55 and 65 by subtracting the mean and dividing by the standard deviation: z1 = (55 - 60) / 5 = -1 z2 = (65 - 60) / 5 = 1 Next, we can find the corresponding cumulative probabilities for these standardized values. The cumulative distribution function (CDF) of the standard normal distribution is denoted by Φ(z). Therefore, we need to calculate P(-1 ≤ Z ≤ 1), where z is a standard normal random variable. Using the properties of the standard normal distribution, we can find these probabilities: P(-1 ≤ Z ≤ 1) = Φ(1) - Φ(-1) You can use a standard normal distribution table or a calculator to find the values of Φ(1) and Φ(-1). Subtracting these values will give you the required probability. b. To find the percentage of observations between 50 and 70, we follow a similar approach as in part a. z1 = (50 - 60) / 5 = -2 z2 = (70 - 60) / 5 = 2 P(-2 ≤ Z ≤ 2) = Φ(2) - Φ(-2) c. To find the percentage of observations between 45 and 75: z1 = (45 - 60) / 5 = -3 z2 = (75 - 60) / 5 = 3 P(-3 ≤ Z ≤ 3) = Φ(3) - Φ(-3) Answer 5 Step 1: a. To find the percentage of observations lying between 55 and 65 in a normal distribution with a mean of 60 and a standard deviation of 5, we can use the cumulative distribution function (CDF) of the normal distribution. The CDF gives the probability of a value being less than or equal to a given point. Let x be a random variable following a normal distribution with mean μ = 60 and standard deviation σ = 5. We want to find P(55 ≤ X ≤ 65). Using the standardization formula, we can transform the values of 55 and 65 into their corresponding z-scores: z1 = (55 - μ) / σ = (55 - 60) / 5 = -1 z2 = (65 - μ) / σ = (65 - 60) / 5 = 1 Next, we can use the standard normal distribution table or a calculator to find the probabilities associated with these z-scores. From the table, we find that P(Z ≤ -1) ≈ 0.1587 and P(Z ≤ 1) ≈ 0.8413. The probability of the observations lying between 55 and 65 is given by: P(55 ≤ X ≤ 65) = P(-1 ≤ Z ≤ 1) = P(Z ≤ 1) - P(Z ≤ - 1) = 0.8413 - 0.1587 = 0.6826 Therefore, approximately 68.26% of the observations lie between 55 and 65. Step 2: b. To find the percentage of observations lying between 50 and 70, we can follow a similar approach as in part (a). Let x be a random variable following a normal distribution with mean μ = 60 and standard deviation σ = 5. We want to find P(50 ≤ X ≤ 70). Using the standardization formula, we calculate the corresponding z-scores: z1 = (50 - μ) / σ = (50 - 60) / 5 = -2 z2 = (70 - μ) / σ = (70 - 60) / 5 = 2 Using the standard normal distribution table or a calculator, we find P(Z ≤ -2) ≈ 0.0228 and P(Z ≤ 2) ≈ 0.9772. The probability of the observations lying between 50 and 70 is: P(50 ≤ X ≤ 70) = P(-2 ≤ Z ≤ 2) = P(Z ≤ 2) - P(Z ≤ -2) = 0.9772 - 0.0228 = 0.9544 Therefore, approximately 95.44% of the observations lie between 50 and 70. Step 3: c. To find the percentage of observations lying between 45 and 75, we follow the same steps as in the previous parts. Let x be a random variable following a normal distribution with mean μ = 60 and standard deviation σ = 5. We want to find P(45 ≤ X ≤ 75). Using the standardization formula, we calculate the z-scores: z1 = (45 - μ) / σ = (45 - 60) / 5 = -3 z2 = (75 - μ) / σ = (75 - 60) / 5 = 3 From the standard normal distribution table or a calculator, we find P(Z ≤ -3) ≈ 0.0013 and P(Z ≤ 3) ≈ 0.9987. The probability of the observations lying between 45 and 75 is: P(45 ≤ X ≤ 75) = P(-3 ≤ Z ≤ 3) = P(Z ≤ 3) - P(Z ≤ -3) = 0.9987 - 0.0013 = 0.9974 Therefore, approximately 99.74% of the observations lie between 45 and 75. Question 5 If two dice are rolled, find the probability that neither die shows a 2 on it. probability= Answer Step 1 When two dice are rolled the total number of possible outcomes are 36. Here the question is to find the probability to have a situation where neither of the dice show two on them. This question can be solved as follows: Probability that neither of the dice show two on it = 1 - (Probability that at least one dice shows two on it) Step 2 Number of out comes where we can have two on any of the dice are 11. They are {(1,2),(3,2),(4,2), (5,2),(6,2),(2,1),(2,3),(2,4),(2,5),(2,6),(2,2)} Probability that at least one of the dice show two on it = (number of possible outcomes) / (total number of outcomes) Probability that at least one of the dice show two on it = 11/36 Step 3 Probability that neither of the dice show two on it = 1 - (Probability that at least one dice shows two on it) Probability that neither of the dice show two on it = 1 - 11/36 = 25/36 Step 4 Therefore, probability that neither of the dice show 25/36