Reiko takes 2 coins at random from the 3 quarters, 5 dimes, and 2 nickels in her pocket. three coins are tossed. what is the probability of 2 heads and 1 tail? what is the probability that she chooses a quarter followed by a dime?

The answers are 3/8 and 1/6.

There are 8 possible outcomes in the sample space for flipping 3 coins.  Of those, 3 include two heads and 1 tail.  This makes the probability 3/8.

Choosing a quarter for the first coin is 3/10, since there are 3 quarters out of 10 coins.  After this, choosing a dime is given by 5/9, since there are 5 dimes out of 9 coins left:
3/10(5/9) = 15/90 = 1/6

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.