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You mean 14 i and ii? There's no (c) and (d). I got 1/25 for i and 1/5 for ii
Oh wait. Q13 (c) and (d)
And how you got that for Q14?
13C I got 10/16 and 13D I got 14/16
Oops meant 1/5 for 14 i
There are 25 possibilities right? 0 then 0, or 0 then 1, or 0 then 2 etc. 5 of them have the same number on both days: (0,0), (1,1), (2,2), (3,3), (4,4). So the probability is 5/25=1/5.
Same idea with 14 ii. There are 25 possibilities. 5 of them have a total of 4: (0,4), (1,3), (2,2), (3,1), (4,0)
I got right for 13(c).
And for (d) too
I don't get 14 one. Sorry. :/
Make a probability tree with all the possibilites if you want. Like this, except there should 0,1,2,3,4 at the end of each of the 5 branches. |dw:1354320893329:dw| You can see there are 25 possibilities. How many of them have the same number on each day?
There's one of them. Except you did the probability wrong, it's 1/5=0.2 not 0.1.
No. In the table it's given 0.1
OOOHHHHHH I completely forgot about the probabilities!! I'm sorry :P I was assuming each day had the same probability! I'll look at it again
No problem. Sure. :)
Yeah so you're on the right track. The probability of 4 on each day is 0.1*0.1=0.01. Now do the same for the other numbers.
Perfect. What about part(ii) ?
Now you want to look at all the possibilites that add up to 4. So: (0,4), (1,3), (2,2), (3,1) and (4,0) Find the probabilities of each of those and add them together.
Got'ya. Thanks a ton.
Ok well that's a really long conversation.Why isn't it closed?