There are 36 possible outcomes (\(\displaystyle 6\times 6=36\)). 10 out of the 36 outcomes are greater than 8: (6 and 3)(6 and 4)(6 and 5)(6 and 6)(5 and 4)(5 and 5)(5 and 6)(4 and 5)(4 and 6)(3 and 6).

\(\displaystyle \small \frac{10}{36}\ =\ \frac{5}{18}\)

Since there are 300 people, \(\displaystyle 300\cdot .21=63\) people play baseball and \(\displaystyle 300\cdot .09=27\) of those people play both baseball and soccer. Therefore, there are \(\displaystyle 63-27=36\) people who play baseball but not soccer.

Probability: \(\displaystyle \frac{36}{300}=\frac{3}{25}\)

probability on one roll: \(\displaystyle \frac{4}{6}=\frac{2}{3}\)

for both times= \(\displaystyle \frac{2}{3}\times \frac{2}{3}= \frac{4}{9}\)

The probability of drawing an ace first is \(\displaystyle \frac{4}{52}\) or \(\displaystyle \frac{1}{13}\).

Assuming an ace is the first card selected, the probability of selecting another ace is \(\displaystyle \frac{3}{51}\) or \(\displaystyle \frac{1}{17}\).

For the third card, the probability is \(\displaystyle \frac{2}{50}\) or \(\displaystyle \frac{1}{25}\).

To calculate the probability of all 3 events happening, you must multiply the probabilities:

\(\displaystyle \frac{1}{13}\times\frac{1}{17}\times\frac{1}{25}=\frac{1}{5525}\)

Since the number must be larger than 4999, the thousand’s digit has to be 5 or 7. We are also told that the number must be even. Thus, the unit’s digit must be 2 or 4. The middle digits can by any of the numbers 2,4,5, or 7. Therefore, we have a total of \(\displaystyle 2\cdot 4\cdot 4\cdot 2=64\) possibilities.

A standard dice has 6 faces numbered \(\displaystyle 1,2,3,4,5,6\).

There are \(\displaystyle 3\) even numbers, \(\displaystyle 2,4,6\), divided by the total number of faces:

\(\displaystyle \frac{3}{6}= 0.5\)

There are three scenarios favorable to this event.

1: He hits a bullseye with his first two shots; the third shot doesn't matter.

The probability of this happening is \(\displaystyle 0.40 \cdot 0.40 = 0.16\)

2: He hits a bullseye with his first shot, misses with his second shot, and hits with his third shot.

The probability of this happening is \(\displaystyle 0.40 \cdot 0.60 \cdot 0.40 = 0.096\)

3: He misses with his first shot and hits a bullseye with his other two shots.

The probability of this happening is \(\displaystyle 0.60 \cdot 0.40 \cdot 0.40 = 0.096\)

Add these probabilities:

\(\displaystyle 0.096 + 0.096 + 0.16 = 0.352\)

The customer gets a discount if he does not hit a pink region. There are ten out of twenty ways to hit a pink region and ten to not hit one - this makes the odds 10 to 10, or, in  lowest terms, 1 to 1 against a discount.

If 1,000 tickets are sold at $10 apiece, then $10,000 will be raised. The prizes are $3,000, $1,000, and $500, so $4,500 will be given back, meaning that the 1,000 ticket purchasers will collectively lose $5,500. This means that on the average, one ticket will be worth

\(\displaystyle \frac{-\$5,500}{1,000} = -\$5.50\)

This is the expected value of one ticket. 

The easiest way to solve this is by using the complement. Only one of the five plates is NOT blue or green. So \(\displaystyle \dpi{100} \small \frac{1}{5}\) of the plates are NOT blue or green.  Therefore \(\displaystyle \dpi{100} \small 1-\frac{1}{5}=\frac{4}{5}\) of the plates are blue or green.