Let's address the problem step-by-step to determine the probability that Alexander will choose two different cookies from a plate of 10 cookies, where 6 cookies are chocolate and 4 cookies are strawberry jam.
1. Determine the number of ways to choose 1 chocolate cookie and 1 strawberry cookie:
- There are 6 chocolate cookies and 4 strawberry cookies.
- To find the number of possible combinations of choosing one chocolate cookie and one strawberry cookie, we multiply the number of choices for each cookie, which is [tex]\(6 \times 4 = 24\)[/tex].
2. Calculate the total number of ways to choose any 2 cookies out of 10:
- We use the combination formula to find the number of ways to choose 2 cookies out of 10, which is given by [tex]\(\binom{10}{2}\)[/tex].
- [tex]\(\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45\)[/tex].
3. Compute the probability of choosing 2 different cookies:
- The probability is the ratio of the number of favorable outcomes (choosing 1 chocolate and 1 strawberry cookie) to the total number of possible outcomes (choosing any 2 cookies out of 10).
- Thus, the probability [tex]\(P\)[/tex] is [tex]\( P = \frac{\text{Number of ways to choose 1 chocolate and 1 strawberry cookie}}{\text{Total number of ways to choose any 2 cookies}} = \frac{24}{45} \)[/tex].
4. Simplify the fraction:
- Simplifying [tex]\(\frac{24}{45}\)[/tex] by dividing both numerator and denominator by their greatest common divisor (3) results in [tex]\( \frac{24 \div 3}{45 \div 3} = \frac{8}{15} \)[/tex].
Therefore, the probability that Alexander will choose two different cookies is [tex]\( \frac{8}{15} \)[/tex].
The correct answer is not given among the provided options (A, B, C, D) because those answers do not match our calculation [tex]\(\frac{8}{15}\)[/tex]. Thus, we infer that either the given options might have an error or there might be a need for re-evaluation based on the provided answer options.
