Question

Part A: Students from 4 towns attend a conference. Each town is assigned a number from 1 to 4. Drag yes or no next to the first sentence.

Yes or No

A number is drawn and replaced 6 times. Is this a fair way to pick 6 members to give each town representation? __________

Part B: What is the probability that Town 1 will have at least 1 representative after 6 number draws, to the nearest percent?

probability = __________ %

Answer 1

Yes, the method of selection is a fair way to give each town representation.

The probability that Town 1 will have atleast 1 representative after 6 draws is 82% .

Part A:

A fair selection is one which is done at random without the presence of bias in our selection. Here, each town has the name number of representation, which is 1. And selection is also done with replacement. This means that each town has an equal chance of being selected for each pick made.

Hence, the selection process is fair.

PART B:

Using binomial probability concept:

• 
  `nCx * p^(x) * (1-p)^(n-x)`
• n = number of trials = 6
• p = probability of success = 1/4 = 0.25
• x = number of successes = 1

P(X >= 1) = P(x = 1) + ... + p(x = 6)

Using a binomial probability calculator;

P(X >= 1) = 0.822

To the nearest percent :

• 0.822 × 100% = 82.2% ≈ 82%

Part B :

0.822

Answer 2

Part A - No

Part B - 100 %

Explanation:

Part A.

This is because, a fair chance requires that each member has the same probability of being picked.

Since there are 4 towns, a fair chance would be 1/4.That is, 1 out of every town.

Part B

The probability of at least 1 representative after 6 number draws is P(at least 1) = 1 - P(none from Town 1)

P(none from town 1) = 0

P(at least 1) = 1 - P(none from Town 1)

P(at least 1) = 1 - 0 = 1

In percentage 1 × 100 % = 100 %