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again probability Q

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Reply 20
I don't know where you're getting a power of 6 from.
Reply 21
Original post by BabyMaths
Then you need

P(LMH or LHM or MLH or MHL or HLM or HML) = 6*0.5*0.35*0.15.


sorry am so thick .

if they ask for days : would that mean you do : 8 * 0.5*0.35*0.15 ?
Reply 22
Original post by BabyMaths
I don't know where you're getting a power of 6 from.


sorry i meant multiplied by 6 .
Reply 23
Original post by charly2012
sorry am so thick .

if they ask for days : would that mean you do : 8 * 0.5*0.35*0.15 ?


Something missing here? But the answer will be no anyway.
Reply 24
Original post by BabyMaths
Something missing here? But the answer will be no anyway.


Sorry , you said for 3 day you time the result by 6

would that mean for 4 days you multiply by 8 ?

how do you figure it out

thank you for your help
Original post by charly2012
Sorry , you said for 3 day you time the result by 6

would that mean for 4 days you multiply by 8 ?

how do you figure it out

thank you for your help


If there was four days, there would be 16 combinations I think.

I'm really bad at probability too so this is really helpful for me too!
Reply 26
Original post by charly2012
Sorry , you said for 3 day you time the result by 6

would that mean for 4 days you multiply by 8 ?

how do you figure it out

thank you for your help


Assuming you mean four days that include at least one of each then you would have various permutations of LLMH, various permutations of MMLH and various permutations of HHLM.
Reply 27
i really dnt get this kind of question. But thanks alot for your support .
Reply 28
You need to take into account all possible combinations.
Draw a tree diagram I say, a lot easier to visualise from there. Pick out the combinations which would give one of each letter from each stem. Then do the same for the two other stem.

Drawing a tree diagram should really help you out, trust me.

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