SAT Math - Probability | Practice Hub

A penny is altered so that the odds are 3 to 2 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of both coins coming up heads?

23 to 12 against

23 to 12 in favor

29 to 6 in favor

29 to 6 against

Even

Explanation

3 to 2 odds in favor of heads is equal to a probability of $\frac{3}{3+2} = \frac{3}{5}$ , which is the probability that the penny will come up heads. Similarly, 4 to 3 odds in favor of heads is equal to a probability of $\frac{4}{4+3} = \frac{4}{7}$ , which is the probability that the nickel will come up heads.

Since the tosses of the two coins are independent, multiply the probabilities. The probability that there will be two heads is

$\frac{3}{5} \times \frac{4}{7} = \frac{12}{35}$

This is to 12 against both coins coming up heads.

Suppose there is a bag of marbles. There are two reds, two greens, and one orange marbles. Without replacement, what is the probability of choosing a red, and another red?

$\frac{1}{10}$

$\frac{13}{20}$

$\frac{3}{20}$

$\frac{3}{25}$

$\frac{2}{25}$

Explanation

There are five marbles altogether.

The first marble is a red, which is two of the five marbles in the bag.

The probability of choosing a red for the first trial is: $\frac{2}{5}$

Without replacement, there are four marbles remaining, and only one red marble remains.

The probability of choosing a red for the second trial is: $\frac{1}{4}$

Multiply the two probabilities.

$\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$

The answer is: $\frac{1}{10}$

Two dice, one red and one blue, are altered. The red die comes up a "1" $\frac{1}{4}$ of the time and a "2" $\frac{1}{4}$ of the time, and the blue die comes up a "1" $\frac{2}{7}$ of the time and a "2" $\frac{1}{7}$ of the time. What are the odds against rolling a "2", a "3", or a "4" with the two dice?

Insufficient information is given to solve the problem.

23 to 5

11 to 3

6 to 1

8 to 3

Explanation

A "2" can only be rolled on the two dice with a double "1";, a "3", with a "1-2" or "2-1", and a "4", with a "1-3", a "2-2", or a "3-1". This makes six favoriable rolls. To answer the question, we must know the probabilities that the red die will come up "1", "2", and "3"; we must also know the same for the blue die. We know the probabilities for "1" and "2" for each die, but not "3". Therefore, insufficent information is given in the problem.

In a carnival game, a player must spin two independent spinners. If the first spinner has a probability of $\frac{2}{7}$ landing on red, and the second spinner has a probability of $\frac{3}{8}$ landing on red, what is the probability of the first spinner NOT landing on red, and the second spinner landing on red?

$\frac{15}{56}$

$\frac{3}{28}$

$\frac{23}{56}$

$\frac{37}{56}$

Explanation

Since the spinners are independent of each other, we will need to multiply the probability of not landing on red on the first spinner and the probability of landing on red on the second spinner together.

The probability of not landing on red for the first spinner is found by subtracting the probability of landing on red from .

$\text{P(not red)}=1-\frac{2}{7}=\frac{5}{7}$

Thus, $\text{P}=(\frac{5}{7})(\frac{3}{8})=\frac{15}{56}$

If an integer is randomly selected such that $0 \leq x\leq20$ , what is the probability that is a prime number?

$\frac{8}{21}$

$\frac{1}{5}$

$\frac{1}{4}$

$\frac{13}{21}$

$\frac{13}{20}$

Explanation

Probability is defined as specified outcomes divided by total possible outcomes.

$probability = \frac{specified}{total }$

Recall that a prime numbered is defined as a number that is divisible only by one and itself, and is not prime. There are prime numbers in the range of possible values:

$\left { 2,3,5,7,11,13,17,19 \right }$

Notice that the range of values, $0 \leq x\leq20$ includes zero , so there are total numbers to select from.

$p=\frac{8}{21}$

I pick a number from to . What's the probability I get a prime number

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{3}{10}$

$\frac{1}{2}$

$\frac{1}{10}$

Explanation

Prime numbers are numbers with factors of and itself. Those numbers are . Our probability:

$\frac{4}{10}=\frac{2}{5}$ The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.

Answer is $\frac{4}{5}$ .

What is the probability of getting an ace in any standard deck of cards?

$\frac{1}{13}$

$\frac{1}{52}$

$\frac{1}{26}$

$\frac{1}{2}$

$\frac{1}{4}$

Explanation

There are four aces in a standard fifty-two card deck. So we set up a fraction.

$\frac{4}{52}=\frac{1}{13}$ The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.

Answer is $\frac{1}{13}$ .

If you draw an ace from and deck then place it back in the deck, what is the probability of drawing another ace?

$\frac{1}{13}$

$\frac{1}{52}$

$\frac{1}{4}$

$\frac{1}{221}$

$\frac{3}{4}$

Explanation

Since the first card was placed back into the deck, it will have no effect on the second draw.

Since there are aces in a deck of cards, the probability would be

$\frac{4}{52}$ or $\frac{1}{13}$ .

If there are blue balls, black balls, green balls, and red balls and I want to pick two balls without replacement. what’s the probability I get a black ball on the first pull and a red ball on the second?

$\frac{5}{102}$

$\frac{5}{108}$

$\frac{2}{81}$

$\frac{5}{51}$

$\frac{7}{102}$

Explanation

Since we are trying to get first black then red balls without replacement, this is a condition we need to meet. This means we need to multiply probabilities. For my first pick I have five balls out of eighteen in which the probability is $\frac{5}{18}$ . With one ball gone, there are only seventeen balls left. Since I want a red ball next, there are three left out of seventeen balls or $\frac{3}{17}$ . Now, we multiply our probabilities: