How many odd numbers greater than 4000 can be formed using the digits 2, 3, 4, 5, and 6, if each digit is used once in each numeral?

Alright Faith — let's do this. 13. a) How many odd numbers greater than 4000 can be formed, using the numbers 2, 3, 4, 5 and 6, if each digit is used once in each numeral? Step 1: Determine the number of digits in the numeral. Since there are 5 distinct digits given (2, 3, 4, 5, 6) and "each digit is used once in each numeral", we are forming 5-digit numbers. Step 2: Apply the condition "greater than 4000". Any 5-digit number formed using these digits will automatically be greater than 4000 (e.g., the smallest possible 5-digit number is 23456). Therefore, this condition is satisfied by all 5-digit numbers formed. Step 3: Apply the condition "odd numbers". For a number to be odd, its last digit must be odd. From the given digits 2, 3, 4, 5, 6, the odd digits are 3, 5. So, there are 2 choices for the last digit (d_5). Step 4: Arrange the remaining digits. After choosing the last digit, there are 4 digits remaining. These 4 digits can be arranged in the first four positions (d_1 d_2 d_3 d_4) in 4! ways. 4! = 4 × 3 × 2 × 1 = 24 Step 5: Calculate the total number of odd numbers. Multiply the number of choices for the last digit by the number of ways to arrange the remaining digits. Total odd numbers = (Choices for d_5) × (Arrangements of d_1 d_2 d_3 d_4) Total = 2 × 24 = 48 There are 48 odd numbers greater than 4000 that can be formed. 13. b) A biased die is thrown 6 times and the probability of getting an odd number is (2)/(5). Find the probability of obtaining: This is a binomial probability problem. Let n = 6 (number of trials). Let p = P(odd number) = (2)/(5). Let q = P(even number) = 1 - p = 1 - (2)/(5) = (3)/(5). The binomial probability formula is P(X=k) = nk p^k q^n-k. b)(i) an even number 5 times; Step 1: Identify the parameters for this case. We want 5 even numbers and 1 odd number. So, k=5 for even numbers. The probability of success (getting an even number) is q = (3)/(5). P(5 even numbers) = 65 ((3)/(5))^5 ((2)/(5))^6-5 P(5 even numbers) = 65 ((3)/(5))^5 ((2)/(5))^1 Step 2: Calculate the binomial coefficient. 65 = (6!)/(5!(6-5)!) = 6!