A painter has 12 tins of paint. Seven tins are red and five tins are yellow.

Question

For my IB Math class, I have a test on probability and statistics next week. I'm reviewing practice problems in the book and working on some to prepare myself for the test. This is a problem I'm having a little trouble with: A painter has 12 tins of paint. Seven tins are red and five tins are yellow. Two tins are chosen at random. Calculate the probability that both tins are the same color. So, you have 12 choices. 1/12 but then you only have 11 choices for your second choice, 1/11. Do you multiply those and get your answer?

Answer 1

Note that there are $$ \binom{12}{2} $$ ways to choose the tins. Note that of these possibilities $$ \binom{7}{1}\binom{5}{1} $$ have different colours. Hence the probability that the tins have the same colour is $$ 1-\frac{\binom{7}{1}\binom{5}{1}}{\binom{12}{2}} $$ Here $$ \binom{n}{k}=\frac{n(n-1)\dotsb(n-k+1)}{k!} $$ where combinatorially $\binom{n}{k}$ represents the number of unordered selections of size $k$ from a set of size $n$ without repetition.

Answer 2

Under the given condition you don't have $12$ and $11$ choices, so what you are saying makes no sense.

Anyway to find the wanted answer you need to count the probability that you select two red and two yellow tins. For the first one you have to choose one of the seven red tins and the probability is $\frac{7}{12}$. For the second tin you have $6$ red tins out of total $11$ and so the probability is $\frac{6}{11}$. Finally the probability you have selected two red tins is $\frac{7}{12} \cdot \frac{6}{11}$.

Do the same for yellow tins and the probability will turn out to be $\frac{5}{12} \cdot \frac{4}{11}$

Finally add them to get that the probability is $$\frac{42 + 20}{132} = \frac{31}{66}$$

Answer 3

There are two ways to get two tins the same color:

1. pick a red tin with probability $\frac{7}{12}$ and then pick another red tin with probability $\frac{6}{11}$. The probability of getting two red tins is $\frac{7}{12}\cdot\frac{6}{11}=\frac{42}{132}$.
2. pick a yellow tin with probability $\frac{5}{12}$ and then pick another yellow tin with probability $\frac{4}{11}$. The probability of getting two yellow tins is $\frac{5}{12}\cdot\frac{4}{11}=\frac{20}{132}$.

Since these two possibilities are mutually exclusive, we can sum the probabilities to get an overall probability of $\frac{42}{132}+\frac{20}{132}=\frac{62}{132}=\frac{31}{66}$

Answer 4

The $132$ green and gray squares represent the equally likely ways to choose two tins of paint.* The tin colors (above and to the left) match for the $62$ green squares, so the probability of matching tins is $\dfrac{62}{132}=\dfrac{31}{66}$.

*The squares represent order-matters selection; just half the picture (upper-right triangle, say) could have been used.