Three bags each contain tokens. The green bag contains 22 round green tokens, each with a different integer from 1 to 22. The red bag contains 15 triangular red tokens, each with a different integer from 1 to 15. The blue bag contains 10 square blue tokens, each with a different integer from 1 to 10.
Any token in a specific bag has the same chance of being selected as any other token from that same bag. There is a total of 22 × 15 × 10 = 3300 different combinations of tokens created by selecting one token from each bag. Note that selecting the 7 red token, the 5 blue token and 3 green token is different than selecting the 5 red token, 7 blue token and the 3 green token. The order of selection does not matter.
You select one token from each bag. What is the probability that two or more of the selected tokens have the number 5 on them?
Am I on the right track? Here is my work
Do we only have 8/3300 with 2 or more tokens with number 5? I feel like something is wrong, please help, thank you!
The green bag has two tokens out of $22$ containing the digit $5$. The red bag has two out of $15$, and the blue bag has one out of ten.
Ways for all three tokens to contain a $5$: $2\cdot 2\cdot 1 = 4$.
Ways for green and red to contain a $5$, but not blue: $2\cdot 2\cdot (10-1) = 36$.
Ways for green and blue to contain a $5$, but not red: $2\cdot (15-2)\cdot 1 = 26$.
Ways for red and blue to contain a $5$, but not green: $(22-2)\cdot 2\cdot 1 = 40$.
Total: $\dfrac{4+36+26+40}{3300} = \dfrac{53}{1650}$