SAT – Probability

Q:  What is the probability of randomly guessing the last four digits of a telephone number correctly?

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A: 1/10,000

Explanation:  Each digit is chosen from the integers 0 through 9, which is ten possible choices for each digit.  There are four digits to choose independently, so there are:

10 x 10 x 10 x 10 = 10,000 ways to choose the four digits.

Then, the one correct sequence of 4 digits, is one out of 10,000.  The probability is 1/10,000.

Another way to arrive at the number 10,000 is as follows:

The 4-digit sequence may be anything from 0000 to 9999.  So, how many is that?  9999 – 0000 +1 = 10,000.  Why the ‘plus one’ on the end?  If you just subtract the numbers, it tells you how many spaces there are between the numbers.  Add one to include either the first number or the last number, whichever one has not been already counted.

 

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SAT – Counting and Ordering

Q:  20 students are to choose a president and vice president from among themselves. How many possible ways can they be chosen?

Answer:  380

Explanation 1:  Using the Counting Principle, set up two blanks to represent the president and vice president, from left to right:

____    ____

There are 20 choices for the president.  Put 20 in the first blank:

_20_    ____

The same person cannot be chosen for both positions, so there are 19 left to choose from for the vice president.  Put 19 in the second blank:

_20_    _19_

Now, multiply these numbers.  20 x 19 = 380.

Explanation 2:  This is a permutation problem since you are choosing 2 people from a total of 20 people, and the order is important.  The order is important because the two people chosen must be given specific positions, and switching those positions results in a different scenario.  Do  20 nPr 2 , or use the formula for nPr.  On the TI-84 Plus, find nPr in MATH-PRB-2.  The result is 380.

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SAT: Counting and Ordering

Q: How many ways are there to go from point A to point B on the rectangular prism below, by traveling only along exactly 3 edges of the prism?

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Answer: 6

Explanation: There are 3 ways to travel away from A, shown below in green. There are 2 ways to continue along each of these paths to B, shown in blue. That’s 3 x 2 = 6 paths to B along three edges of the prism. This is easier than making a mess of the diagram trying to draw all 6 paths (and making it hard to count them).

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SAT – Counting and Ordering

Q: Four people are to stand side by side for a photo. The shortest two people must stand at the ends, while the tallest two people must stand in the center. How many ways are there to arrange the people for the photo?
 
Answer: 4
Explanation: Use the Counting Principle.  Set up four blanks:
___  ___  ___  ___ to represent the four positions.
Decide on the number of possibilites for each position.
The first position has 2 possibilities (the 2 shortest people):
_2_  ___  ___  ___.  Whichever of the two shortest people isn’t in the first spot, will have to be in the fourth spot:
_2_  ___  ___  _1_.
Do the same for the middle two positions, using the two tallest people:
_2_  _2_  _1_  _1_.  Now, multiply these numbers together: 4.
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SAT

Q: If seven squeeks equal one blip, and four blips equal one grrr, how many squeeks are equal to three grrrs?
 
Answer: 84
Explanation: One blip is 7 squeeks.  One blip is also 1/4 of a grrr.  So, 7 squeeks = 1/4 grrr.  Multiply both sides by 4 and you will get 28 squeeks = 1 grrr.  Multiply both sides by 3 and now 84 squeeks = 3 grrrs.
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