The total number of balls in the box will be 

\(\displaystyle 1 + 2 + 3 + ... + 10\).

Since 

\(\displaystyle 1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}\),

it follows that the number of balls is

 \(\displaystyle 1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55\).

The even numbers are 2, 4, 6, 8, and 10. According to the pattern, there are nine balls marked "2", seven marked "4", five marked "6", three marked "8", and one marked "10". Therefore, there are

\(\displaystyle 9+5+7+3+1 = 25\)

balls marked with even numbers. 

The probability of drawing a ball with an even number is therefore

\(\displaystyle \frac{25}{55} = \frac{5}{11}\)

The total number of balls in the box will be 

\(\displaystyle 1 + 2 + 3 + ... + 10\).

Since,

\(\displaystyle 1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}\),

it follows that the number of balls is

 \(\displaystyle 1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55\).

The prime numbers in the range of \(\displaystyle \textup{1 to 10}\) are \(\displaystyle \textup{ 2, 3, 5, and 7}\). The number of balls with each number corresponds to the number itself, so the number of balls with prime numbers will be

\(\displaystyle 2+3+5+7 = 17\).

The probability of drawing one of them is \(\displaystyle \frac{17}{55}\).

The radii of the halves of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count. In order to compare the probabilities of the arrow stopping in regions of the three colors, it suffices to compare the totals of the measures of the central angles of the different colors - we will call them \(\displaystyle G', R', Y'\).

Each sector of the larger semicircle is one-sixth of a \(\displaystyle 180^{\circ }\) sector and has central angle

\(\displaystyle \frac{1}{6} \times 180^{\circ } = 30^{\circ }\).

Each sector of the smaller semicircle is a quarter of a circle, and has central angle 

\(\displaystyle \frac{1}{4} \times 360^{\circ } = 90^{\circ }\).

The green regions comprise three sectors of the top semicircle, so the total of their central angles is 

\(\displaystyle G' = 3 \times 30^{\circ } = 90^{\circ }\).

The yellow regions comprise one sector of the top semicircle and one sector of the bottom one, so the total of their central angles is 

\(\displaystyle Y' =30^{\circ } +90^{\circ } = 120^{\circ }\).

The red regions comprise two sectors of the top semicircle and one sector of the bottom one, so the total of their central angles is 

\(\displaystyle R' = 2 \times 30^{\circ } + 90 ^{\circ } = 60^{\circ } +90^{\circ } = 150^{\circ }\).

\(\displaystyle G'< Y'< R'\); it follows that \(\displaystyle G< Y< R\).

The total number of balls in the box will be 

\(\displaystyle 1 + 2 + 3 + ... + 26\).

Since 

\(\displaystyle 1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}\),

it follows that the number of balls is

 \(\displaystyle 1+ 2 + 3 + ...+ 26 = \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351\).

Since "O" is the fifteenth letter of the alphabet, there are \(\displaystyle 15\) balls out of \(\displaystyle 351\) marked with "O". Therefore, the probability that a randomly-drawn ball will be one of these is

\(\displaystyle \frac{15}{351} = \frac{15\div 3 }{351 \div 3 } = \frac{5}{117}\)

Since the areas of the sectors are what are important here, we need only rank the total areas of the different colors, which we will call \(\displaystyle G', R', Y'\).

For the sake of simplicity, we will assume the smaller semicircle of the target has radius \(\displaystyle 1\); consequently, the larger semicircle has radius \(\displaystyle 2\). 

The larger semicircle has total area

\(\displaystyle A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi\)

The smaller semicircle has total area

\(\displaystyle A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi\)

Each of the six sectors of the larger semicircle has area \(\displaystyle \frac{1}{6}\) of its area, or 

\(\displaystyle \frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi\)

Each of the two sectors of the smaller semicircle has area \(\displaystyle \frac{1}{2}\) of its area, or 

\(\displaystyle \frac{ 1}{2}\cdot \frac{ 1}{2} \pi = \frac{1}{4} \pi\)

The total of the areas of the green sectors - three sectors of the larger semicircle - is

\(\displaystyle G' = 3 \cdot \frac{1}{3} \pi = \pi\)

The total of the areas of the red sectors - two sectors of the larger semicircle and one of the smaller - is

\(\displaystyle R' = 2 \cdot \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{2}{3} \pi + \frac{1}{4} \pi = \frac{8}{12} \pi + \frac{3}{12} \pi = \frac{11}{12} \pi\)

The total of the areas of the yellow sectors - one sector of the larger semicircle and one of the smaller - is

\(\displaystyle Y' = \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{4}{12} \pi + \frac{3}{12} \pi = \frac{7}{12} \pi\)

\(\displaystyle Y' < R' < G'\); it follows that \(\displaystyle Y < R < G\).

