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From the Ross book ex.13 chapter 4: A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to sale with probability $0.3$ and the second will lead independently to a sale of probability $0.6$. Any sale made is equally likely to be either for the deluxe model $1000$ dollars and the standard model $500$ dollars. Determine the probability mass function of $X$, the total dollar value of all sales.

So $X = 0, 500, 1000, 1500, 2000$; if $A_{1} = \{first \space sale\}$, $A_{2}=\{second \space sale\}, B_1=\{standard \space model\}$ and $B_2=\{deluxe \space model\}$, $P\{X=0\} = P(A_{1}^cA_{2}^c)=P(A_{1}^c)P(A_{2}^c)=(1-0.3)(1-0.6)=(0.7)(0.4)=0.28$, because the events are independent; now I have problems to calculate the followings events: $P\{X=500\} = P(A_1B_1A_2^c)+P(A_2B_1A_1^c)$??

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Note the following: there are two ways to get $X=500$: sell standard model in deal1 AND nothing in deal two OR nothing in deal1 AND standard in deal2. The probability of this is $0.3 \cdot \frac{1}{2} \cdot 0.4 + 0.7 \cdot 0.6 \cdot \frac{1}{2}$.

It is possible to get $X=1000$ in the following ways: either sell two standard in BOTH deals or 1 advanced in deal1 AND nothing in deal two or the other way around.

It is possible to get $X=1500$ in the following way: either advanced in deal1 AND standard in deal2 or the other way around.

Can you sort out the probabilities for these two outcomes?

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Is $1/2$ the probability to sell standard model? –  blob Apr 23 '13 at 16:40
    
is it correct to say $P\{X=500\} = P(A_1B_1A_2^c)+P(A_2B_1A_1^c)$? –  blob Apr 23 '13 at 16:47
    
Yes, but correct notation is $A_1 \cap B_1 \cap A^c_2$ –  Alex Apr 23 '13 at 16:57
    
'Any sale made is equally likely' –  Alex Apr 23 '13 at 16:57
    
so, $P\{X=2000\} = P(A_1 \cap B_2 \cap A_2 \cap B_2)$ ? –  blob Apr 23 '13 at 17:03

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