Question

A marketing executive is studying Internet habits of married men and women during the 8am – 10pm time period on weekends ("prime Internet time"). Based on past data, he has determined that husbands are on the Internet 10% of the time during prime Internet time. It has also been determined that when the husband is on the Internet during prime Internet time, 40% of the time the wife is also on the Internet. When the husband is not on the Internet during prime Internet time, the wife is on the Internet 20% of the time. If the wife is on the Internet during prime Internet time, what is the probability that the husband is also on the Internet?

Answer 1

Using Bayes' theorem, the probability that the husband is on the Internet during prime time weekends given that the wife is on the Internet is approximately 18.18%.

Step-by-step explanation:

To find the probability that the husband is on the Internet given that the wife is on the Internet during prime Internet time, we can use Bayes' theorem. The problem gives us the following information:

• The probability that the husband is on the Internet during prime time (P(H)) is 10% or 0.10.
• The probability that the wife is on the Internet given the husband is also on the Internet during prime time (P(W|H)) is 40% or 0.40.
• The probability that the wife is on the Internet given the husband is not on the Internet during prime time (P(W|H')) is 20% or 0.20.

We want to find the probability that the husband is on the Internet given that the wife is on the Internet (P(H|W)). According to Bayes' theorem:

P(H|W) = (P(W|H) * P(H)) / P(W)

Where P(W) (the probability that the wife is on the Internet) can be calculated as:

P(W) = P(W|H)*P(H) + P(W|H')*P(H')

P(W) = (0.40)(0.10) + (0.20)(0.90)

P(W) = 0.04 + 0.18

P(W) = 0.22

Now we can calculate P(H|W):

P(H|W) = (0.40 * 0.10) / 0.22

P(H|W) = 0.04 / 0.22

P(H|W) ≈ 0.1818 or 18.18%

The probability that the husband is also on the Internet given the wife is on the Internet is approximately 18.18%.

Answer 2

Answer: Our required probability is 0.18.

Step-by-step explanation:

Since we have given that

Probability that husband is on the internet = 10% = 0.10

Probability that husband is not on the internet = 1-0.10 = 0.9

Probability that wife is on internet given that husband is on internet = 40% = 0.40

Probability that wife is on internet given that husband is not on internet = 20% = 0.20

Probability that wife is on internet is given by

`0.1* 0.4+0.9* 0.2\\\\=0.04+0.18\\\\=0.22`

So, Probability that the husband is also on internet given that wife is on internet is given by

`\frac{P(\text{wife and husband both on internet)}}{P(wife\ on \ internet)}\\\\=(0.4* 0.1)/(0.22)\\\\=(0.04)/(0.22)\\\\=0.18`

Hence, our required probability is 0.18.