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A box has 8 hard centred and 9 soft centred chocolates. two are selected at random.what is the probability that one is hard centred and other is soft centred
 one year ago
 one year ago
A box has 8 hard centred and 9 soft centred chocolates. two are selected at random.what is the probability that one is hard centred and other is soft centred
 one year ago
 one year ago

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Chelsea04Best ResponseYou've already chosen the best response.1
you find the probability of getting a hard centred: 1/8 then the probability of getting a soft centred: 1/9 then multiply them together
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
I think you may have to double your answer @Chelsea04 as you could get one hard, then one soft, or one soft, then one hard.
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
i'm thinking of a different question. oh, right, yea. that i think makes much more sense
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
no! no! I got it now! hard centred: 8/17 soft centred: 9/17 multiply these together 72/289
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
Yeah that sounds a bit better... been a while since i've done one of these.
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
do you multiply that by 2?
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
yea, me too! been like 3 months
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
Hmm i'm not sure if that's correct. It may be easier to do it using combinations...
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
use a tree diagram, i'll draw one
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
There's four possible outcomes: (h, h), (s,s), (h,s), (s,h)
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
(note not all are equally likely as there's 8 hard and 9 soft)
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
\[\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) \] I think this covers all possibilities...
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
so basically in the end, you did need to multiply it by 2 ;)
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
Yep, but your original equation had them both over 17, not one over 17 and one over 16.
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
right, no replacement! Sorry, it's been a while
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.1
agent's is the right answer, follow his equation
 one year ago

agent0smithBest ResponseYou've already chosen the best response.1
\[\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) = 2 \times \left(\frac{ 9 \times 8 }{ 17 \times 16 } \right) = 0.529\] So the probability is about 50%, which makes sense given our possible outcomes when picking two chocolates (h for hard, s for soft): (h, h), (s,s), (h,s), (s,h) If all options were equally likely, the probability would be exactly 50%.
 one year ago
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