STA1DCT Data Based Critical Thinking Assignment Sample

STA1DCT Data Based Critical Thinking

When you are asked to calculate an answer by hand, you may still use your calculator for basic calculations (e.g. multiplication, division, taking a square-root etc.) Your workings should show that you know how a formulae or process works.

1. Consider an 21-ball lottery game. In total there are 21 balls numbered 1 through to 21 inclusive. 6 balls are drawn (chosen randomly), one at a time, without replacement (so that a ball cannot be chosen more than once).

To win the grand prize, a lottery player must have the same numbers selected as those that are drawn. Order of the numbers is not important so that if a lottery player has chosen the combination

16, 17, 18, 19, 20, 21

and, in order, the numbers

18, 16, 21, 20, 17, 19

are drawn, then the lottery player will win the grand prize (to be shared with other grand prize winners). You can assume that each ball has exactly the same chance of being drawn as each of the others.

(a) Consider a population of size N = 21. How many different random samples of size n = 6 are possible from a population of N = 21?

(b) Suppose that you choose the numbers 16, 17, 18, 19, 20 and 21 ahead of the next lottery draw. As a fraction, what is the exact probability that you will win the grand prize in the lottery in the next draw with these numbers?

(c) Continuing on part (b) and again as a fraction, what is the exact probability that you will not win the lottery in the next draw with these numbers? Show workings.

(d) This question is tougher and you need to think carefully about the answer. Recall that the order of the numbers chosen is not important and that each number can only be chosen once. In total, how many combinations are there available that include the numbers 20 and 21 but not the numbers 16, 17, 18 and 19? Explain and show workings. (Hint: The available combinations must include the numbers 20 and 21 and they may also include the numbers 16, 17, 18 or 19, but they must not include the numbers 16, 17, 18 and 19 in one combination at the same time.)

2. The total number of cards in a deck of Italian playing cards is 40. Out of those 40 playing cards there are ten cards that are Cups. Consider an experiment where the deck is properly shuffled and a single card is dealt.

(a) What is the probability that the card dealt is a Cup?

(b) Now suppose that the card is returned to the deck and that the deck is once again properly shuffled. The experiment is repeated another 47 times (i.e. in total 48 trials of the experiment are conducted – including the initial trial in (a) above) where the card that is dealt is returned to the deck after each trial and the deck is properly shuffled again before the next card is dealt. Out of the 48 trials, let X equal the number of Cup cards that are observed.

i. What are all of the possible values that can be observed for X?

ii. What is the expected frequency for the number of Cup cards dealt? Show workings.

iii. Suppose that X = 16 is observed. What is the relative frequency for this many Cup cards dealt?

iv. Now suppose that the experiment is to be repeated but this time with a very large n (number of trials). For a very large n, what should the relative frequency for the number of Cup cards dealt be approximately equal to?

3. Consider an experiment that firstly involves rolling a fair seven-sided die once and then secondly rolling a fair five-sided die once. Let X denote the number rolled on the first roll and let Y denote the number rolled on the second roll.

(a) How many possible outcomes are there? That is, how many different observed values for (X, Y ) can eventuate from this experiment? Explain how you got your answer.

(b) For the remainder of this question you may assume that the different observed values (outcomes) for (X, Y ) are all equally likely to occur. What is the probability associated with each outcome occurring?

(c) Let A denote the event that the number rolled on the first roll is greater than the number rolled on the second roll (i.e. X > Y ). What is the probability of event A occurring. Show workings.

(d) What is the probability of A not occurring?

(e) Now, let B denote the event that the sum of the numbers rolled on the two dice is at least 6 (i.e. X+Y ≥ 6). What is the probability of event B occurring. Show workings.

(f) Are events A and B mutually exclusive? Explain.

(g) What is the probability that event A occurs and event B occurs?

(h) What is the probability that event A occurs and event B does not occur?

(i) What is the probability that event A does not occur and event B occurs?

Solution

1.

i.

a.

