HiSET: Math : Probability models and how to find the probabilities of events

Study concepts, example questions & explanations for HiSET: Math

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Example Questions

Example Question #1 : Probability Models And How To Find The Probabilities Of Events

How many ways can eight people be seated at two tables if four people should be at each table, and it only matters at which table each person sits (ass opposed to which seat)?

Possible Answers:

\displaystyle 576

\displaystyle 1,680

\displaystyle 70

\displaystyle 4,900

\displaystyle 24

Correct answer:

\displaystyle 70

Explanation:

The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is \displaystyle C(8,4).

Since

\displaystyle C(n, r) = \frac{n!}{(n-r)!r!},

then, setting \displaystyle n= 8, r= 4,

\displaystyle C(8,4) = \frac{8!}{(8-4)!4!}

\displaystyle = \frac{8!}{4!4!}

\displaystyle = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 }{1 \times 2 \times 3 \times 4 \times 1 \times 2 \times 3 \times 4}

\displaystyle = \frac{ 5 \times 6 \times 7 \times 8 }{ 1 \times 2 \times 3 \times 4}

\displaystyle = \frac{1680}{24}

\displaystyle = 70,

the correct number of arrangements.

Example Question #2 : Probability Models And How To Find The Probabilities Of Events

The spades and the aces are removed from a standard deck of 52. Give the probability that a card drawn at random will be a red three.

Possible Answers:

\displaystyle \frac{1}{18}

\displaystyle \frac{1}{13}

\displaystyle \frac{1}{12}

\displaystyle \frac{1}{16}

\displaystyle \frac{1}{26}

Correct answer:

\displaystyle \frac{1}{18}

Explanation:

In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of

\displaystyle 13 + 4 - 1 = 16

cards, to leave a modified deck of 

\displaystyle n(C))= 52- 16 = 36

cards.

None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is

\displaystyle P(R3) = \frac{n(R3)}{n(C)} = \frac{2}{36} = \frac{1}{18}.

 

Example Question #1 : Probability Models And How To Find The Probabilities Of Events

A standardized test comprises five examinations: reading, writing, math, science, and social studies, and these examinations may be taken in any order; however, in one state, there is one restriction: neither the writing nor the math test may be taken last.

In how many different orders can the five tests be taken so that this restriction is met?

Possible Answers:

\displaystyle 24

\displaystyle 96

\displaystyle 72

\displaystyle 120

\displaystyle 48

Correct answer:

\displaystyle 72

Explanation:

Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:

First, the test to be taken last must be one of three tests: reading, science, or social studies;

Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.

By the multiplication principle, this makes

\displaystyle 3 \times 4 \times 3 \times 2 \times 1 = 72

possible schedules.

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