The total number of balls in the box will be 

\(\displaystyle 1 + 2 + 3 + ... + 26\).

Since 

\(\displaystyle 1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}\),

it follows that the number of balls is

 \(\displaystyle 1+ 2 + 3 + ...+ 26 = \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351\).

Since "I" is the ninth letter of the alphabet, there are nine balls marked "I", and \(\displaystyle 351-9 = 342\) other balls. Therefore, the odds against drawing a ball marked "I" are 342 to 9, or

\(\displaystyle \frac{342}{9} = \frac{342 \div 9}{9 \div 9} = \frac{38}{1}\)

The correct choice is 38 to 1.

For the sake of simplicity, we will assume the smaller semicircle of the target has radius 1, and, consequently, the larger semicircle has radius 2. 

The larger semicircle has total area

\(\displaystyle A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi\)

The smaller semicircle has total area

\(\displaystyle A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi\)

The total area of the target is

\(\displaystyle A = A_{1}+ A_{2} = 2 \pi + \frac{1}{2} \pi = \frac{5}{2} \pi\)

The purple sector has area one-sixth that of the larger semicircle, or

\(\displaystyle \frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi\)

The area that is not purple is

\(\displaystyle \frac{5}{2} \pi - \frac{1}{3} \pi = \frac{15}{6} \pi - \frac{2}{6} \pi = \frac{13}{6} \pi\)

The odds against the dart landing in the purple region are 

\(\displaystyle \frac{\frac{13}{6} \pi}{\frac{1}{3} \pi} = \frac{\frac{13}{6} \pi \cdot \frac{6}{\pi}}{\frac{1}{3} \pi \cdot \frac{6}{\pi}} = \frac{13}{2}\) - that is, 13 to 2.

Eduardo has a standard deck of cards plus two jokers (54 cards in all). What are the odds that he draws a joker, followed by a red card that isn't a king? Assume replacement and that there is one red joker and one black joker.

This is a probability question. Probability can be found via the following:

\(\displaystyle P=\frac{\#\ of\ desired\ outcomes}{total\ number\ of\ outcomes}\)

To find the probability of multiple events, we find the probability of each event and multiply them together.

The probability of the first event is as follows:

\(\displaystyle P_1=\frac{2}{54}=\frac{1}{27}\)

Note that there are 54 cards and only two jokers.

The second event will still have the same denominator as the first event. The numerator however, will need to change. 

"...a red card that isn't a king..." So, our deck would normally be half red. However, two of those cards will be kings, so the number of desired outcomes should be 25

\(\displaystyle P_2=\frac{25}{54}\)

Next, we need to multiple our two probabilities together.

\(\displaystyle P_{tot}=\frac{25}{54}*\frac{1}{27}=\frac{25}{1458}\approx0.017\)

So our answer is:

\(\displaystyle 0.017\)

The total number of balls in the box will be 

\(\displaystyle 1 + 2 + 3 + ... + 26\).

Since 

\(\displaystyle 1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}\),

it follows that the number of balls is

 \(\displaystyle 1+ 2 + 3 + ...+ 26= \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351\).

The number of balls \(\displaystyle M\) with a given letter of the alphabet is equal to the number of its position in the alphabet; the probability of a ball with that letter being drawn is that number divided by the total number of balls, 351. Therefore, for this probability to exceed \(\displaystyle \frac{1}{30}\), we must have the relation

\(\displaystyle \frac{M}{351} > \frac{1}{30}\)

\(\displaystyle \frac{M}{351} \cdot 351 > \frac{1}{30} \cdot 351\)

\(\displaystyle M > \frac{351}{30} = 11 \frac{7}{10}\).

Therefore, \(\displaystyle M > 11\). 

The 11th letter of the alphabet is "K", so in order to answer this question, it suffices to know whether the letter comes after "K" in the alphabet.

From Statement 1 alone, the question can be answered, since all of the letters in the word "lousy" appear after "K" in the alphabet. From Statement 2 alone, however, the question cannot be answered, since the letter "K" itself appears in the word "skunk".

The radii of the portions of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count.

The purple sector is one-sixth of the larger semicircle, so it is one-twelfth of a circle. This means that the probability of a spinner stopping inside that sector is \(\displaystyle \frac{1}{12}\), and the odds against this are

\(\displaystyle \frac{\frac{11}{12} }{\frac{1}{12}} = \frac{11}{1}\) - that is, 11 to 1.