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: $\dfrac{13}{52} = \dfrac{1}{4}$.

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

Total outcomes: $6$.

Probability $= \dfrac{3}{6} = \dfrac{1}{2}$.

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

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

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

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

So:

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

Expanding: $E = \tfrac{1}{2} + \tfrac{1}{2}E + \tfrac{1}{2} + \tfrac{1}{4}E + \tfrac{1}{2} = \tfrac{3}{2} + \tfrac{3}{4}E$. Therefore $\tfrac{1}{4}E = \tfrac{3}{2}$, 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 $\dfrac{2}{3}$.

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 $= \dfrac{9}{90} = \dfrac{1}{10}$.