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morpeh [17]
2 years ago
13

Paul needs to randomly select one of his four children to go first in a board game. which simulation tools could he use in this

situation?
SAT
1 answer:
Aliun [14]2 years ago
3 0

The simulation tools that could be used by Paul in this situation is :

  • A spinner divided evenly into 12 sections, with three sections each of four different colors
  • A standard deck of cards.
  • Two coins

<h3>Description of the simulator tools </h3>

A spinner divided into 12 sections with three sections each having four different colors can be used by calculating the prob of each subsection which simulates the probability of picking one of four children

Probability of each subsection

=   number subsection / number of section

= 3 / 12 =  1/4

Two coins can also simulate the picking of one of four children

The outcome of the two coins = ( HH, HT, TT, TH )

Therefore the probability of picking a H or T

= 1 / 4

Also deck of 52 cards can also simulate the probability of picking one out of four children to go first in a board game.

Hence we can conclude that the simulator tools that can be used are A spinner divided evenly into 12 sections, with three sections each of four different colors A standard deck of cards. and Two coins.

Learn more about Probability : brainly.com/question/24756209

#SPJ1

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Among the seven nominees for two vacancies on the city council are three men and four women. In how many ways may these vacancie
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Answer:

This question is incomplete

Complete Question

Among the seven nominees for two vacancies on the city council are three men and four women. In how many ways may these vacancies be filled

a) with any two of the nominees?

b) with any two of the women?

c) with one of the men and one of the women?

Answer:

a) 21 ways

b) 6 ways

c) 12 ways

Step-by-step explanation:

We solve this question using combination formula

C(n, r) = nCr = n!/r! (n - r)!

a) with any two of the nominees?

Probability (two of the nominees) = 7C2

= 7!/2! ×(7 - 2)!

= 7!/ 2! × 5!

= 7 × 6 × 5 × 4 × 3 × 2 × 1/2 × 1 ×(5 × 4 × 3 × 2 × 1)

= 21

b) with any two of the women?

We have a total of 4 women

Hence, the probability of any two of the four women, filling the vacancies =

P(any two of the women) = 4C2

= 4!/2! ×( 4 - 2)!

= 4!/ 2! × 2!

= 4 × 3 × 2 × 1/ 2 × 1 ×( 2 × 1)

= 6

c) with one of the men and one of the

Total number of men = 3

Total number of women = 4

= 3C1 × 4C1

= [3!/1! ×(3 - 1)! ] × [4!/1! ×(4 - 1)! ]

= [3!/1! × 2!] × [4!/1! ×3!]

= [3 × 2 × 1/ 1 × 2 × 1] × [4 × 3 × 2 × 1/ 1 × 3 × 2 × 1]

= 3 × [24/6]

= 3 × 4

= 12 ways

6 0
3 years ago
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