PRO Practice Questions

There are 2 possible outcomes on each flip: heads or tails. Since the coin is flipped three times, there are \( 2 \times 2 \times 2 = 8 \) total possibilities: HHH, HHT, HTH, HTT, TTT, TTH, THT, THH.
Of these 8 possibilities, how many involve exactly two heads? We can simply count these up: HHT, HTH, THH. We see that there are 3 outcomes that involve exactly two heads. Thus, the correct answer is \( \frac{3}{8} \).
Alternatively, we can draw an anagram table to calculate the number of outcomes that involve exactly 2 heads.

$$\begin{array}{ccc} A & B & C \\ H & H & T \end{array}$$

The top row of the anagram table represents the 3 coin flips: A, B, and C. The bottom row of the anagram table represents one possible way to achieve the desired outcome of exactly two heads. The top row of the anagram yields \( 3! \), which must be divided by \( 2! \) since the bottom row of the anagram table contains 2 repetitions of the letter H. There are \( \frac{3!}{2!} = 3 \) different outcomes that contain exactly 2 heads.
The probability of the coin landing on heads exactly twice is \( \frac{\text{number of two-head results}}{\text{total number of outcomes}} = \frac{3}{8} \).
The correct answer is B.

The period from July 4 to July 8, inclusive, contains \( 8 - 4 + 1 = 5 \) days, so we can rephrase the question as “What is the probability of having exactly 3 rainy days out of 5?”
Since there are 2 possible outcomes for each day (R = rain or S = shine) and 5 days total, there are \( 2 \times 2 \times 2 \times 2 \times 2 = 32 \) possible scenarios for the 5 day period (RRRSS, RSRSS, SSRRR, etc...). To find the probability of having exactly three rainy days out of five, we must find the total number of scenarios containing exactly 3 R’s and 2 S’s, that is the number of possible RRRSS anagrams:
\( \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10 \)
The probability then of having exactly 3 rainy days out of five is \( \frac{10}{32} \) or \( \frac{5}{16} \).
Note that we were able to calculate the probability this way because the probability that any given scenario would occur was the same. This stemmed from the fact that the probability of rain = shine = \( 50\% \). Another way to solve this question would be to find the probability that one of the favourable scenarios would occur and to multiply that by the number of favorable scenarios. In this case, the probability that RRRSS (1st three days rain, last two shine) would occur is \( (\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{32} \). There are 10 such scenarios (different anagrams of RRRSS) so the overall probability of exactly 3 rainy days out of 5 is again \( \frac{10}{32} \). This latter method works even when the likelihood of rain does not equal the likelihood of shine.
The correct answer is C.

The simplest way to solve the problem is to recognize that the total number of gems in the bag must be a multiple of 3, since we have \( \frac{2}{3} \) diamonds and \( \frac{1}{3} \) rubies. If we had a total number that was not divisible by 3, we would not be able to divide the stones into thirds. Given this fact, we can test some multiples of 3 to see whether any fit the description in the question.
The smallest number of gems we could have is 6: 4 diamonds and 2 rubies (since we need at least 2 rubies). Is the probability of selecting two of these diamonds equal to \( \frac{5}{12} \)?
\( \frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5} \). Since this does not equal \( \frac{5}{12} \), this cannot be the total number of gems.
The next multiple of 3 is 9, which yields 6 diamonds and 3 rubies:

\( \frac{6}{9} \times \frac{5}{8} = \frac{30}{72} = \frac{5}{12} \). Since this matches the probability in the question, we know we have 6 diamonds and 3 rubies. Now we can figure out the probability of selecting two rubies:
\( \frac{3}{9} \times \frac{2}{8} = \frac{6}{72} = \frac{1}{12} \)
The correct answer is C.