An Introduction to Probability

A fair die has \(6\) faces and exactly one of them shows a \(3\), so \(1\) of the \(6\) equally likely outcomes is favorable. The probability is \(\boxed{\frac{1}{6}}\). Favorable over total only works because the faces are equally likely, which is what fair means.

The odd faces are \(1, 3, 5\), so \(3\) of the \(6\) equally likely faces are favorable. That gives \(\frac{3}{6} = \boxed{\frac{1}{2}}\). A probability of \(\frac{1}{2}\) is an even chance, so odd and even are equally likely on a die.

A fair die shows \(1, 2, 3, 4, 5, 6\), and none of those \(6\) faces is a \(7\). The favorable count is \(0\), so the probability is \(\frac{0}{6} = \boxed{0}\). An event that cannot happen sits at \(0\), the far left end of the probability scale.

The die has \(8\) sides and \(5\) of them are yellow, so \(5\) of the \(8\) equally likely outcomes are favorable, giving \(\boxed{\frac{5}{8}}\). Count the sides, not the colors. There are only three colors but eight equally likely sides, and the sides are what carry equal chances.

A shuffled deck puts any of the \(52\) cards on top with the same chance, and \(13\) of them are spades. That gives \(\frac{13}{52} = \boxed{\frac{1}{4}}\). It matches what you would expect, since spades are one of four equal suits and take up a quarter of the deck.

A face card is a Jack, a Queen, or a King, so there are \(3\) face ranks in each of the \(4\) suits, which is \(3 \times 4 = 12\) face cards. Every one of the \(52\) cards is equally likely on top, so the probability is \(\frac{12}{52}\), and dividing top and bottom by \(4\) gives \(\boxed{\frac{3}{13}}\).

Each die has \(6\) faces, so there are \(6 \times 6 = 36\) equally likely pairs. The pairs that add to \(6\) are \((1,5), (2,4), (3,3), (4,2), (5,1)\), which is \(5\) of them, so the probability is \(\boxed{\frac{5}{36}}\). Order matters here, so \((1,5)\) and \((5,1)\) are two different rolls while \((3,3)\) happens only one way.

Each coin has \(2\) results, so four coins give \(2 \times 2 \times 2 \times 2 = 16\) equally likely outcomes, and only \(1\) of them is all heads. The probability is \(\boxed{\frac{1}{16}}\). Multiplying the separate chances gives the same thing, since \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}\).

The first card is red with probability \(\frac{26}{52} = \frac{1}{2}\). That card is not put back, so only \(51\) cards remain and only \(25\) of them are red, giving \(\frac{25}{51}\) for the second. Both stages have to happen, so multiply them, $$\frac{1}{2} \times \frac{25}{51} = \boxed{\frac{25}{102}}.$$ Without replacement both counts drop after the first draw, so the second fraction is never a copy of the first.

There are \(6 \times 6 = 36\) equally likely pairs. The only multiple of \(5\) on a die is \(5\) itself, so the product is a multiple of \(5\) exactly when at least one die shows a \(5\). Counting the opposite is faster. One die avoids a five with probability \(\frac{5}{6}\), so neither die shows a five with probability \(\frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\), and subtracting gives $$1 - \frac{25}{36} = \boxed{\frac{11}{36}}.$$ At-least-one questions are usually easier from the complement, since no fives at all is one clean case.

The first card is black with probability \(\frac{26}{52} = \frac{1}{2}\). One card is gone, so \(51\) remain, and all \(26\) red cards are still there, which gives \(\frac{26}{51}\). Multiplying the two stages, $$\frac{1}{2} \times \frac{26}{51} = \frac{26}{102} = \boxed{\frac{13}{51}}.$$ Only the black count dropped, because the card removed was black, so the red count stayed at \(26\).

Each die has \(8\) faces, so there are \(8 \times 8 = 64\) equally likely ordered pairs. The ones whose product is greater than \(36\) are \((5,8),(6,7),(6,8),(7,6),(7,7),(7,8),(8,5),(8,6),(8,7),(8,8)\), which is \(10\) of them, so the probability is \(\frac{10}{64} = \boxed{\frac{5}{32}}\). Work down from the largest face and find its partners, and keep order in mind so \((6,7)\) and \((7,6)\) both get counted.

The die has \(8\) equally likely sides and \(1\) of them is blue, so the probability is \(\boxed{\frac{1}{8}}\). The equally likely outcomes are the sides, not the three colors.

The faces that are \(4\) or less are \(1, 2, 3, 4\), so \(4\) of the \(6\) equally likely faces are favorable. That gives \(\frac{4}{6} = \boxed{\frac{2}{3}}\).

The faces that are not a \(1\) are \(2, 3, 4, 5, 6\), which is \(5\) of the \(6\) equally likely faces, so the probability is \(\boxed{\frac{5}{6}}\). The complement gets there too, since one face is a \(1\) and \(1 - \frac{1}{6} = \frac{5}{6}\).

There are \(4\) tens and \(4\) jacks, and no card is both, so \(8\) of the \(52\) equally likely cards are favorable. That gives \(\frac{8}{52} = \boxed{\frac{2}{13}}\). Adding the two counts is only safe because the groups do not overlap.

There are \(6 \times 6 = 36\) equally likely ordered pairs. The ones that add to \(4\) are \((1,3),(2,2),(3,1)\), which is \(3\), so the probability is \(\frac{3}{36} = \boxed{\frac{1}{12}}\). Count ordered pairs, since \((1,3)\) and \((3,1)\) are separate rolls while \((2,2)\) happens only one way.

There are \(6 \times 6 = 36\) equally likely ordered pairs, and the only ones that add to \(11\) are \((5,6)\) and \((6,5)\). Two favorable out of \(36\) gives \(\frac{2}{36} = \boxed{\frac{1}{18}}\).

There are \(6 \times 6 = 36\) equally likely ordered pairs. Each die shows at least a \(1\), so \((1,1)\) is the only way to total \(2\), and the probability is \(\boxed{\frac{1}{36}}\). A sum at the edge of the range can happen just one way, which is why \(2\) and \(12\) are the rarest totals.

The first card is a spade with probability \(\frac{13}{52} = \frac{1}{4}\). That spade is gone, so \(12\) spades remain among \(51\) cards, giving \(\frac{12}{51}\). Multiplying the two stages, $$\frac{1}{4} \times \frac{12}{51} = \frac{12}{204} = \boxed{\frac{1}{17}}.$$ Both counts drop after the first draw, so the second fraction is never a repeat of the first.

The first card is an ace with probability \(\frac{4}{52} = \frac{1}{13}\). One ace is gone, so \(3\) aces remain among \(51\) cards, giving \(\frac{3}{51}\). Multiplying the two stages, $$\frac{1}{13} \times \frac{3}{51} = \frac{3}{663} = \boxed{\frac{1}{221}}.$$

The first card is a face card with probability \(\frac{12}{52} = \frac{3}{13}\). That card is set aside, so \(11\) face cards remain among \(51\) cards, giving \(\frac{11}{51}\). Multiplying the two stages, $$\frac{3}{13} \times \frac{11}{51} = \frac{33}{663} = \boxed{\frac{11}{221}}.$$

Four coins give \(2 \times 2 \times 2 \times 2 = 16\) equally likely outcomes, and only one of them, TTTT, has no heads at all. So the probability of at least one head is \(1 - \frac{1}{16} = \boxed{\frac{15}{16}}\). Counting head on would mean handling one head, two heads, three heads, and four heads, while the complement is a single outcome.

Four coins give \(2^4 = 16\) equally likely outcomes. Exactly two heads means choosing which \(2\) of the \(4\) coins are the heads, and each coin pairs with the other \(3\) while every pair gets counted twice, so there are \(\frac{4 \times 3}{2} = 6\) ways. That gives \(\frac{6}{16} = \boxed{\frac{3}{8}}\).

A product is odd only when both numbers are odd. Each die is odd with probability \(\frac{3}{6}\), so both dice odd is \(\frac{3}{6} \times \frac{3}{6} = \frac{1}{4}\), and even is everything left over, $$1 - \frac{1}{4} = \boxed{\frac{3}{4}}.$$ One even factor is enough to make the whole product even, so the odd case is the small one worth counting.

The bag holds \(4 + 3 + 5 = 12\) marbles and \(5\) of them are green, so the probability is \(\boxed{\frac{5}{12}}\).