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Flashcards: 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)?

$\displaystyle 70$

$\displaystyle 24$

576 $\displaystyle 576$

4,900 $\displaystyle 4,900$

1,680 $\displaystyle 1,680$

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Flashcards: 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)?

$\displaystyle 70$

$\displaystyle 24$

576 $\displaystyle 576$

4,900 $\displaystyle 4,900$

1,680 $\displaystyle 1,680$

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Flashcards: 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)?

Your answer:

$\displaystyle 24$

Your answer:

576 $\displaystyle 576$

Your answer:

4,900 $\displaystyle 4,900$

Your answer:

1,680 $\displaystyle 1,680$

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 C(8,4) $\displaystyle C(8,4)$ .

Since

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

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

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

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

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

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

$= \frac{1680}{24}$ $\displaystyle = \frac{1680}{24}$

= 70 $\displaystyle = 70$ ,

the correct number of arrangements.

Flashcards: 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.

$\frac{1}{13}$ $\displaystyle \frac{1}{13}$

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

$\frac{1}{26}$ $\displaystyle \frac{1}{26}$

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

$\frac{1}{18}$ $\displaystyle \frac{1}{18}$

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Flashcards: 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.

Your answer:

$\frac{1}{13}$ $\displaystyle \frac{1}{13}$

Your answer:

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

Your answer:

$\frac{1}{26}$ $\displaystyle \frac{1}{26}$

Your answer:

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

Correct answer:

$\frac{1}{18}$ $\displaystyle \frac{1}{18}$

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

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

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