SAT II Math I : Probability

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #151 : Data Analysis And Statistics

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

Possible Answers:

\(\displaystyle \frac{1}{221}\)

\(\displaystyle \frac{3}{4}\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle \frac{1}{13}\)

\(\displaystyle \frac{1}{52}\)

Correct answer:

\(\displaystyle \frac{1}{13}\)

Explanation:

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

Since there are \(\displaystyle 4\) aces in a deck of \(\displaystyle 52\) cards, the probability would be

 \(\displaystyle \frac{4}{52}\) or \(\displaystyle \frac{1}{13}\).

Example Question #152 : Data Analysis And Statistics

A fair 6 sided die is rolled twice. What is the probability of observing a 6 both times?

Possible Answers:

\(\displaystyle \frac{1}{12}\)

\(\displaystyle \frac{1}{46656}\)

\(\displaystyle \frac{1}{36}\)

\(\displaystyle \frac{1}{6}\)

\(\displaystyle \frac{1}{216}\)

Correct answer:

\(\displaystyle \frac{1}{36}\)

Explanation:

The probability of rolling a 6 once is \(\displaystyle \small \frac{1}{6}\) because 6 is one of 6 different possible outcomes.

The probability of two independent events happening can be multiplied together. The probability of getting 2 6's is:

\(\displaystyle Pr(6)*Pr(6)=\frac{1}{6}*\frac{1}{6}=\frac{1}{36}\)

Example Question #11 : Probability

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

Possible Answers:

\(\displaystyle \frac{13}{21}\)

\(\displaystyle \frac{8}{21}\)

\(\displaystyle \frac{13}{20}\)

\(\displaystyle \frac{1}{5}\)

\(\displaystyle \frac{1}{4}\)

Correct answer:

\(\displaystyle \frac{8}{21}\)

Explanation:

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

\(\displaystyle probability = \frac{specified}{total }\)

Recall that a prime numbered is defined as a number that is divisible only by one and itself, and \(\displaystyle 1\) is not prime. There are \(\displaystyle 8\) prime numbers in the range of possible \(\displaystyle x\) values: 

\(\displaystyle \left \{ 2,3,5,7,11,13,17,19 \right \}\)

Notice that the range of values, \(\displaystyle 0 \leq x\leq20\) includes zero , so there are \(\displaystyle 21\) total numbers to select from.

\(\displaystyle p=\frac{8}{21}\)

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