Introduction to Probability: Additional Practice (Set 2)

Each coin has $2$ outcomes, so there are $2 \times 2 = 4$ equally likely results: HH, HT, TH, TT. Only HH is two heads.

$P(\text{both heads}) = \frac{1}{4}$

One favorable outcome out of four.

The bag holds $5 + 7 + 8 = 20$ marbles, all equally likely, and $8$ are green.

$P(\text{green}) = \frac{8}{20} = \frac{2}{5}$

Reduce $\frac{8}{20}$ by dividing top and bottom by $4$.

There are $6 \times 6 = 36$ equally likely outcomes. The pairs that add to $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, which is $6$ of them.

$P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6}$

Reduce $\frac{6}{36}$ to $\frac{1}{6}$.

Equal sectors are equally likely, so there are $8$ equally likely outcomes, $2$ of them shaded.

$P(\text{shaded}) = \frac{2}{8} = \frac{1}{4}$

Reduce $\frac{2}{8}$ to $\frac{1}{4}$.

A sum of $2$ needs both dice to show $1$, which is only the single outcome $(1,1)$ out of $36$.

$P(\text{sum } 2) = \frac{1}{36}$

This is the smallest possible non-zero probability with two dice.

Three coins give $2 \times 2 \times 2 = 8$ equally likely outcomes, and only HHH is all heads.

$P(\text{all heads}) = \frac{1}{8}$

One favorable outcome out of eight.

The primes from $1$ to $10$ are $\{2, 3, 5, 7\}$, which is $4$ of the $10$ cards.

$P(\text{prime}) = \frac{4}{10} = \frac{2}{5}$

Reduce $\frac{4}{10}$ to $\frac{2}{5}$.

There are $6$ factors of $12$ in the list, out of the $12$ equally likely cards.

$P(\text{factor of } 12) = \frac{6}{12} = \frac{1}{2}$

Reduce $\frac{6}{12}$ to $\frac{1}{2}$.

Two spins give $4 \times 4 = 16$ equally likely outcomes. Since each of the $4$ sectors is a different color, the matching pairs are one per color, so $4$ of them.

$P(\text{same color}) = \frac{4}{16} = \frac{1}{4}$

Reduce $\frac{4}{16}$ to $\frac{1}{4}$.

Convert the fraction to a decimal by dividing $1$ by $8$.

$\frac{1}{8} = 0.125$

One eighth of a whole is $0.125$.

Sums less than $4$ are $2$ and $3$. A sum of $2$ comes from $(1,1)$ and a sum of $3$ from $(1,2), (2,1)$, which is $3$ outcomes out of $36$.

$P(\text{sum} < 4) = \frac{3}{36} = \frac{1}{12}$

Reduce $\frac{3}{36}$ to $\frac{1}{12}$.

Two spins give $5 \times 5 = 25$ equally likely outcomes. The matching pairs are $(1,1), (2,2), (3,3), (4,4), (5,5)$, which is $5$ of them.

$P(\text{same number}) = \frac{5}{25} = \frac{1}{5}$

Reduce $\frac{5}{25}$ to $\frac{1}{5}$.