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?

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.

$$4! = 4 \times 3 \times 2 \times 1 = 24$$

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 $d_5$) $\times$ (Arrangements of $d_1 d_2 d_3 d_4$)

$$\text{Total} = 2 \times 24 = 48$$

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 $\frac{2}{5}$. Find the probability of obtaining:

This is a binomial probability problem.
Let $n = 6$ (number of trials).
Let $p = P(\text{odd number}) = \frac{2}{5}$.
Let $q = P(\text{even number}) = 1 - p = 1 - \frac{2}{5} = \frac{3}{5}$.
The binomial probability formula is $P(X=k) = \binom{n}{k} p^k q^{n-k}$.

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 $q = \frac{3}{5}$.

$$P(\text{5 even numbers}) = \binom{6}{5} \left(\frac{3}{5}\right)^5 \left(\frac{2}{5}\right)^{6-5}$$ $$P(\text{5 even numbers}) = \binom{6}{5} \left(\frac{3}{5}\right)^5 \left(\frac{2}{5}\right)^1$$

Step 2: Calculate the binomial coefficient.

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