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 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
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

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Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1you find the probability of getting a hard centred: 1/8 then the probability of getting a soft centred: 1/9 then multiply them together

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1I think you may have to double your answer @Chelsea04 as you could get one hard, then one soft, or one soft, then one hard.

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1i'm thinking of a different question. oh, right, yea. that i think makes much more sense

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1no! no! I got it now! hard centred: 8/17 soft centred: 9/17 multiply these together 72/289

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Yeah that sounds a bit better... been a while since i've done one of these.

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1do you multiply that by 2?

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1yea, me too! been like 3 months

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Hmm i'm not sure if that's correct. It may be easier to do it using combinations...

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1use a tree diagram, i'll draw one

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1There's four possible outcomes: (h, h), (s,s), (h,s), (s,h)

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1(note not all are equally likely as there's 8 hard and 9 soft)

agent0smith
 one year ago
Best 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...

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1so basically in the end, you did need to multiply it by 2 ;)

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Yep, but your original equation had them both over 17, not one over 17 and one over 16.

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1right, no replacement! Sorry, it's been a while

Chelsea04
 one year ago
Best ResponseYou've already chosen the best response.1agent's is the right answer, follow his equation

agent0smith
 one year ago
Best 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%.
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