Two normal fair dice are rolled, one red and the other blue, and the two numbers obtained are added together. Event A is that obtaining an even number with the blue die. Event B is that obtaining an even total. Event C is that of the total obtained being 12. For each possible pair of events determine, with justification, whether the events are independent or not.
\(P(A)\) is straight forward, it is \(\frac{1}{2}\) because half the numbers on one die are even
\(P(B)=\frac{1}{2}\) as well, clearly (i hope it is clear, out of the possible 36 combinations half are even)
yeap i got these two~
only one way to get a 12, namely \((6,6)\) so that probability is \(\frac{1}{36}\) so far so good?
yeap~
got that as well~
for independence you need to check if \(P(B|A)=P(A)\) we already have the right hand side, what about the left?
you can do it with a table, check all the ones where the first die is even, and see if half of those will give an even total
i am pretty sure it will because it means you know the first die is even, what is the probability that the second die is also even (since that is the only way to get an even total)
is it clear what you need to check and how to check it?
hmm~ I use the P(AnB)=P(A)xP(B) if the even it independent~ and i got 0.25=0.5 x 0.5
yes, that works too
i believe the next pair will not be that is \(P(C)\neq P(C|B)\)
we already know \(P(C)=\frac{1}{36}\) but if we have rolled an even number, there are only 18 elements in our sample space, so the probability that you get a total of 12 given the total is even is \(\frac{1}{18}\)
yea~ got that too~ so event C is not independent?
hold the phone you cannot say one event is dependent you have to be comparing two events so you would say B and C are DEPENDENT (don't say not independent, your teacher will think you are not paying attention)
haha~ i see~ thank you xD
and am fairly sure A and C are dependent as well, and the reasoning is clear if you roll an odd number on the first die, there is no way for you to get a total of 12
whereas if you roll an even number, you have some chance of getting a 12 electric shock and all
i see~ yea~ using the calculation it is dependent~ thank you so much~
yw, keep up the good videos (or at least keep watching them)
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