A bank has $1500$ guests. From them the $1000$ has taken home loan and $700$ consumer loan.
a) Calculate the probability that a guest has taken home loan given that he has taken consumer loan.
b) The bank want to make an offer for consumer loan for those who took a home loan but not a consumer loan. To how many people will they send the offer?
c) Each day at the bank come average $4$ guests that want to take a loan. What is the probability that one day between 8 in the morning till 4 in the afternoon, that $1$ or $2$ guests get served?
d) for the guests that took consumer loan we know that $40\%$ are women and $60\%$ are men. From the men the $50\%$ are of age $45-55$ and from the women $40\%$. A guest of age $45-55$ is in the bank to take a loan. Which is the probability that it is a woman of age $45-55$ ?
e) For the guests that took consumer loan we know that the probability one of them took a loan less than $5000$ euros is $0.14$. If we take randomly $7$ guests, then which is probability at most $2$ of them have taken a loan less than $5000$ euros ?
f) If we check these $7$ guests whici is the probability only the fourth one has taken loan less than $7000$ euros?
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a) Does it hold that $P(H\cap C)=\frac{200}{1500}$ since $1000+700-1500=200$ ?
Then the probability is equal to $$P(H\mid C)=\frac{P(H\cap C)}{P(C)}=\frac{\frac{200}{1500}}{\frac{700}{1500}}=\frac{200}{700}\approx 28,57\%$$
Is that correct?
b) We have to calculate the number for $H\cap C^c$. Is this equal to $1000-700 =300$ ?
c) Do we have Poisson distribution since we have an average?
We have that $\lambda=4$. Then $$P(X=1)+P(X=2)=e^{-4}\cdot \frac{4^1}{1!}+e^{-4}\cdot \frac{4^2}{2!}=\frac{12}{e^4}\approx 0.21979$$ Is that correct?
d) Is the probability equal to \begin{align*}P(W\mid A)&=\frac{P(A\mid W)\cdot P(W)}{P(A)}\\ & =\frac{P(A\mid W)\cdot P(W)}{P(A\mid W)\cdot P(W)+P(A\mid M)\cdot P(M)}\\ & =\frac{0.40\cdot 0.40}{0.40\cdot 0.40+0.50\cdot 0.60}\\ & =34.78\%\end{align*}
e) Is the probability equal to \begin{align*}P(X\leq 2)&=P(X=0)+P(X=1)+P(X=2)\\ & =\binom{7}{0}\cdot 0.14^0\cdot (1-0.14)^{7-0}+\binom{7}{1}\cdot 0.14^1\cdot (1-0.14)^{7-1}+\binom{7}{2}\cdot 0.14^2\cdot (1-0.14)^{7-2}\\ & \approx 0.9380\end{align*}
f) Is there a typo and it must be $5000$ instead of $7000$ ? Or is it possible to calculate the probability with $7000$ without knowing the probability for "sucess" for that?
a) CORRECT. To realize this, consider the Venn's diagram representation
b) WRONG: looking at the same diagram: 800 persons
c) poisson distribution is correct. The poisson parameter is 4 if the bank opening time is 8:00 am till 4:00 pm
d) Correct but it is easier to represent the situation in a table like the following
thus the answer is simply $\frac{112}{322}=\frac{8}{23}$
e) Correct! it is a binomial $Bin(7;0.14)$
f) Yes, surely a typo...substituting 7000 with 5000 the answer is $0.14\times 0.86^6$. The probability that only one took a loan less then 5000€ is the same for any combination; it is irrelevant that the success comes out at the first, second, fourth or seventh trial