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I need help with part (1) of this problem and confirmation I've done the right thing for (2).

1) A survey carried out by a firm found 60% of clients buy products every month and 20% buy high-tech products. Of those who buy products every month, 30% buy high-tech products.

(i) Are the events 'buy products every month' and 'buy high-tech products' independent? Justify your answer.

(ii) what is the probability that a customer who buys high-tech products buys product every month?

2) It is known that 25% of all individuals in a large group are smokers.

(i) Suppose 4 individuals are selected at random. What is the probability that exactly 2 of them are smokers?

(ii) Suppose 400 individuals are selected at random. What is the probability that less than 113 of them are smokers?

I'm completely clueless about (1) but for (2i) I worked it out to be 4C2 x 0.25$^2$ x 0.75$^2$ =0.2109

and for (2ii) I did a normal distribution Y~N(100,0.75) which gave me P(1.50 < Z)=0.9332.

Hope I've done it right for 2 and need help for 1. Cheers

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I'm refraining from posting this as an answer since it seems too basic: for 1.i, what is the definition of independent events and why does it apply/not apply? For 1.ii, given your understanding of the independence of the two events from part i, what is the probability that a client buys every month given that they buy high tech products? –  Rafe Kettler May 15 '12 at 2:21
    
Independent events are 2 events where the probability that 1 occurs does not affect the probability of the other occurring so the events are independent. –  Ricky Rozay May 15 '12 at 2:28
    
is the probability for (ii) 20%+30%=50% or 0.5? –  Ricky Rozay May 15 '12 at 2:31
    
no, that's not right. I recommend you read this wiki article en.wikipedia.org/wiki/Independence_(probability_theory) –  Rafe Kettler May 15 '12 at 2:35
    
using the independence formula it should be 0.6x0.3 then which is 0.18 –  Ricky Rozay May 15 '12 at 2:38
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1 Answer 1

up vote 4 down vote accepted

Question 1: Let $H$ be the event "buys high-tech" and $M$ the event "buys every month."

(i) We have been told that $P(H|M)=0.3$. If $H$ and $M$ were independent, we would have $P(H|M)=P(H)$. But we have been told that $P(H)=0.2$. So $H$ and $M$ are not independent.

Much more informally, the proportion of high-tech shoppers among monthly shoppers is $0.3$, substantially more than the proportion of high-tech shoppers in the general population. If we know that someone is a monthly shopper, our estimate that she is a high-tech shopper is different (and bigger) that if we do not know about the monthly shopping habit.

(ii) We want $P(M|H)$, the probability of $M$ given $H$. But $$P(M|H)P(H)=P(M\cap H)=P(H|M)P(M).$$ We know $P(H|M)$, and $P(M)$, and $P(H)$, so we can compute $P(M|H)$. The answer turns out to be $0.27$.

We can also use the fact that $P(H\cap M)=P(H|M)P(M)$ to find that $P(H\cap M)=(0.3)(0.6)=0.18$. But $P(H)P(M)=(0.2)(0.6)=0.12$. So $P(H\cap M)\ne P(H)P(M)$, which is another (and in this case more complicated) way of seeing that $H$ and $M$ are not independent.

Question 2: The procedure for (i) is right.

For (ii), if $Y$ is the number of smokers in a sample of $400$, then $Y$ has binomial distribution, mean $(400)(0.25)$ and variance $(400)(0.25)(0.75)=75$. The probability that $Y \le 112$ is, approximately, the probability that a normal with mean $100$ and variance $75$ is $\le 112.5$. (We have made the continuity correction. If you do not, and use $112$ instead, the approximation is likely to be less good.)

So our probability is approximately the same as the probability that $Z\le \frac{112.5-100}{\sqrt{75}}$, where $Z$ is standard normal.

Added Remarks: I do not think that "less than $113$" can be interpreted as meaning $113$ or fewer, which is where your $1.5$ comes from. By the way, the inequality should go the other way, we want (with your interpretation) $P(Z\lt 1.5)$.

The probability that $Y \le 112$ is (Wolfram Alpha) approximately $0.924184$. With the continuity correction, the normal approximation gives probability roughly $0.925543$. Not bad. The probability that the normal is $\le 112$ (so no continuity correction) is about $0.917072$, respectable, but not nearly as accurate.

Note that with the availability of good computing tools, we can evaluate binomial probabilities directly, so the normal approximation to the binomial is of diminishing practical importance.

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