Question

a study showed that 60 percent of the wall street journal subscribers watch cnbc every day. of these, 70 percent watch it outside the home. only 20 percent of those who dont watch cnbc every day watch it outside the home. let o be the event \watches cnbc outside the home.\ let h be the event \watches cnbc at home.\ let d be the event \watches cnbc daily.\ (a) calculate the probability of all joint probabilities. note: round your answers to 2 decimal places. (b) do the joint probabilities sum to 1? yes no

Explanation:

Step1: Identify given probabilities

We know that 60% of subscribers watch CNBC daily. Let \(P(D) = 0.6\) and \(P(ND)=1 - 0.6=0.4\). Among daily - watchers, 70% watch outside the home, so \(P(H|D)=0.7\) and \(P(O|D)=1 - 0.7 = 0.3\). Among non - daily watchers, assume the probabilities of watching at home or outside are 0. Let's use the formula for joint probability \(P(A\cap B)=P(A|B)P(B)\).

Step2: Calculate \(P(H\cap D)\)

Using the formula \(P(A\cap B)=P(A|B)P(B)\), we have \(P(H\cap D)=P(H|D)P(D)=0.7\times0.6 = 0.42\).

Step3: Calculate \(P(O\cap D)\)

\(P(O\cap D)=P(O|D)P(D)=0.3\times0.6 = 0.18\).

Step4: Calculate \(P(H\cap ND)\) and \(P(O\cap ND)\)

Since non - daily watchers (\(P(ND) = 0.4\)) don't watch CNBC daily, \(P(H\cap ND)=0\) and \(P(O\cap ND)=0\).

The joint probabilities are:
\(P(H\cap D)=0.42\), \(P(O\cap D)=0.18\), \(P(H\cap ND)=0\), \(P(O\cap ND)=0\)

Answer:

\(P(H\cap D)\)	\(P(O\cap D)\)	\(P(H\cap ND)\)	\(P(O\cap ND)\)

For part (b), the sum of the joint probabilities is \(P(H\cap D)+P(O\cap D)+P(H\cap ND)+P(O\cap ND)=0.42 + 0.18+0 + 0=0.6
eq1\), so the answer is No.