STA1DCT Data Based Critical Thinking Assignment 5 Sample

STA1DCT Data based critical thinking Assignment 5 Sample

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.), however your workings should show that you now how a formulae or process works.

1. Suppose that a friend gives us the chance to participate in a game. The game is as follows. There is a bucket containing 22 red tokens, 18 green tokens and 10 purple tokens. We are to blindly reach into the bucket and randomly choose one token. Keeping this token (i.e. the chosen token is not returned to the bucket) we are to once again reach in blindly and randomly retrieve another token. If the first token we have chosen is red and the second is green then we win the game. Otherwise we lose. Define the two events below as follows:

• R1 = first token selected is red.

• G2 = second token selected is green.

(a) What is P(R1)?

(b) In words, explain what the event G2|R1 is.

(c) What is P(G2|R1)?

(d) What is P(R1 and G2)? Show your workings.

(e) Suppose that you would like to decrease the probability that you win the game where you propose the same number of red tokens (22 red tokens) and green tokens (18 green tokens) to be used but with a new total number of purple tokens. You would like your probability of winning the game to be at most 0.10 (i.e. P(R1 and G2) ≤0.10). Use Excel to calculate the smallest number of purple tokens needed to achieve this? What is the smallest number of purple tokens needed to achieve this? Explain very clearly for Management Assignment Expert how you obtained your answer.

2. Consider a standard 52 card deck of playing cards. In total there are four cards that are Aces, four cards that are Kings, four cards that are Queens and four cards that are Jacks. The remaining 36 cards are four each of the numbers 2, 3, . . . , 10. That is there are four cards that are twos, four cards that are threes etc. For this question, suppose that we reduce the number of cards in the deck by

• removing one of the Aces

• removing four other cards that are not Aces

The cards that are removed are discarded and are not used for the remainder of this question. As such we now have a deck that consists of just 47 cards. Suppose that a card is randomly drawn from this reduced sized deck. Let A1 denote the event that this card is an Ace. This card that was drawn from the deck of cards is now discarded and we continue with a deck of just 46 cards. Suppose that a second card is now randomly drawn from this 46-card deck and let A2 denote the event that this card is an Ace. Answer the following questions.

(a) What is P(A1)?

(b) Given that the first card drawn was an Ace, what is the probability that the second card drawn is not an Ace?

(c) What is the probability that the first card drawn is an Ace and the second card drawn is not an Ace? That is, what is P(A1 and AC2)? Show your workings.

(d) Given that the first card drawn was not an Ace, what is the probability that the second card drawn is an Ace?

(e) What is the probability that exactly one of the cards, out of the two cards drawn, is an ace?

Solution

Question 1:

P (R1) = 22/50 = 0.44
The event of getting green token in the second draw provided the first draw is a red token
P (G2|R1) = (18/49) = 0.367
P (R1 * G2) = (22/50) * (18/49) = 0.162

Thus, the smallest number of purple tokens needed to achieve P(R1 and G2) ≤ 0.10 = 23
On increasing the number of purple tokens to 15, we see that
P (R1 and G2) = (22/55) * (18/54) = 0.133
On increasing the number of purple tokens to 20, we see that
P (R1 and G2) = (22/60) * (18/59) = 0.112
On increasing the number of purple tokens to 23, we see that
P (R1 and G2) = (22/63) * (18/62) = 0.10

Thus, the smallest number of purple tokens needed to achieve P(R1 and G2) ≤ 0.10 = 23

Question 2:

P (A1) = 3/47 = 0.064
P (A2C|A1) = 43/46 = 0.935
P (A1 and A2C) = 3/47 * (43/46) = 0.064 * 0.935 = 0.06
P (A2) = (44/47) * 3/46 = 0.936 * 0.065 = 0.061
P (A1 and A2C) + (A1c and A2) = 3/47 * (43/46) + 44/47) * 3/46 = 0.06 + 0.061 = 0.121