Question

 

Suppose a professor of probability is tired of reading the depressing news and so he decides that he will quickly scan the first 5 headlines in the New Yorks Times and the first 5 headlines in the Boston Globe and if at most 3 of the articles in each are depressing, he will read the news that day. Further suppose that the probability of a NYTs headline being depressing is 0.6 and for the Globe the probability of a headline being depressing is 0.55.

(a) What is the probability that he will read the news the first day he tries this?

(b) In order to be "well-informed" he needs to read the news at least half the time; what is the probability that he will be well-informed after doing this for a week?

Hint: This is another problem where there are two independent parts of the random experiment. You might want to phrase it as three different random variables, all three binomial.

Answer 1

The professor decides that he will quickly scan the first 5 headlines of the New York Times and the first 5 headlines of the Boston Globe. If at most 3 in each are depressing, he will read the news that day.

Given, the probability of a NYT headline being depressing is 0.6 = (p1 , say)

          the probability of a Globe headline being depressing is 0.55 = (p2 , say)

Let X denote the number of headlines among the first 5 headlines of the NYT that are depressing.

Let Y denote the number of headlines among the first 5 headlines of the Globe that are depressing.

Clearly both X and Y follow Binomial distributions as in each case the headline can be one of exactly 2 independent outcomes ie depressing or not depressing.

Hence, X ~ Bin(5,p1)    and Y ~ Bin(5,p2)

The probability mass function of X is defined as :

                                                 ; x=0,1,2,3,4,5

Accordingly, the Probability mass function of Y is defined as :

                                                     ; y=0,1,2,3,4,5

So, the probability of atmost 3 news from NYT being depressing is P(X<=3) and the probability of atmost 3 news from the Globe being depressing is P(Y<=3) .

Now,     P(X<=3)    

                        

                        

                        

                        

And, P(Y<=3)

                        

                        

                      

                        

a.)    Hence the probability that he reads the news the first day he tries

= Probability that NYT has atmost3 depressing news and Probability that Global has atmost 3 depressing news

= P( X<=3 ) * P( Y<=3 )   {Since, each newspaper having depressing news is independent of the other}

= 0.66304 * 0.74678

= 0.49515    (approx)

b.) In order to be well informed, he needs to read the newspaper atleast half the time , which means atleast 4 days a week.

The chances of him reading the newspaper on a certain day is completely independent of the chances of him reading the newspaper on any other day.

Let Z denote the number of days of the week he reads the newspaper.

Clearly, Z follows a Binomial distribution.

The chances of him reading the newswpaper on a certain day is obtained in part (a.) as 0.49515 (=p ,say).

Then , Z~ Bin(7,p=0.49515)

The Probability mass function of Z is given as :

                        ; z=0,1,2,3,4,5,6,7

The required probability is :

   P( Z >= 4)

                      

                       

                       

                              (approx)