Probability practice questions — SAT Math

There are $8$ green counters out of $20$ in total: $$P(\text{green})=\frac{8}{20}$$ Simplify the fraction by dividing top and bottom by $4$: $$\frac{8}{20}=\frac{2}{5}$$

The deck has $13$ hearts out of $52$ cards: ${\displaystyle \frac{13}{52}}={\displaystyle \frac{1}{4}}$.

Even outcomes on a die: $\{2,4,6\}$ — that is $3$ outcomes.

Total outcomes: $6$.

Probability $={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$.

Even outcomes: $\{2,4,6\}$, three out of six. Probability $={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$.

A fair coin has two equally likely outcomes; the probability of heads is ${\displaystyle \frac{1}{2}}$.

$P(\bar{A})=1-P(A)=1-0.3=0.7$.

A random selection makes every member equally likely, so the probability is the favorable count over the total count.

$$P(\text{plays})=\frac{20}{50}=\frac{2}{5}$$

Let $E$ be the expected number of flips to reach two heads in a row from the start. Condition on outcomes:

- With probability $\frac{1}{2}$, the first flip is T. We've used one flip and are back at the start: contributes $\frac{1}{2}(1+E)$. - With probability $\frac{1}{4}$, the first two flips are HT. We've used two flips and are back at the start: contributes $\frac{1}{4}(2+E)$. - With probability $\frac{1}{4}$, the first two flips are HH — done in two flips: contributes $\frac{1}{4}\cdot 2$.

So:

$$E=\frac{1}{2}(1+E)+\frac{1}{4}(2+E)+\frac{1}{4}\cdot 2$$

Expanding: $E=\frac{1}{2}+\frac{1}{2}E+\frac{1}{2}+\frac{1}{4}E+\frac{1}{2}=\frac{3}{2}+\frac{3}{4}E$. Therefore $\frac{1}{4}E=\frac{3}{2}$, i.e. $E=6$.

Sanity check: "flip until first head" has expected value $2$; requiring a *consecutive* second head triples it.

Even outcomes: $\{2,4,6\}$. Among these, $\{4,6\}$ are $>3$: probability ${\displaystyle \frac{2}{3}}$.

There are $90$ two-digit numbers. Digit sums divisible by $11$ in range $[1,18]$: only $11$. Numbers whose digits sum to $11$: $29,38,47,56,65,74,83,92$ — nine numbers. Probability $={\displaystyle \frac{9}{90}}={\displaystyle \frac{1}{10}}$.