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QUESTION 10 Previous studies have found 28% of all people prefer Coke to all other cola drinks. A random sample of 17 people
QUESTION 11 A Melbourne real estate agent lists two houses for lease in the same street, one built of brick and the other of
QUESTION 10 Previous studies have found 28% of all people prefer Coke to all other cola drinks. A random sample of 17 people is selected. What is the probability exactly three people in this sample prefer Coke? (3 decimal places
QUESTION 11 A Melbourne real estate agent lists two houses for lease in the same street, one built of brick and the other of timber. The agent estimates that the probability of being leased within one week is 0.72 for the brick and 0.69 for the timber house, and the probability of both being leased within one week is 0.51. If the timber house is leased within one week, what is the probability that the brick house is also leased within one week? Give your answer correct to three decimal places. QUESTION 12 The number showing on the upper faces of two, fair, six-sided dice are observed when rolled. Consider events A and B defined as: .A B the sum of the two values observed on the upper faces is at most 5 atleast one dice has a 3 or 4 on the upper face 1. P(B') 2. P(A or B) (3 decimal places)
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Solution:

Q10) People who prefer coke, p = 28% = 0.28

Sample size, n = 17. Let X be the variable describing number of people preferring coke over other drinks (here, X = r (= 3)). Then, using binomial distribution, we can find the required probability as follows:

Prob(X = 3) = nCr*pr*(1-p)n-r ; C denoting the combinations

Prob(X = 3) = [17!/(3!*(17-3)!)]*(0.28)3*(1 - 0.28)17-3

Prob(X = 3) = 680*0.022*0.01 = 0.151 (approx)

So, probability that in this sample, exactly 3 people prefer coke is 0.151.

Q11) P(lease of bricked house) = 0.72 ; P(lease of timber house) = 0.69

P(lease of bricked house and timber house) = 0.51

P(bricked house leased | timber house leased) = P(both leased)/P(timber house leased)

= 0.51/0.69 = 0.739

Q12) According to what is given, the possibilities for event

A = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

B = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}

So, n(A) = 10 and n(B) = 12

For two fair dices rolled together, note that total number of observations = 62 = 36

1. So, P(B) = 12/36 = 0.333

P(B') = 1 - P(B) = 1 - 0.333 = 0.667

2. A and B both = {(3,1), (3,2), (4,1)} so, n(A and B) = 3

P(A or B) = P(A) + P(B) - P(A and B)

= (10/36) + (12/36) - (3/36) = 19/36

P(A or B) = 0.528

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