Merita has decided to play in The Clothing Combo Contest. First, she will randomly choose from a pair of brown, purple, blue, green, or black pants. Next, she will randomly choose from a black, brown, or green shirt. If both the shirt and pants are a color that starts with a "B", she will win 10 dollars. If only one of the pieces of clothing is a color that starts with a "B", she will break even. Under any other outcome, she will lose 20 dollars.
What is Merita’s expected value of playing The Clothing Combo Contest? Round your answer to the nearest cent.
This is the question , and I can easily figure out the odds of her winning which is 6/15. How can I figure out the probability of her breaking even without writing up a table and counting the number of times there is only 1 "B" in the possible outcomes?
You don't need to consider the probability of one B, in order to compute the expected result, because having one B results in no profit or loss.
If you play the game 15 times, you can expect one occurrence of each of the $~5 \times 3~$ possibilities.
This implies that at the end of these 15 games, you will have won $~3 \times 2 \times 10\$ = 60\$~$ and lost $~2 \times 1 \times 20\$ = 40\$.~$
So, your expectation is (in dollars) $~\dfrac{60 - 40}{15}.$