answer:To solve this problem, we can use the concept of conditional probability and the binomial theorem. Let's denote the probability of making a free throw after a successful attempt as P(S) = 0.8 and the probability of making a free throw after a missed attempt as P(M) = 0.6. We need to find the probability of making exactly 7 out of 10 free throw attempts. There are multiple ways this can happen, and we need to consider all possible sequences of 7 successes (S) and 3 misses (M). We can represent these sequences as a string of S and M, for example, SSSSSSSMMM. There are C(10, 7) = 10! / (7! * 3!) = 120 ways to arrange 7 successes and 3 misses in a sequence of 10 attempts. For each of these sequences, we can calculate the probability of that specific sequence occurring and then sum up the probabilities of all sequences. Let's calculate the probability of a specific sequence. For example, consider the sequence SSSSSSSMMM. The probability of this sequence occurring is: P(SSSSSSSMMM) = P(S) * P(S|S) * P(S|S) * P(S|S) * P(S|S) * P(S|S) * P(S|S) * P(M|S) * P(M|M) * P(M|M) Since the first attempt has an 80% chance of success, P(S) = 0.8. For the next 6 successful attempts, the probability is P(S|S) = 0.8. For the first miss after a successful attempt, the probability is P(M|S) = 0.2. For the next two misses after a miss, the probability is P(M|M) = 0.4. So, P(SSSSSSSMMM) = 0.8 * 0.8^6 * 0.2 * 0.4^2 = 0.8^7 * 0.2 * 0.4^2. Now, we need to calculate the probability of all 120 possible sequences and sum them up. Since each sequence has the same probability structure (7 successes and 3 misses), we can multiply the probability of one sequence by the number of sequences: P(7 out of 10) = 120 * (0.8^7 * 0.2 * 0.4^2) ≈ 0.2466 So, the probability that the player makes exactly 7 out of their 10 free throw attempts is approximately 24.66%.

answer:Let's use a transition matrix to represent the probabilities of users switching between light and heavy user categories. The matrix will be: ``` | 0.7 0.4 | | 0.3 0.6 | ``` Where the first row represents the probabilities of a light user remaining a light user (0.7) and switching to a heavy user (0.3), and the second row represents the probabilities of a heavy user switching to a light user (0.4) and remaining a heavy user (0.6). Now, let's represent the current distribution of users as a vector: ``` | 0.8 | | 0.2 | ``` Where 0.8 represents the percentage of light users and 0.2 represents the percentage of heavy users. To find the distribution of users in one month, we multiply the transition matrix by the current distribution vector: ``` | 0.7 0.4 | | 0.8 | | 0.7*0.8 + 0.4*0.2 | | 0.56 + 0.08 | | 0.64 | | 0.3 0.6 | * | 0.2 | = | 0.3*0.8 + 0.6*0.2 | = | 0.24 + 0.12 | = | 0.36 | ``` So, in one month, 64% of users will be light users and 36% will be heavy users. To find the distribution in two months, we multiply the transition matrix by the distribution vector from one month: ``` | 0.7 0.4 | | 0.64 | | 0.7*0.64 + 0.4*0.36 | | 0.448 + 0.144 | | 0.592 | | 0.3 0.6 | * | 0.36 | = | 0.3*0.64 + 0.6*0.36 | = | 0.192 + 0.216 | = | 0.408 | ``` So, in two months, 59.2% of users will be light users and 40.8% will be heavy users.

answer:Let S represent a sunny day and R represent a rainy day. We are given the following probabilities: P(S) = 0.7 P(R) = 0.3 P(S|S) = 0.6 P(R|R) = 0.7 We want to find the probability P(S_3|S_1), where S_1 represents that it is sunny today, and S_3 represents that it is sunny three days from now. We can use the law of total probability to find P(S_3|S_1): P(S_3|S_1) = P(S_3|S_1, S_2) * P(S_2|S_1) + P(S_3|S_1, R_2) * P(R_2|S_1) We know the following probabilities: P(S_2|S_1) = 0.6 (given) P(R_2|S_1) = 1 - P(S_2|S_1) = 1 - 0.6 = 0.4 P(S_3|S_1, S_2) = P(S_3|S_2) = 0.6 (given) P(S_3|S_1, R_2) = P(S_3|R_2) = 1 - P(R_3|R_2) = 1 - 0.7 = 0.3 Now we can plug these probabilities into the equation: P(S_3|S_1) = P(S_3|S_1, S_2) * P(S_2|S_1) + P(S_3|S_1, R_2) * P(R_2|S_1) P(S_3|S_1) = (0.6 * 0.6) + (0.3 * 0.4) P(S_3|S_1) = 0.36 + 0.12 P(S_3|S_1) = 0.48 So, the probability that it will be sunny three days from now, given that it is sunny today, is 0.48 or 48%.

answer:To solve this problem, we need to find the expected number of each type of car sold tomorrow. We can do this by multiplying the probability of selling each type of car by the total number of cars sold today. Expected number of type A cars sold tomorrow = 0.3 * 50 = 15 Expected number of type B cars sold tomorrow = 0.4 * 50 = 20 Expected number of type C cars sold tomorrow = 0.3 * 50 = 15 So, we can expect the company to sell 15 type A cars, 20 type B cars, and 15 type C cars tomorrow.