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*p^{r}*(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 = 6^{2} = 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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