Question

1. you randomly choose blocks from the set. if you choose two blocks at random, what is the probability of choosing a t and a p block?

options: \\(\frac{1}{33}\\), \\(\frac{1}{72}\\), \\(\frac{1}{36}\\), \\(\frac{1}{16}\\)

Explanation:

Step1: Count total blocks

There are 3 rows with 4 blocks each, so total blocks \( n = 3\times4 = 12 \).

Step2: Count T and P blocks

Number of T blocks: 2, number of P blocks: 1.

Step3: Calculate probability of first T then P

Probability of choosing T first: \( \frac{2}{12} \), then P: \( \frac{1}{11} \). So \( \frac{2}{12}\times\frac{1}{11}=\frac{2}{132}=\frac{1}{66} \)? Wait, no, wait. Wait, actually, the number of ways to choose T and P: either T then P or P then T.
Wait, total number of ways to choose 2 blocks: \( C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66 \).
Number of favorable outcomes: number of T blocks (2) times number of P blocks (1), and since order doesn't matter? Wait, no, when choosing two blocks, the favorable cases are (T1,P), (T2,P), (P,T1), (P,T2)? Wait, no, actually, the number of ways to choose one T and one P is \( 2\times1\times2 \)? Wait, no. Wait, the number of T is 2, number of P is 1. So the number of ways to pick one T and one P is \( 2\times1 = 2 \) (since for each T, we can pair with P). Wait, no, if we consider unordered pairs, the number of favorable pairs is \( 2\times1=2 \)? Wait, no, T1 and P, T2 and P: that's 2 unordered pairs. Wait, total number of unordered pairs is \( C(12,2)=66 \). So probability is \( \frac{2}{66}=\frac{1}{33} \). Wait, but let's check again.
Wait, first, total number of blocks: 12. Number of T: 2, number of P: 1.
Probability of choosing T then P: \( \frac{2}{12}\times\frac{1}{11}=\frac{2}{132}=\frac{1}{66} \).
Probability of choosing P then T: \( \frac{1}{12}\times\frac{2}{11}=\frac{2}{132}=\frac{1}{66} \).
Total probability: \( \frac{1}{66}+\frac{1}{66}=\frac{2}{66}=\frac{1}{33} \).
Yes, so the probability is \( \frac{1}{33} \).

Answer:

\(\frac{1}{33}\) (the first option: \(\frac{1}{33}\))