The number of different random samples of size n = 6 that can be selected from a population of N = 21 without replacement can be calculated as: 21C6 = 21!/6! (21-6)! = 54264

Therefore, there are 54264 different random samples are possible.

b.

The probability of correctly choosing all six numbers in order is: 1/21C6 = 1/54264 = 0.0000184

c.

Exact probability of not winning is: 1 – 1/54264 = .99999

So, the exact probability of not winning the lottery in the next draw with these numbers is .99999

d.

Total combination 21C6

Combination where 20 and 21 is selected by not 16, 17, 18 and 19 is (21-6)C4 = 15C4 = 1365

Therefore, there are 1365 combinations available that include the numbers 20 and 21 but not the numbers 16, 17, 18, and 19.

2.

a.

Since there are 10 Cup cards in the deck and a total of 40 cards, the probability of drawing a Cup card can be calculated as:
P(Cup) = number of Cup cards / total number of cards = 10/40 = ¼ = .25
Therefore, the probability of drawing a Cup card from the deck is 0.25.

b.

i.

The possible values of X, the number of Cup cards observed in 48 trials, can range from 0 to 48, in increments of 1. Specifically, X can take on the values 0, 1, 2, 3, ..., 46, 47, 48.

ii.

The expected frequency of Cup cards dealt can be found using the mean or expected value of the binomial distribution E(X) = np
Here, n = 48 and p is .25
So, E(X) = np = 48 * .25 = 12

Therefore, we expect the number of Cup cards dealt in the 48 trials to be around 12.

iii.

Expected relative frequency can be calculated using binomial distribution, where n is 48 and p is .25.
P (X = 16) = (48 choose 16) × (1/4)^16 × (3/4)^32 = .0527

Hence, the expected relative frequency is .0527

iv.

For a very large number of trials, the relative frequency for the number of Cup cards dealt should be approximately equal to the probability of drawing a Cup card in a single trial. As the number of trials approaches infinity, the relative frequency should converge to the true probability, which is the probability of drawing a Cup card in a single trial as per the law of large numbers.

3.

a.

There are 35 possible outcomes when rolling a seven-sided die and a five-sided die.
Multiplication principle for MBA assignment expert can be used here to explain that states if are m ways of doing one thing and n ways of doing another thing, then there are m x n ways of doing both things together.

b.

Since we are assuming that each outcome is equally likely, the probability of any given outcome (X, Y) is 1/35. This is because there are 35 possible outcomes, and they are all equally likely.

c.

By counting the events where X is greater than Y, it can be seen that there are 21 outcomes where X>Y
Probability of event A occurring is: P(A) = number of outcomes where X>Y/ total outcome
= 21/35
= .6
Therefore, the probability of rolling a greater number on the first die than the second die is 0.6 or 60%.

d.

Probability of not occurring event A = 1 – P(A) = 1 - .6 = .4
Hence, the probability of A not occurring is 0.4 or 40%.

e.

Outcomes where X+Y ≥ 6 is 25
Therefore, the probability of event B occurring is:
P(B) = number of outcomes where X+Y ≥ 6 / total number of outcomes
= 20/35
= .57
Therefore, the probability of rolling a sum of at least 6 on the two dice is approximately 0.57 or 57%.

f.

Events A and B are not mutually exclusive.
There exist outcomes where both A and B occur. For instance, if X=4 and Y=2, then A (X > Y) and B (X+Y ≥ 6) are both true. Hence, A and B cannot be mutually exclusive.

g.

Outcome where both X > Y and X+Y ≥ 6 is 6
So, P (A∩B) = 6/35 = .1714

Hence, probability of occurring both event A and B is .1714 or 17.14%.

h.

Outcome where X>Y occur and X+Y ≥ 6 does not occur is 12
So, P (A and Not B) = 12/35 = .35
So, the probability is approximately 0.35.

i.

Outcome where X>Y does not occur and X+Y ≥ 6 occur is 24
So, P (Not A and B) = 24/35 = .6857
Hence, probability is approximately .6857 or 68.57%