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I was working on a probability question when I got stuck. The question goes as follows:

On any day, the probability that a boy eats his prepared lunch is 0.5. The probability that his sister eats her lunch i 0.6. The probability that the girl eats her lunch given that the boy eats his is 0.9. Determine the probability that: a) both eat their lunch b) the boy eats his lunch given that the girl eats hers c) at least one of them eats their lunch.

The first one I got right, but I am not completely sure why it is right. Apparently the answer to a) is $9\over 20$ because since the boy ate his lunch, the girls probability of eating is 0.9 and if you multiply the two probabilities you get $9\over 20$. However this does not seem to take into account the possibility that the girl ate her lunch first, in which case the probability of both eating is $3\over 10$.

As for question b) I did understand, but c) made me rather confused. My reasoning was that the probability that at least one of them ate their lunch was complementary to neither eating their lunch, which thus leads to the answer 0.8. However, the answer which the book gives is 0.65. I therefore though that $P($ at least one eats $)=P(both)+P(B|G')+P(G|B')$, but this gave me some strange number. Does anyone have any suggestions on how to continue?

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up vote 1 down vote accepted

For(a), $P(A \cap B) = P(A|B)P(B) = P(B|A)P(A) = P(B \cap A)$.

For (b) the answer is $P(B|A) = 0.45/0.6$.

For (c), the answer is $P(A \cup B) = P(A)+P(B)-P(A \cap B)$.

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I believe that you first point is incorrect as $P(A|B)P(B)=9/20$ and $P(B|A)P(A)=3/10$ – E.O. Jan 8 '12 at 3:03
@EmileOkada: How did you get 3/10? – Thomas Jan 8 '12 at 3:08
I see my mistake now. Srry :P – E.O. Jan 8 '12 at 3:10

c) $P< at\ least \ one \ of\ them\ eat\ their\ lunch>=1-P<A^c \cap B^c>$

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Not quite. The event that both eat their lunch would be $A\cap B$ while what is asked is the probability that at least one of them eats their lunch, that is, $P(A \cup B)$. Your expression on the right is in fact equal to $P(A\cup B)$, not $P(A \cap B) = P(\text{both eat their lunch})$. – Dilip Sarwate Jan 8 '12 at 3:56
this is the ans to part c,which is at least one of them will eat their lunch – johnny Jan 8 '12 at 6:59
Is there any difference between what you have written on the left side of the equation, viz. "both of them eat their lunch" and "at least one of them will eat their lunch"? I have already agreed that the right side of your equation is (one way of correctly expressing) the answer to part c. that at least one of them will eat their lunch. What I am complaining about is that the left side says that the right side stands for something different. – Dilip Sarwate Jan 8 '12 at 12:41
o, i c,i will change it – johnny Jan 9 '12 at 4:04

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