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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

Probability
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1/8*1/9 so 1/72
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
it's wrong isn't it

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Other answers:

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.
i'm thinking of a different question. oh, right, yea. that i think makes much more sense
no! no! I got it now! hard centred: 8/17 soft centred: 9/17 multiply these together 72/289
Yeah that sounds a bit better... been a while since i've done one of these.
do you multiply that by 2?
yea, me too! been like 3 months
Hmm i'm not sure if that's correct. It may be easier to do it using combinations...
use a tree diagram, i'll draw one
There's four possible outcomes: (h, h), (s,s), (h,s), (s,h)
(note not all are equally likely as there's 8 hard and 9 soft)
\[\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) \] I think this covers all possibilities...
yea, it does
so basically in the end, you did need to multiply it by 2 ;)
Yep, but your original equation had them both over 17, not one over 17 and one over 16.
right, no replacement! Sorry, it's been a while
is it the rite ans
agent's is the right answer, follow his equation
\[\left(\frac{ 8 }{ 17 } \times \frac{ 9 }{ 16 }\right) + \left(\frac{ 9 }{ 17 } \times \frac{ 8 }{ 16 }\right) = 2 \times \left(\frac{ 9 \times 8 }{ 17 \times 16 } \right) = 0.529\] 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%.

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